Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap16 Circular motion of a rigid object 18 Time-varying basis vectors

Chapter 17 General planar motion of a single rigid object

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17.0Introductionstudy PDF
17.1Rigid object kinematicsstudy PDF + solutions
17.2Mechanics of a rigid-objectstudy PDF + solutions
17.3Kinematics of rolling and slidingstudy PDF + solutions
17.4Mechanics of contactstudy PDF + solutions
17.5Collisionsstudy PDF + solutions

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The main goal here is to generate equations of motion for general planar motion of a (planar) rigid object that may roll, slide or be in free flight. Multi-object systems are also considered so long as they do not involve other kinematic constraints between the bodies. Features of the solution that can be obtained from analysis are discussed, as are numerical solutions.

Many machine and structural parts move in straight-lines (Chapter 14) or circles (Chapters 13). But other things have more general motions, like a plane in unsteady flight or a connecting rod in a car engine. Keeping track of such motion is a bit more difficult.

In this chapter we will use these two modeling approximations:

•

The objects are planar, or symmetric with respect to a plane; and
•

They have planar motions in that plane.

Filename:tfigure8-rel-ang-vel

Figure 17.1: Planar motion of a 3D car. If the car is symmetrical it can be studied by the means of this chapter.

A planar object is one where the whole object is flat and all its matter is confined to one plane, say the $x y$ plane. This is a palatable approximation for a piece cut out of flat sheet metal. For more substantial real objects, like a full car, the approximation seems at a glance to be terrible. But it turns out that so long as the motion is planar and the car is reasonably idealized as symmetrical (left to right) that treating the car as equivalent to it being squished into a plane does not introduce any more approximation. Thus, even in this 3-D world we live in with 3-D objects, it is fruitful to do 2-D analysis of the type you will learn in this chapter.

A planar motion is one where the velocities of all points are in the same constant plane, say a fixed $x y$ plane, at all times and where points with, say, the same $z$ coordinate have the same velocity. The positions of the points do not have to be in the same plane for a planar motion. Each point stays in a plane, but different points can be in different planes, with each plane parallel to the others.

Example: A car going over a hill

Assume the road is straight in map view in, say, the $x$ direction. Assume the whole width of the road has the same hump. Although the car is clearly not planar, the car motion is probably close to planar, with the velocities of all points in the car in the $x y$ plane (see fig. 17.1)

Filename:tfigure8-ang-vel-ex

Figure 17.2: Planar motion of a skewered sphere. This can be studied by the means in this chapter.

Example: Skewered sphere

A sphere skewered and rotating about a fixed axis in the $\hat{\boldsymbol{k}}$ direction has a planar motion (see fig. 17.2). The points on the object do not all lie in a common plane. But all of the velocities are orthogonal to $\hat{\boldsymbol{k}}$ and thus in the $x y$ plane. This problem does fit in with the methods of this chapter. The symmetry of the sphere with respect to the $x y$ plane makes it so that the three-dimensional mass distribution does not invalidate the two-dimensional analysis.

Filename:tfigure8-ang-vel

Figure 17.3: Planar motion of a planar object. But the plane of the motion is not the plane of the object. This is not a natural topic for this chapter.

Example: Skewered plate

A flat rectangular plate with normal $\hat{\boldsymbol{n}}$ has a fixed axis of rotation in the direction $\hat{\boldsymbol{\lambda}}$ that makes a $45^{{}^{\circ}}$ angle to $\hat{\boldsymbol{n}}$ (see fig. 17.3). This is a planar object (a plane normal to $\hat{\boldsymbol{n}}$ ) in planar motion (all velocities are in the plane normal to $\hat{\boldsymbol{\lambda}}$ ). But the plane of motion is not the plane of the mass distribution, the object is not symmetric with respect to a motion plane, so this example does not fit into the discussion of this chapter^†^†margin: Actually, a two-dimensional analysis of the plate in this example would be legitimate in this sense. Project all the plate’s mass into the plane normal to the $\hat{\boldsymbol{\lambda}}$ direction. The projections of the forces on this plane would be correctly predicted, but three-dimensional effects, like those associated with dynamic imbalance, would be lost in this projection. .

No real object is exactly planar and no real motion is exactly a planar motion. But many objects are relatively flat and thin or symmetrical and many motions are approximately planar motions. Thus many, if not most, simple engineering analyzes assume planar motion. For bodies that are approximately symmetric about the $x y$ plane of motion (such as a car, if the asymmetrically placed driver’s mass etc. is neglected), there is no loss in doing a two-dimensional planar rather than full three dimensional analysis.

The plan of this chapter. We start with planar kinematics. Then we evaluate and use expressions for the rates of change of linear and angular momentum for planar bodies. Finally we discuss rolling, sliding and collisions.

17.1 Description of motion: planar rigid-object kinematics

As a rigid object moves, how do the points on it move? There are two reasons to ask this question. First, velocities and accelerations of mass points are needed to apply the momentum-balance equations. Second, formulas for positions, velocities and accelerations of points are useful to understand mechanisms, machines where various parts (each one usually idealized as a rigid object) are connected to each other with hinges and bearings of one type or another.

The central observation in all rigid-object kinematics is the definition of a rigid object:

All pairs of points on a single rigid object keep constant distance from each other as the object moves.

In this section you will learn how to use rigidity to calculate positions, velocities and accelerations of all points (millions and billions of them) on a rigid object given only a few numbers (9 of them). This goal is achieved by putting together the ideas from Chapter 11 (arbitrary motion of one particle), Chapter 14 (straight-line motion), and Chapter chapter 15 (circular motion of a rigid object in a plane).

Displacement and rotation

When a planar object (say a machine part) $\mathcal{B}$ moves from one configuration in the plane to another it has a displacement and a rotation. It starts in some reference configuration *. We mark a reference point on the body that, in the reference configuration, coincides with a fixed reference point, say 0

^†^†margin: The body never has to pass through this reference position, however. For example, the position of an airplane flying from New York to Mumbai is measured relative to a point in the Gulf of Guinea 1000 miles west of Gabon ( the location of

O^{{}^{\circ}}

longitude and

O^{{}^{\circ}}

latitude) even though the airplane never goes there (nor does anyone want it to).

We also mark a (directed) line on the body that, in the reference configuration, coincides with a fixed reference line, say the positive $x$ axis.

Filename:tfigure8-ang-accel

Figure 17.4: Rotation of object

\mathcal{B}

is measured by the rotation of real or imagined lines marked on the object. The lines make different angles:

\theta_{1}\neq\theta_{2}

\theta_{2}\neq\theta_{3}

etc, but

\dot{\theta}_{1}=\dot{\theta}_{2}=\dot{\theta}_{3}=\dots

. Angular velocity is defined as

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}

with

\omega\equiv\dot{\theta}_{1}=\dot{\theta}_{2}=\dot{\theta}_{3}=\dots

Filename:tfigure8-ang-accel-ex

Figure 17.5: The displacement and rotation of a planar object relative to a reference configuration.

We could measure rotation by measuring the rotation of any line segment connecting any pair of points fixed to the object. For each line we keep track of the angle that line makes with a line fixed in space, say the positive $x$ or $y$ axis. It’s simplest to stick to the convention that counter-clockwise rotations are positive (fig. 17.4).

The angles $\theta_{1}$ , $\theta_{2}\dots$ , all change with time and are all different from each other. But all the angles change the same amount, just like in section 16.1. We can pick any one line we like for definiteness and measure the object rotation by the rotation of that line. So

The net motion of a rigid planar object is described by translation, the vector displacement of a reference point from a reference position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{o^{\prime}/o}}={% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{oo^{\prime}}}$ , and a scalar rotation $\theta$ of the reference line from its reference orientation.

The position of a point on a moving rigid object.

Given that P on the object is at ${{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}}^{*}$ in the reference (*) configuration, where is it (What is ${{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}}$ ?) after the object has been displaced by ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{\prime}/0}}$ and rotated an angle $\theta$ ?

The base-independent or direct vector notation. An easy way to treat this is to write the new position of P as (see fig. 17.5), in the base-independent or direct vector representation of the position of P,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{\prime}/0}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{\prime}}}.

(17.1)

The vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{\prime}/0}}$ describes translation, that’s half the story. The other term ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{\prime}}}$ we find by rotating ${{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}}^{*}$ as we did in Section 16.2. This ‘base-independent’ formula is correct no matter what base vectors are used to represent the vectors in the formula (e.g., $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ or $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ ).

Fixed basis or component representation. For a given basis, say $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ associated with $x$ and $y$ , we can write the coordinates of a point as,

\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}\right]_{% xy}=\underbrace{\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {0^{\prime}/0}}\right]_{xy}}_{\rm displacement}+\underbrace{\left[R(\theta)% \right]\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{% \prime}}}\right]_{x^{\prime}y^{\prime}}}_{\rm rotation}

(17.2)

or, writing out all the components of the vectors and matrices, We get the fixed basis or component representation of the motion:

\left[\begin{array}[]{cc}x_{P}\\ y_{P}\end{array}\right]=\left[\begin{array}[]{cc}x_{0^{\prime}/0}\\ y_{0^{\prime}/0}\end{array}\right]+\left[\begin{array}[]{rr}\cos\theta&-\sin% \theta\\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{cc}x_{P/0}^{*}\\ y_{P/0}^{*}\end{array}\right].

(17.3)

As the motion progresses the displacement $\left[\begin{array}[]{cc}x_{0^{\prime}/0}\\ y_{0^{\prime}/0}\end{array}\right]$ changes with time as does the rotation angle $\theta$ . In this fixed basis or component representation of the motion of eqn. (17.3) we get the components of the position in terms of base vectors that are fixed in space.

Example:

If in the reference position a particle on a rigid object is at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}=(1\hat{% \boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}})\,\mathrm{m}$ and the object displaces by ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{\prime}/0}}=(3\hat% {\boldsymbol{\imath}}+4\hat{\boldsymbol{\jmath}})\,\mathrm{m}$ and rotates by $\theta=\pi/3\,{\rm rad}=60^{\circ}$ relative to that configuration, then its new position is:

$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/% 0}}\right]_{xy}$	$\displaystyle=$	$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^% {\prime}/0}}\right]_{xy}+\left[R(\theta)\right]\left[{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0^{\prime}}}\right]_{x^{\prime}% y^{\prime}}$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{cc}x_{0^{\prime}/0}\\ y_{0^{\prime}/0}\end{array}\right]+\left[\begin{array}[]{rr}\cos\theta&-\sin% \theta\\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{cc}x_{\rm P/0}^{% }\\ y_{\rm P/0}^{}\end{array}\right]$
	$\displaystyle=$	$\displaystyle\left\{\left[\begin{array}[]{cc}3\\ 4\end{array}\right]+\left[\begin{array}[]{rr}\cos\pi/3&-\sin\pi/3\\ \sin\pi/3&\cos\pi/3\end{array}\right]\left[\begin{array}[]{cc}1\\ 2\end{array}\right]\right\}\,\mathrm{m}$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{cc}3.5-\sqrt{3}\\ 5+\sqrt{3}/2\end{array}\right]\,\mathrm{m}$
$\displaystyle\Rightarrow{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/0}}$	$\displaystyle=$	$\displaystyle\left((3.5-\sqrt{3})\hat{\boldsymbol{\imath}}+(5+\sqrt{3}/2)\hat{% \boldsymbol{\jmath}}\right)\,\mathrm{m}$

Changing-base representation. Finally, the changing base representation uses base vectors $\hat{\boldsymbol{\imath}}^{\prime},\hat{\boldsymbol{\jmath}}^{\prime}$ that are aligned with $\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}}$ in the reference configuration but which are glued to the rotating object. If we define $x^{\prime}$ and $y^{\prime}$ as the $x$ and $y$ components of P in the reference (*) configuration we have that

	$\displaystyle\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P% /0}}}^{*}\right]_{xy}$	$\displaystyle=$	$\displaystyle\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P% /0^{\prime}}}}^{*}\right]_{x^{\prime}y^{\prime}}=\left[\begin{array}[]{cc}x^{% \prime}\\ y^{\prime}\end{array}\right]$		(17.9)
	$\displaystyle\quad{\rm so}\quad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{P/0}}$	$\displaystyle=$	$\displaystyle\left(x_{0^{\prime}/0}\hat{\boldsymbol{\imath}}+y_{0^{\prime}/0}% \hat{\boldsymbol{\jmath}}\right)+\left(x^{\prime}\hat{\boldsymbol{\imath}}^{% \prime}+y^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}\right).$		(17.10)

The second equation above is the changing base representation.

Although all three notations are in some sense equivalent, they have their best uses depending on context. The base-independent or direct-vector notation eqn. (17.1) is the most direct, least ornate and shortest to write. The component or fixed base representation eqn. (17.3) is the best for computer calculations. And the changing-base notation eqn. (17.10) is the most explicit for pencil and paper work.

Angular velocity

Filename:tfigure8-ang-accel1

Figure 17.6: It is generally best to take positive

\omega

to be counterclockwise when viewed from the positive

z

axis.

Because all lines on object $\mathcal{B}$ rotate at the same rate (at a given instant) $\mathcal{B}$ ’s rotation rate is the single number we call $\omega_{\cal B}$ (‘omega b’). As for the case of pure rotation, we define a vector angular velocity with magnitude $\omega_{\cal B}$ which is perpendicular to the $x y$ plane:

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}=\underbrace{\omega% _{\cal B}}_{\dot{\theta}}\hat{\boldsymbol{k}}

where $\dot{\theta}$ is the rate of change of the angle of any line marked on object $\mathcal{B}$ .

So long as you are careful to define angular velocity by the rotation of line segments and not by the motion of individual particles, the concept of angular velocity in general motion is defined exactly as for an object rotating about a fixed axis. The key kinematic fact is, in words:

On a rigid object in 2D any given point is moving in circles about any other given point (relative to that point).

So you can think of the general planar motion of a rigid object as the general motion of a point plus uniform circular motion about that point.

When a rigid object moves it always has an angular velocity (possibly zero). If we call the object $\cal B$ (script B for body), we then call the object’s angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ . Again, it is generally best to use the CCW sign convention that when $\omega_{\cal B}>0$ the object is rotating counterclockwise when viewed looking in from a point outside the object on the positive $z$ axis (that is, ‘looking down’, see fig. 17.6).

The angular velocity vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}$ of an object $\cal B$ describes its rate and direction of rotation. For planar motions ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}=\omega_{\cal B}% \hat{\boldsymbol{k}}$ .

Relative velocity of two points on a rigid object

For any two points A and B glued to a rigid object $\cal B$ the relative velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}$ of the points (‘the velocity of B relative to A’) is given by the cross product of the angular velocity of the object with the relative position of the two points (fig. 17.7):

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}\equiv{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {v}}}_{A}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}.

(17.11)

Filename:sfig8-2-3

Figure 17.7: The relative velocity of points A and B is in the

x y

plane and perpendicular to the line segment AB.

That is, the relative velocity of two points on a rigid object is the same as would be predicted for one of the points if the other were stationary. The derivation of this formula is the same as for planar circular motion.

Note that we use three-dimensional cross products even though we are doing planar 2D kinematics. Generally the plane of motion is the $x y$ plane and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ will be in the $z$ direction. Because ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ must be perpendicular to ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ it is perpendicular to the $z$ axis. So the three-dimensional cross product ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ gives a vector that is perpendicular to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ and is in the $x y$ plane .

We can also represent the relative velocity in the changing base notation as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}$	$\displaystyle=$	$\displaystyle{d\over dt}\left(x_{\textrm{B/A}}^{\prime}\hat{\boldsymbol{\imath% }}^{\prime}+y_{\textrm{B/A}}^{\prime}\hat{\boldsymbol{\jmath}}^{\prime}\right)$
	$\displaystyle=$	$\displaystyle x_{\textrm{B/A}}^{\prime}{d\over dt}\hat{\boldsymbol{\imath}}^{% \prime}+y_{\textrm{B/A}}^{\prime}{d\over dt}\hat{\boldsymbol{\jmath}}^{\prime}$
	$\displaystyle=$	$\displaystyle x_{\textrm{B/A}}^{\prime}\left({\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times\hat{\boldsymbol{\imath}}^{\prime}\right)+% y_{\textrm{B/A}}^{\prime}\left({\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}_{\cal B}\times\hat{\boldsymbol{\jmath}}^{\prime}\right).$

That is, the relative velocity is the same as the relative position, but with the base vectors substituted with their rates of change.

Finally, we can use the fixed-base or component notation for position and differentiate the terms (see box 16.2 on page 16.2):

$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B/A}}\right]_{xy}\par$	$\displaystyle=$	$\displaystyle{d\over dt}\left[\begin{array}[]{cc}x_{B/A}\\ y_{B/A}\end{array}\right]$
	$\displaystyle=$	$\displaystyle{d\over dt}\left\{\left[\begin{array}[]{rr}\cos\theta&-\sin\theta% \\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{cc}x_{\rm B/A}^{% }\\ y_{\rm B/A}^{}\end{array}\right]\right\}$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}-\dot{\theta}\sin\theta&-\dot{\theta}% \cos\theta\\ \dot{\theta}\cos\theta&-\dot{\theta}\sin\theta\end{array}\right]\left[\begin{% array}[]{cc}x_{\rm B/A}^{}\\ y_{\rm B/A}^{}\end{array}\right]$
	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}\phantom{-}0&-\omega\\ \omega&0\end{array}\right]\left[\begin{array}[]{rr}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{cc}x_{\rm B/A}^{% }\\ y_{\rm B/A}^{}\end{array}\right]$

where $x_{B/A}^{*}$ and $y_{B/A}^{*}$ are the components of the position of B with respect to A in the reference configuration and hence do not change with time.

Absolute velocity of a point on a rigid object

Filename:sfig8-2-3a

Figure 17.8: Absolute velocity of a point on a rigid object:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}+\omega% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}

If one knows the velocity of one point on a rigid object and one also knows the angular velocity of the object, then one can find the velocity of any other point. How? By addition. Say we know the velocity of point A, the angular velocity of the object, and the relative position of A and B, then

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+({\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {v}}}_{A})$	(17.16)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B/A}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}}_{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}}.$

That is, the absolute velocity of the point $B$ is the absolute velocity of the point $A$ plus the velocity of the point $B$ relative to the point $A$ (fig. 17.8).

Equation (17.16) is a 2D vector equation, and thus equivalent to 2 scalar equations, in 4 vectors which have 7 independent components ( ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ only has one independent component). Generally one could use eqn. (17.16) to solve for any 2 components in terms of the other 5.

Filename:sfig8-2-3b

Figure 17.9: The center of rotation C is that point ‘on’ the object that has zero velocity. So long as

\omega\neq 0

there always is such a point.

Instantaneous Center of Rotation. Interestingly, for any planar object with non-zero $\omega$ there is some point C ‘on’ the object that has exactly zero velocity (fig. 17.9). By ‘on’ the object we mean a point that moves with the object in a rigid way. Imagine a giant sheet of clear plexiglass glued to the object and extending way beyond the object. By ‘on’ the object we mean a point on the plexiglass. Why must there be such a point? Consider a point A on the object with some velocity. Relative to A the other points are going in circles. Each radial line extending from A has a relative velocity orthogonal to that line. Thus, considering all radial lines, every velocity direction is represented. And the relative velocities have magnitudes from 0 to infinity as points on a radial line are scanned from A outwards. Thus each possible relative velocity is had by some point ‘on’ the body. And one of those exactly cancels the velocity of A. Thus some point C has no velocity and all other points on the body rotate around it. This point is called the center of rotation. For other than purely circular motion, as an object moves, its center of rotation changes with time. See Sample 17.1 on page 17.1 for more details.

When trying to understand a mechanism it is often useful to think about the locus of locations of the center of rotation of each of the parts.

Angular acceleration

The angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ (‘alpha’) of a rigid object is the rate of change of angular velocity, ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$ . The angular acceleration of a object $\cal B$ is ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{\cal B}$ .

As for angular velocity, in 2D angular acceleration is perpendicular to the plane of motion:

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}% \qquad\mbox{and}\qquad{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=% \alpha\hat{\boldsymbol{k}}=\dot{\omega}\hat{\boldsymbol{k}}.

In 2D the angular acceleration is only due to the speeding up or slowing down of the rotation rate; i.e., $\alpha=\dot{\omega}=\ddot{\theta}$ .

Relative acceleration of two points on a rigid object

For any two points A and B glued to a rigid object $\cal B$ , the acceleration of B relative to A is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}$	$\displaystyle=$	$\displaystyle{d\over dt}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0% .5mu\mathrm{B/A}}$	(17.17)
	$\displaystyle=$	$\displaystyle{d\over dt}\left\{{\accentset{\rightharpoonup}{\boldsymbol{\omega% }}}_{\mathcal{B}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B/A}}\right\}$
	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times({\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/A}}),$
	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}),$
	$\displaystyle=$	$\displaystyle\alpha_{\cal B}\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}+(-\omega_{\mathcal{B}}^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}),$

This is the base-independent or direct-vector expression for relative acceleration. If point A has no acceleration, this formula for relative acceleration reduces to the total acceleration of a point undergoing circular motion from Chapter 15.

Equation (17.17) could also be derived, with some algebra, by taking two time derivatives of the relative position coordinate expression

\left[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}\right]_{% xy}=\left[R(\theta)\right]\left[{{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{\!\mathrm{B/A}}}^{*}\right]_{x^{\prime}y^{\prime}}\par

or by taking two time derivatives of the changing base vector expression

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}=x^{\prime}_{B/A% }\hat{\boldsymbol{\imath}}^{\prime}+y^{\prime}_{B/A}\hat{\boldsymbol{\jmath}}^% {\prime}.

Absolute acceleration of a point on a rigid object

If one knows the acceleration of one point on a rigid body and the angular velocity and acceleration of the body, then one can find the acceleration of any other point. How? Add the absolute acceleration of one point to the acceleration of the second relative to the first.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+({\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B}-{\accentset{\rightharpoonup}{\boldsymbol% {a}}}_{A})={\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B/A}$	(17.18)
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A})+\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal B}\times{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{B/A}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}-\omega_{% \mathcal{B}}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}+\alpha_{% \mathcal{B}}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{B/A}$

This is the base-independent or direct-vector expression for acceleration. The fixed-base (component) and changing-base notations are somewhat more complex.

Equation 17.18 is often called ‘the three term acceleration formula’ because the acceleration of B is the sum of three terms. First is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ , the acceleration of some point A on the object. Second is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}$ ), the centripetal acceleration of B going in circles relative to A. It is directed from B towards A. Third is $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}$ , due to the change in the magnitude of the angular velocity. The third term is in the direction normal to the line from A to B (tangent to the circle of B going around A).

Example: Robot arm

Filename:sfig8-3-1

Figure 17.10: A two link robot arm.

Given the configuration shown in fig. 17.10 the acceleration of point B can be found by thinking of link AB as the object $\cal B$ in eqn. (17.18) and using what you know about circular motion to find the acceleration of A:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}-\omega_{% \mathcal{B}}^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}+\alpha_{% \mathcal{B}}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{B/A}$
	$\displaystyle=$	$\displaystyle\left(-\omega_{\rm OA}^{2}\ell\hat{\boldsymbol{\jmath}}-\dot{% \omega}_{\rm OA}\ell\hat{\boldsymbol{\imath}}\right)-\left(\omega_{\rm AB}^{2}% \ell\hat{\boldsymbol{\imath}}\right)+\left(\dot{\omega}_{\rm AB}\hat{% \boldsymbol{k}}\times(\ell\hat{\boldsymbol{\imath}})\right)$
	$\displaystyle=$	$\displaystyle-\left(\dot{\omega}_{\rm OA}\ell+\omega_{\rm AB}^{2}\ell\right)% \hat{\boldsymbol{\imath}}+\left(-\omega_{\rm OA}^{2}\ell+\dot{\omega}_{\rm AB}% \ell\right)\hat{\boldsymbol{\jmath}}$

[Note that $\omega_{\rm AB}\neq\dot{\theta}$ where $\theta$ is the angle between the links. Rather $\omega_{\rm AB}=\omega_{\rm 0A}+\dot{\theta}$ .]

Computer graphics

Given one point given by the $x y$ pair $\left[\begin{array}[]{cc}x_{0}\\ y_{0}\end{array}\right]$ we can find out what happens to it by rotation $[R]$ as

\left[\begin{array}[]{cc}x\\ y\end{array}\right]=[R]\left[\begin{array}[]{cc}x_{0}\\ y_{0}\end{array}\right].

For example the point $\left[\begin{array}[]{cc}0\\ 2\end{array}\right]$ gets changed by a $45\deg$ rotation to

	$\displaystyle\left[\begin{array}[]{cc}x\\ y\end{array}\right]=[R]\left[\begin{array}[]{cc}x_{0}\\ y_{0}\end{array}\right]$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{rr}\cos{\pi\over 4}&-\sin{\pi\over 4}\\ \sin{\pi\over 4}&\cos{\pi\over 4}\end{array}\right]\left[\begin{array}[]{cc}0% \\ 2\end{array}\right]$
		$\displaystyle\approx$	$\displaystyle\left[\begin{array}[]{rr}.7&-.7\\ .7&.7\end{array}\right]\left[\begin{array}[]{rr}0\\ 2\end{array}\right]\approx\left[\begin{array}[]{rr}-1.4\\ 1.4\end{array}\right].$

A translation is just a vector addition. For example the point $\left[\begin{array}[]{rr}-1.4\\ 1.4\end{array}\right]$ gets translated a distance 2 in the $y$ direction by the addition of $\left[\begin{array}[]{cc}x_{t}\\ y_{t}\end{array}\right]=\left[\begin{array}[]{cc}0\\ 2\end{array}\right]$ like this

\left[\begin{array}[]{cc}x\\ y\end{array}\right]_{\rm translated}=\left[\begin{array}[]{cc}x\\ y\end{array}\right]+\left[\begin{array}[]{cc}x_{t}\\ y_{t}\end{array}\right]=\left[\begin{array}[]{cc}-1.4\\ 1.4\end{array}\right]+\left[\begin{array}[]{cc}0\\ 2\end{array}\right]=\left[\begin{array}[]{cc}-1.4\\ 3.4\end{array}\right].

Putting these together the point $\left[\begin{array}[]{cc}x_{0}\\ y_{0}\end{array}\right]$ gets rotated and translated by first multiplying by the rotation matrix and then adding the translation:

\displaystyle\left[\begin{array}[]{cc}x\\ y\end{array}\right]=[R]\left[\begin{array}[]{cc}x_{0}\\ y_{0}\end{array}\right]+\left[\begin{array}[]{cc}x_{t}\\ y_{t}\end{array}\right]

\displaystyle\approx

\displaystyle\left[\begin{array}[]{rr}.7&-.7\\ .7&.7\end{array}\right]\left[\begin{array}[]{cc}0\\ 2\end{array}\right]+\left[\begin{array}[]{cc}0\\ 2\end{array}\right]\approx\left[\begin{array}[]{cc}-1.4\\ 3.4\end{array}\right].

A collection of points all rotated the same amount and then all translated the same amount keep their relative distances.

A picture is a set of points on a plane. If all the points are rotated and translated the same amount the picture is rotated and translated. Thus a picture of a rigid object described by points is rigidly rotated and translated. On a computer line drawings are often represented as a connect-the-dots picture. The picture is represented by the $x$ and $y$ coordinates of the reference dots at the corners. These can be stored in an array with the first row being the $x$ coordinates and the second row the $y$ coordinates as explained on page 16.1. Each column of this matrix represents one point of the connect-the-dots picture. Thus a primitive picture of a house at the origin is given by the array

[P_{0}]\equiv\left[\mbox{$xy$ points originally}\right]=\left[\begin{array}[]{% cccccc}0&2&2&1&0&0\\ 0&0&2&3&2&1\end{array}\right]

with the lower left corner of the house at the origin.

To rotate this picture we rotate each of the columns of the matrix $P_{0}]$ . But this is exactly what is accomplished by the matrix multiplication $[R][P_{0}]$ . To translate the points you add the translation vector to each of the columns of the resulting matrix. Thus the whole picture rotated by $45^{{}^{\circ}}$ and translated up by 1 is given by

Filename:sfig8-3-1a

Figure 17.11: (a) A house drawn as 6 dots connected by line segments. The first and last point are the same. (b) The same house but rigidly rotated and translated.

\displaystyle[P_{\rm new}]=[R][P_{0}]+\left[\begin{array}[]{cc}x_{t}\\ y_{t}\end{array}\right]

\displaystyle\approx

\displaystyle\left[\begin{array}[]{cc}.7&-.7\\ .7&.7\end{array}\right][P_{0}]+\left[\begin{array}[]{cc}0\\ 1\end{array}\right]

which gives a new array of points that, when connected give the picture shown. We have allowed the informal notation of adding a column matrix to a rectangular matrix, by which we mean adding to each column of the rectangular matrix.

To animate the motion of a house flying in, say, the Wizard of Oz, you would first define the house as the set of points $[P_{0}]$ . Then define, maybe by means of numerical solution of differential equations, a set of rotations and translations. Then for each rotation and translation the picture of the house should be drawn, one after the other. The sequence of such pictures is an animation of a flying and spinning house.

Summary of the kinematics of one rigid object in general 2D motion

You can use the position of one reference point and the rotation of the object as simple kinematic measures of the entire motion of the object. If you know the position, velocity, and acceleration of one point on a rigid object (represented by 6 scalars, say) , and you know the angular rate and angular acceleration (3 scalars) then you can find the position, velocity and acceleration of any point on the object (also given its initial position). In 2D, just 9 numbers tell you the position, velocity, and acceleration of any of the billions of points whose initial positions you know^†^†margin: In 1D it takes just 3 numbers and in 3D just 18. The unusual pattern (3,9,18) comes from rotation being characterized by 0, 1, and 3 numbers in 1, 2, and 3 dimensions, respectively. .

SAMPLE 17.1

Filename:sfig8-5-wiper

Figure 17.12: The rigid triangular plate ABC rotates counterclockwise at

5\,\mathrm{rad}/\mathrm{s}

. Velocity of point B is given. We have to find the velocity of points A and C.

Velocity of a point on a rigid body in planar motion. An equilateral triangular plate ABC is in motion in the $x$ - $y$ plane. At the instant shown in the figure, point B has velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}=0.3\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+0.6\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\jmath}}$ and the plate has angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=2\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ .

1.

Find the velocity of point A using vector calculations.
2.

Find the velocity of point C using matrix calculations.

Solution

1.

We are given ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}$ and $\omega$ , and we need to find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}$ , the velocity of point A on the same rigid body. We know that,

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}$

Thus, to find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}$ , we need to find ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}$ . Let us take an $x$ - $y$ coordinate system whose origin coincides with point A of the plate at the instant of interest and the $x$ -axis is along AB. Then,

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}-{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}-% (0.2\,\mathrm{m}\hat{\boldsymbol{\imath}})=-0.2\,\mathrm{m}\hat{\boldsymbol{% \imath}}$

Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}$

$\displaystyle=$ $\displaystyle(0.3\hat{\boldsymbol{\imath}}+0.6\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}+2\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}\times(-0.% 2\hat{\boldsymbol{\imath}})\,\mathrm{m}$

$\displaystyle=$ $\displaystyle(0.3\hat{\boldsymbol{\imath}}+0.2\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{A}}=(0.3% \hat{\boldsymbol{\imath}}+0.2\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}$
2.

Let $x y$ axes with origin at A be fixed in space with $x$ -axis along AB at the instant of interest. Let $x^{\prime}y^{\prime}$ axes be fixed to the plate with origin at O and the $x^{\prime}$ -axis along AC as shown in fig. 17.13. Thus the coordinates of point C in the rotating axes are

Filename:sfig8-5-wiper-a

Figure 17.13: Velocity of point A is known. The axes $x y$ are fixed in space and the axes $x^{\prime}y^{\prime}$ are fixed to the rotating plate in the orientation shown. Thus in the $x^{\prime}y^{\prime}$ axes, the coordinates of point C will always be $x^{\prime}_{C}=0.2\,\mathrm{m},y^{\prime}_{C}=0$ .

$\left[\begin{array}[]{c}x^{\prime}_{C}\\ y^{\prime}_{C}\end{array}\right]=\left[\begin{array}[]{c}0.2\,\mathrm{m}\\ 0\end{array}\right]$

Clearly, at the instant of interest, the rotating axes make an angle of $\theta=\pi/3$ with the fixed axes. Now, we can calculate the velocity of point C as follows.

$\left[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}% \right]_{xy}=\left[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A}}\right]_{xy}+\left[\begin{array}[]{rr}0&-\omega\\ \omega&0\end{array}\right]\left[\begin{array}[]{rr}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right]\left[\begin{array}[]{c}x^{\prime}_{C/A% }\\ y^{\prime}_{C/A}\end{array}\right].$

Using $\theta=\pi/3$ , the angle that the rotating axes make with the fixed axes at the instant of interest, and $\omega=2\,\mathrm{rad}/\mathrm{s}$ , we get

$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}\right]_{xy}$ $\displaystyle=$ $\displaystyle\left[\begin{array}[]{c}0.3\,\mathrm{m}/\mathrm{s}\\ 0.2\,\mathrm{m}/\mathrm{s}\end{array}\right]+\left[\begin{array}[]{rr}0&-2\,% \mathrm{rad}/\mathrm{s}\\ 2\,\mathrm{rad}/\mathrm{s}&0\end{array}\right]\left[\begin{array}[]{rr}\frac{1% }{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2}\end{array}\right]\left[\begin{array}[]{c}0.2\,% \mathrm{m}\\ 0\end{array}\right]$

$\displaystyle=$ $\displaystyle\left[\begin{array}[]{c}0.3\,\mathrm{m}/\mathrm{s}\\ 0.2\,\mathrm{m}/\mathrm{s}\end{array}\right]+\left[\begin{array}[]{c}-0.35\,% \mathrm{m}/\mathrm{s}\\ 0.2\,\mathrm{m}/\mathrm{s}\end{array}\right]=\left[\begin{array}[]{c}-0.05\,% \mathrm{m}/\mathrm{s}\\ 0.4\,\mathrm{m}/\mathrm{s}\end{array}\right].$

Thus, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=-0.05\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+0.4\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\jmath}}$ . We can also verify this answer by computing ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}+{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=-0.05\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+0.4\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\jmath}}$

SAMPLE 17.2

The instantaneous center of rotation. A rigid body is in planar motion. At some instant $t$ , the angular velocity of the body is ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=5\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$ and the linear velocity of a point C on the body is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=2\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}-5\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\jmath}}$ . Find the instantaneous center of rotation.

Solution

Let a point O be the instantaneous center of rotation. We need to find the location of O using the given information. Whether O lies inside the object or outside is irrelevant here since the object boundary is not specified. So, let us consider an abstract rigid object as shown in fig. 17.14. What we know about O is that it has zero velocity (since it is the center of rotation).

Filename:sfig8-5-2disks

Figure 17.14:

Let point O be a distance $r$ away from C in some unknown direction $\hat{\boldsymbol{n}}$ (to be found). Thus, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}=r\hat{% \boldsymbol{n}}$ is the position vector of point O with respect to point C. Now,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0% .5mu\mathrm{O}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+\,{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}\times\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/O}}}_{-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}}$
		$\displaystyle=$	$\displaystyle-\omega\hat{\boldsymbol{k}}\times r\hat{\boldsymbol{n}}=-\omega r% (\hat{\boldsymbol{k}}\times\hat{\boldsymbol{n}}).$

Thus, $\hat{\boldsymbol{k}}\times\hat{\boldsymbol{n}}$ must be parallel to and in the opposite direction of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}$ . Let ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=v_{C}% \hat{\boldsymbol{\lambda}}$ . Then, we must have $\hat{\boldsymbol{k}}\times\hat{\boldsymbol{n}}=-\hat{\boldsymbol{\lambda}}$ , which says that $\hat{\boldsymbol{n}}$ must be normal to $\hat{\boldsymbol{\lambda}}$ . Since $\hat{\boldsymbol{\lambda}}={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{C}}/|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}|=(2% \,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}-5\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\jmath}})/(\sqrt{2^{2}+5^{2}}\,\mathrm{m}/\mathrm{s})=0.37\hat{% \boldsymbol{\imath}}-0.93\hat{\boldsymbol{\jmath}}$ , we can immediately write $\hat{\boldsymbol{n}}=0.93\hat{\boldsymbol{\imath}}+0.37\hat{\boldsymbol{\jmath}}$ . So, now,

$\displaystyle v_{C}\hat{\boldsymbol{\lambda}}$	$\displaystyle=$	$\displaystyle-\omega r(\overbrace{\hat{\boldsymbol{k}}\times\hat{\boldsymbol{n% }}}^{-\hat{\boldsymbol{\lambda}}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{C}$	$\displaystyle=$	$\displaystyle\omega r$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}r$	$\displaystyle=$	$\displaystyle v_{C}/\omega$
	$\displaystyle=$	$\displaystyle\sqrt{29}\,\mathrm{m}/\mathrm{s}/(5\,\mathrm{rad}/\mathrm{s})=1.0% 8m.$

Thus

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}=r\hat{% \boldsymbol{n}}=1.08\,\mathrm{m}(0.93\hat{\boldsymbol{\imath}}+0.37\hat{% \boldsymbol{\jmath}})=1\,\mathrm{m}\hat{\boldsymbol{\imath}}+0.4\,\mathrm{m}% \hat{\boldsymbol{\jmath}}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}=1\,\mathrm{m}% \hat{\boldsymbol{\imath}}+0.4\,\mathrm{m}\hat{\boldsymbol{\jmath}}$

Alternatively,

We can find ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}$ purely by vector algebra. Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}=(x\hat{% \boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}})\,\mathrm{m}$ . So,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{O}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}$
$\displaystyle\hbox to36.135pt{or\hfil}{\accentset{\rightharpoonup}{\boldsymbol% {0}}}$	$\displaystyle=$	$\displaystyle 2\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}-5\,\mathrm{m}/% \mathrm{s}\hat{\boldsymbol{\jmath}}+(5\,\mathrm{rad}/\mathrm{s})\hat{% \boldsymbol{k}}\times(x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}})\,% \mathrm{m}$
	$\displaystyle=$	$\displaystyle(2-5y)\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+(-5+5x)\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}.$

Dotting this equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ respectively, we get

	$\displaystyle 2-5y$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle-5+5x$	$\displaystyle=$	$\displaystyle 0.$

Solving these two equations simultaneously, we get, $x=1$ and $y=0.4$ . Thus,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}=1\,\mathrm{m}% \hat{\boldsymbol{\imath}}+0.4\,\mathrm{m}\hat{\boldsymbol{\jmath}}

as obtained above.

SAMPLE 17.3

Filename:sfig8-4-1

Figure 17.15:

A cheerleader throws her baton up in the air in the vertical $x y$ -plane. At an instant when the baton axis is at $\theta=60^{{}^{\circ}}$ from the horizontal, the velocity of end A of the baton is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=2\,\mathrm{m}/\mathrm{s}\hat% {\boldsymbol{\imath}}+\sqrt{3}\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$ . At the same instant, end B of the baton has velocity in the negative $x$ -direction (but $|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}|$ is not known). If the length of the baton is $\ell=\frac{1}{2}\,\mathrm{m}$ and the center-of-mass is in the middle of the baton, find the velocity of the center-of-mass.

Solution

	$\displaystyle\mbox{We are given:}\quad{\accentset{\rightharpoonup}{\boldsymbol% {v}}}_{A}$	$\displaystyle=$	$\displaystyle(2\hat{\boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}$
	$\displaystyle\mbox{and}\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle-v_{B}\hat{\boldsymbol{\imath}}$

where $v_{B}=|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}|$ is unknown. We need to find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}.$ Using the relative velocity formula for two points on a rigid body, we can write:

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{A}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/A}

(17.25)

Filename:sfig8-4-1a

Figure 17.16:

Here, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/A}$ are known. Thus, to find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}$ , we need to find ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , the angular velocity of the baton. Since the motion is in the vertical $x y$ -plane, let ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . Then,

Filename:sfig8-6-2

Figure 17.17:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}={\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+% \omega\hat{\boldsymbol{k}}\times\underbrace{\ell(\cos\theta\hat{\boldsymbol{% \imath}}-\sin\theta\hat{\boldsymbol{\jmath}})}_{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}}$
$\displaystyle\mbox{or}\quad-v_{B}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle(2\hat{\boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}+\omega\ell(\cos\theta\hat{\boldsymbol{\jmath}}+\sin% \theta\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle(2\hat{\boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}+\omega\mathord{\cdot}\frac{1}{2}\,\mathrm{m}\mathord{% \cdot}\left(\frac{1}{2}\hat{\boldsymbol{\jmath}}+\frac{\sqrt{3}}{2}\hat{% \boldsymbol{\imath}}\right)$

Dotting both sides of this equation with $\hat{\boldsymbol{\jmath}}$ we get:

	$\displaystyle 0$	$\displaystyle=$	$\displaystyle\sqrt{3}\,\mathrm{m}/\mathrm{s}+\frac{\omega}{2}\,\mathrm{m}% \mathord{\cdot}\frac{1}{2}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega$	$\displaystyle=$	$\displaystyle-\sqrt{3}\frac{\cancel{\,\mathrm{m}}}{\,\mathrm{s}}\mathord{\cdot% }\frac{4}{1\cancel{\,\mathrm{m}}}=-4\sqrt{3}\,\mathrm{rad}/\mathrm{s}.$

Now substituting the appropriate values in Eqn (17.25) we get:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+\omega\hat{% \boldsymbol{k}}\times\underbrace{\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{% \imath}}-\sin\theta\hat{\boldsymbol{\jmath}})}_{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{G/A}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+\frac{\omega% \ell}{2}(\cos\theta\hat{\boldsymbol{\jmath}}+\sin\theta\hat{\boldsymbol{\imath% }})$
	$\displaystyle=$	$\displaystyle(2\hat{\boldsymbol{\imath}}+\sqrt{3}\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}-\sqrt{3}\,\mathrm{m}/\mathrm{s}\mathord{\cdot}\left(% \frac{1}{2}\hat{\boldsymbol{\jmath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{\imath% }}\right)$
	$\displaystyle=$	$\displaystyle\left(2-\frac{3}{2}\right)\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol% {\imath}}+\left(\sqrt{3}-\frac{\sqrt{3}}{2}\right)\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+\frac% {\sqrt{3}}{2}\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}=(0.5\hat{\boldsymbol{\imath}% }+0.87\hat{\boldsymbol{\jmath}})\,\mathrm{m}/\mathrm{s}$

SAMPLE 17.4

Filename:sfig8-6-2a

Figure 17.18: A board in the back of an accelerating truck.

A board in the back of an accelerating truck. A $10\,\mathrm{ft}$ long uniform board AB rests in the back of a flat-bed truck as shown in Fig. 17.18. End A of the board is hinged to the bed of the truck. The truck is going on a level road at 55 $\,\mathrm{mph}$ . In preparation for overtaking a vehicle in the front, the trucker accelerates at a rate of 3 $\,\mathrm{mph}/s$ . At the instant when the speed of the truck is 60 $\,\mathrm{mph}$ , the magnitude of the relative velocity and relative acceleration of end B with respect to the bed of the truck are $10\,{\mathrm{ft/s}}$ and $12\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ , respectively. There is wind and at this instant, the board has lost contact with point C. If the angle $\theta$ between the board and the bed is $45^{{}^{\circ}}$ at the instant mentioned, find

1.

the angular velocity and angular acceleration of the board,
2.

the absolute velocity and absolute acceleration of point B, and
3.

the acceleration of the center-of-mass G of the board.

Solution At the instant of interest, the quantities given are:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle\mbox{ velocity of the truck }=60\,\mathrm{mph}~{}\hat{% \boldsymbol{\imath}}=88\,{\mathrm{ft/s}}~{}\hat{\boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle=$	$\displaystyle\mbox{ acceleration of the truck }=3\,\mathrm{mph}/s=4.4\,{% \mathrm{ft}}/{\mathrm{s}}^{2}~{}\hat{\boldsymbol{\imath}}$
$\displaystyle\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}\|$	$\displaystyle=$	$\displaystyle v_{B/A}=\mbox{ magnitude of relative velocity of B }=10\,{% \mathrm{ft/s}}$
$\displaystyle\|{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}\|$	$\displaystyle=$	$\displaystyle a_{B/A}=\mbox{ magnitude of relative acceleration of B }=12\,{% \mathrm{ft}}/{\mathrm{s}}^{2}.$

Let ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ be the angular velocity and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\omega}\hat{% \boldsymbol{k}}$ be the angular acceleration of the board.

The relative velocity of end B of the board with respect to end A is

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}=\omega\hat{\boldsymbol{k}}% \times\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle\omega\ell(\cos\theta\hat{\boldsymbol{\jmath}}-\sin\theta\hat{% \boldsymbol{\imath}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega$	$\displaystyle=$	$\displaystyle\frac{\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}\|}{\ell% }=\frac{v_{B/A}}{\ell}=\frac{10\,{\mathrm{ft/s}}}{10\,\mathrm{ft}}=1\,\mathrm{% rad}/\mathrm{s}.$

Note that we have taken the positive value for $\omega$ because the board is rotating counterclockwise at the instant of interest (since it is given that the board has lost contact with point C).

Similarly, we can compute the angular acceleration:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B/A}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}$
	$\displaystyle=$	$\displaystyle\dot{\omega}\hat{\boldsymbol{k}}\times\ell(\cos\theta\hat{% \boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})-\omega^{2}\ell(\cos% \theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle\dot{\omega}\ell(\cos\theta\hat{\boldsymbol{\jmath}}-\sin\theta% \hat{\boldsymbol{\imath}})-\omega^{2}\ell(\cos\theta\hat{\boldsymbol{\imath}}+% \sin\theta\hat{\boldsymbol{\jmath}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\|{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B/A}\|$	$\displaystyle=$	$\displaystyle\sqrt{(\dot{\omega}\ell)^{2}+(\omega^{2}\ell)^{2}}=a_{B/A}\mbox{ % (given) }$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a_{B/A}^{2}$	$\displaystyle=$	$\displaystyle(\dot{\omega}\ell)^{2}+(\omega^{2}\ell)^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\omega}$	$\displaystyle=$	$\displaystyle\sqrt{\frac{a_{B/A}^{2}}{\ell^{2}}-\omega^{4}}=\sqrt{\left(\frac{% 12\,{\mathrm{ft}}/{\mathrm{s}}^{2}}{10\,\mathrm{ft}}\right)^{2}-(1\,\mathrm{% rad}/\mathrm{s})^{4}}$
	$\displaystyle=$	$\displaystyle\pm 0.663\,\mathrm{rad}/\mathrm{s}^{2}.$

Once again, we select the positive value for $\dot{\omega}$ since we assume that the board accelerates counterclockwise.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=1\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}},\;\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=% 0.663\,\mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}$

The absolute velocity and the absolute acceleration of the end point B can be found as follows.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B/A}$
	$\displaystyle=$	$\displaystyle v_{A}\hat{\boldsymbol{\imath}}+v_{B/A}(\cos\theta\hat{% \boldsymbol{\jmath}}-\sin\theta\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 88\,{\mathrm{ft/s}}\hat{\boldsymbol{\imath}}+10\,{\mathrm{ft/s}}% (\frac{1}{\sqrt{2}}\hat{\boldsymbol{\jmath}}-\frac{1}{\sqrt{2}}\hat{% \boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 80.93\,{\mathrm{ft/s}}\hat{\boldsymbol{\imath}}+7.07\,{\mathrm{% ft/s}}\hat{\boldsymbol{\jmath}}.$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B/A}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+\dot{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{B/A}$
	$\displaystyle=$	$\displaystyle a_{A}\hat{\boldsymbol{\imath}}+\dot{\omega}\hat{\boldsymbol{k}}% \times\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{% \jmath}})-\omega^{2}\ell(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(a_{A}-\dot{\omega}\ell\sin\theta-\omega^{2}\ell\cos\theta)\hat{% \boldsymbol{\imath}}+(\dot{\omega}\ell\cos\theta-\omega^{2}\ell\sin\theta)\hat% {\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\left(4.4\,{\mathrm{ft}}/{\mathrm{s}}^{2}-\frac{0.66}{\,\mathrm{s% }^{2}}\cdot 10\,\mathrm{ft}\cdot\frac{1}{\sqrt{2}}-\frac{1}{\,\mathrm{s}^{2}}% \cdot 10\,\mathrm{ft}\cdot\frac{1}{\sqrt{2}}\right)\hat{\boldsymbol{\imath}}$
		$\displaystyle+\left(\frac{0.66}{\,\mathrm{s}^{2}}\cdot 10\,\mathrm{ft}\cdot% \frac{1}{\sqrt{2}}-\frac{1}{\,\mathrm{s}^{2}}\cdot 10\,\mathrm{ft}\cdot\frac{1% }{\sqrt{2}}\right)\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-7.34\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}-2.% 40\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=(80.93\hat{\boldsymbol{% \imath}}+7.07\hat{\boldsymbol{\jmath}})\,{\mathrm{ft/s}},\quad{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B}=-(7.34\hat{\boldsymbol{\imath}}+2.40\hat% {\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}.$

Now, we can easily calculate the acceleration of the center-of-mass as follows.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{G/A}$
	$\displaystyle=$	$\displaystyle a_{A}\hat{\boldsymbol{\imath}}+\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/A}% -\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/A}$
	$\displaystyle=$	$\displaystyle a_{A}\hat{\boldsymbol{\imath}}+\dot{\omega}\hat{\boldsymbol{k}}% \times\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}})-\omega^{2}\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{% \imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle a_{A}\hat{\boldsymbol{\imath}}+\dot{\omega}\frac{\ell}{2}(\cos% \theta\hat{\boldsymbol{\jmath}}-\sin\theta\hat{\boldsymbol{\imath}})-\omega^{2% }\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol% {\jmath}})$
	$\displaystyle=$	$\displaystyle 4.4\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}+0.6% 63\,\mathrm{rad}/\mathrm{s}^{2}\cdot\frac{10\,\mathrm{ft}}{2}\cdot(\frac{1}{% \sqrt{2}}\hat{\boldsymbol{\jmath}}-\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}})$
		$\displaystyle-(1\,\mathrm{rad}/\mathrm{s})^{2}\cdot\frac{10\,\mathrm{ft}}{2}% \cdot(\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}}+\frac{1}{\sqrt{2}}\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-1.48\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\imath}}-1.% 19\,{\mathrm{ft}}/{\mathrm{s}}^{2}\hat{\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=-(1.48\hat{\boldsymbol{% \imath}}+1.19\hat{\boldsymbol{\jmath}})\,{\mathrm{ft}}/{\mathrm{s}}^{2}$

Comments: This problem is admittedly artificial. We, however, solve this problem to show kinematic calculations.

SAMPLE 17.5

Filename:efig1-2-28

Figure 17.19:

Tracking motion. A cart moves along a suspended curved path. A rod AB of length $\ell=1\,\mathrm{m}$ hangs from the cart. End A of the rod is attached to a motor on the cart. The other end, B, hangs freely. The motor rotates the rod such that $\theta(t)=\theta_{0}\sin(\lambda t)$ ( $\theta$ is measured with respect to a fixed vertical line) while the cart moves along the path such that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}=t\hat{\boldsymbol{\imath}}+% \frac{t^{3}}{18}\hat{\boldsymbol{\jmath}}$ , where all variables ( $r$ , $t$ , etc.) are nondimensional.

1.

Find the velocity and acceleration of point B as a function of nondimensional time $t$ .
2.

Take $\theta_{0}=\pi/3$ and $\lambda=6$ . Find and plot the position of the bar at $t=0,0.1,0.3,0.9,1,1.1,1.2$ , and $1.5$ . Find and draw ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B}}$ at the specified $t$ .

Solution

The velocity and acceleration of point B are given by

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B/A}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{A}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+\dot{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{B/A}-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{B/A}.$

Thus, in order to find the velocity and acceleration of point B, we need to find the velocity and acceleration of point A and the angular velocity and angular acceleration of the bar. We are given the position of point A and the angular position of the bar as functions of $t$ . We can, therefore, find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ , ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ , and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ by differentiating the given functions with respect to $t$ .

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}$	$\displaystyle=$	$\displaystyle t\hat{\boldsymbol{\imath}}+\frac{t^{3}}{18}\hat{\boldsymbol{% \jmath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle\equiv$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}=\hat{% \boldsymbol{\imath}}+(t^{2}/6)\hat{\boldsymbol{\jmath}}$	(17.26)
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle\equiv$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=(t/3)\hat{% \boldsymbol{\jmath}}$	(17.27)
and
$\displaystyle\theta\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle\theta_{0}\sin(\lambda t)\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle\equiv$	$\displaystyle\dot{\theta}\hat{\boldsymbol{k}}=\theta_{0}\lambda\cos(\lambda t)% \hat{\boldsymbol{k}}$	(17.28)
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle\equiv$	$\displaystyle\ddot{\theta}\hat{\boldsymbol{k}}=-\theta_{0}\lambda^{2}\sin(% \lambda t)\hat{\boldsymbol{k}}.$	(17.29)

So,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times\ell(\sin\theta\hat{\boldsymbol{% \imath}}-\cos\theta\hat{\boldsymbol{\jmath}})$	(17.30)
	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\imath}}+(t^{2}/6)\hat{\boldsymbol{\jmath}}+\ell% \dot{\theta}(\sin\theta\hat{\boldsymbol{\jmath}}+\cos\theta\hat{\boldsymbol{% \imath}})$
	$\displaystyle=$	$\displaystyle(1+\ell\dot{\theta}\cos\theta)\hat{\boldsymbol{\imath}}+(t^{2}/6+% \ell\dot{\theta}\sin\theta)\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}+\ddot{\theta}% \hat{\boldsymbol{k}}\times\ell(\sin\theta\hat{\boldsymbol{\imath}}-\cos\theta% \hat{\boldsymbol{\jmath}})-\dot{\theta}^{2}\ell(\sin\theta\hat{\boldsymbol{% \imath}}-\cos\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(t/3)\hat{\boldsymbol{\jmath}}+\ell\ddot{\theta}\sin\theta\hat{% \boldsymbol{\jmath}}+\ell\ddot{\theta}\cos\theta\hat{\boldsymbol{\imath}}-\ell% \dot{\theta}^{2}\sin\theta\hat{\boldsymbol{\imath}}+\ell\dot{\theta}^{2}\cos% \theta\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\ell(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta)\hat{% \boldsymbol{\imath}}+[t/3+\ell(\ddot{\theta}\sin\theta+\dot{\theta}^{2}\cos% \theta)]\hat{\boldsymbol{\jmath}}$

where $\theta=\theta_{0}\sin(\lambda t),\,\,\dot{\theta}=\theta_{0}\lambda\cos(% \lambda t)$ , and $\ddot{\theta}=-\theta_{0}\lambda^{2}\sin(\lambda t)=-\lambda^{2}\theta.$ Thus ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$ are functions of $t$ .

2.

The position of the rod at any time $t$ is specified by the position of the two end points A and B (or alternatively, the position of A and the angle of the rod $\theta$ ). The position of point A is easily determined by substituting the value of $t$ in the given expression for ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}$ . The position of end B is given by

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{B/A}=t\hat{\boldsymbol{\imath}}+(t^{3}/18)% \hat{\boldsymbol{\jmath}}+\ell(\sin\theta\hat{\boldsymbol{\imath}}-\cos\theta% \hat{\boldsymbol{\jmath}})$

$\displaystyle=$ $\displaystyle(t+\ell\sin\theta)\hat{\boldsymbol{\imath}}+(t^{3}/18-\ell\cos% \theta)\hat{\boldsymbol{\jmath}}.$

To compute the positions, velocities, and accelerations of end points A and B at the given instants, we first compute $\theta$ , $\dot{\theta}$ , and $\ddot{\theta}$ , and then substitute them in the expressions for ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A},{\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{B},{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A},{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{B},{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{A}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$ . A pseudocode for computer calculation is given below.
```
t = [0 0.1 0.3 0.9 1.0 1.1 1.2 1.5]
theta0=pi/3, L=.5, lam=6
for each t, compute
  theta = theta0*sin(lam*t)
  w = lam*theta0*cos(lam*t)
  wdot = -lam^2*theta
       % Position of A and B
  xA=t, yA=t^3/18
  xB = xA + L*sin(theta)
  yB = yA - L*cos(theta)
       % Velocity of A and B
  uA = 1,  vA = t^2/6
  uB = uA + L*w.*cos(theta)
  vB = vA + L*w.*sin(theta)
       % Acceleration of A and B
  axA = 0,  ayA = t/3
  axB = L*wdot*cos(theta) - L*w^2*sin(theta)
  ayB = ayA + L*wdot*sin(theta) + L*w^2*cos(theta)
```
From the above calculation, we get the desired quantities at each $t$ . For example, at $t=0$ we get,
```
xA = 0, yA = 0, xB = 0, yB = -0.5
uA = 1, vA = 0, uB = 4.14, vB = 0, axB = 0, ayB = 19.74
```
which means,

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{0}}},\quad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}=-0.5% \hat{\boldsymbol{\jmath}},\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% A}=\hat{\boldsymbol{\imath}},\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}% }_{B}=4.14\hat{\boldsymbol{\imath}},\quad{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{B}=19.74\hat{\boldsymbol{\jmath}}.$

The position of the bar, the velocity vectors at points A and B, and the acceleration vector at B, thus obtained, are shown in fig. 17.20 graphically. ^†^†margin: We can take several values of $t$ , say 400 equally spaced values between $t=0$ and $t=4$ , and draw the bar at each $t$ to see its motion and the trajectory of its end points. fig. 17.21 shows such a graph.

Filename:pfigure-blue-118-2

Figure 17.20: Position, velocity of the end points A and B, and acceleration of point B at various time instants.

Filename:pfigure-s95f3a

Figure 17.21: Graph of closely spaced configuration of the bar between

t=0

t=4

Problems for 17.1 Rigid object kinematics

Preparatory Problems

17.1.1 A disk of radius R is hinged at point O at the edge of the disk, approximately as shown. It rotates counterclockwise with angular velocity $\dot{\theta}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . A bolt is fixed on the disk at point P at a distance $r$ from the center of the disk. A frame $x^{\prime}y^{\prime}$ is fixed to the disk with its origin at the center C of the disk. The bolt position P makes an angle $\phi$ with the $x^{\prime}$ -axis. At the instant of interest, the disk has rotated by an angle $\theta$ .

(a)

Write the position vector of point P relative to C in the $x^{\prime}y^{\prime}$ coordinates in terms of given quantities.
(b)

Write the position vector of point P relative to O in the $x y$ coordinates in terms of given quantities.
(c)

Write the expressions for the rotation matrix $R(\theta)$ and the angular velocity matrix $\cal{S}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}})$ .
(d)

Find the velocity of point P relative to C using $R(\theta)$ and the angular velocity matrix $\cal{S}({\accentset{\rightharpoonup}{\boldsymbol{\omega}}})$ .
(e)

Using $R=30\,\mathrm{cm}$ , $r=25\,\mathrm{cm}$ , $\theta=60^{\circ}$ , and $\phi=45^{\circ}$ , find $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/0}}]_{xy}$ , and $[{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/0}}]_{xy}$ at the instant shown.
(f)

Assuming that the angular speed is $\omega=10\,\mathrm{rad}/\mathrm{s}$ at the instant shown, find $[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C/0}}]_{xy}$ and $[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P/0}}]_{xy}$ taking other quantities as specified above.

Filename:Danef94s1q2

Figure 17.22:

17.1.2 A uniform rigid rod AB of length $\ell=1\,\mathrm{m}$ rotates at a constant angular speed $\omega$ about an unknown fixed point. At the instant shown, the velocities of the two ends of the rod are ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=-1\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=1\,\mathrm{m}/\mathrm{s}\hat% {\boldsymbol{\jmath}}$ .

(a)

Find the angular velocity of the rod.
(b)

Find the center of rotation of the rod.

Filename:pfigure-blue-123-1

Figure 17.23:

17.1.3 A square plate ABCD rotates at a constant angular speed about an unknown point in its plane. At the instant shown, the velocities of the two corner points A and D are ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=-2\,{\mathrm{ft/s}}(\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}=-(2\,{\mathrm{ft/s}})\hat{% \boldsymbol{\imath}}$ , respectively.

(a)

Find the center of rotation of the plate.
(b)

Find the acceleration of the center of mass of the plate.

Filename:pfigure-blue-119-2

Figure 17.24:

More-Involved Problems

17.1.4 A ten foot ladder is leaning between a floor and a wall. The top of the ladder is sliding down the wall at one foot per second. (The foot is simultaneously sliding out on the floor). When the ladder makes a 45 degree angle with the vertical what is the speed of the midpoint of the ladder?

17.1.5 The slender rod AB rests against the step of height $h$ , while end “A” is moved along the ground at a constant velocity $v_{o}$ . Find $\dot{\phi}$ and $\ddot{\phi}$ in terms of $x$ , $h$ , and $v_{o}$ . Is $\dot{\phi}$ positive or negative? Is $\ddot{\phi}$ positive or negative?

Filename:pfigure-s95q14

Figure 17.25:

17.1.6 Consider the motion of a rigid ladder which can slide on a wall and on the floor as shown in the figure. The point A on the ladder moves parallel to the wall. The point B moves parallel to the floor. Yet, at a given instant, both have velocities that are consistent with the ladder rotating about some special point, the center of rotation (COR). Define appropriate dimensions for the problem.

(a)

Find the COR for the ladder when it is at some given position (and moving, of course). [Hint: if a point A is ‘going in circles’ about another point C, that other point C must be in the direction perpendicular to the motion of A.
(b)

As the ladder moves, the COR changes with time. What is the set of points on the plane that are the COR’s for the ladder as it falls from straight up to lying on the floor?

Filename:Danef94s3q2

Figure 17.26:

17.1.7 A car driver on a very boring highway is carefully monitoring her speed. Over a one hour period, the car travels on a curve with constant radius of curvature, $\rho=25\,{\rm mi}$ , and its speed increases uniformly from $50\,\mathrm{mph}$ to $60\,\mathrm{mph}$ . What is the acceleration of the car half-way through this one hour period, in path coordinates?

17.1.8 The bar AB shown in the figure is 1 m long. The end A of the bar is dragged to the left on the horizontal floor at a constant speed $v_{0}=0.2\,\mathrm{m}/\mathrm{s}$ . The other end of the bar drags along the inclined plane that makes an angle $\phi=60^{\circ}$ with the horizontal.

(a)

Does the end B of the rod also have a constant speed? Does it imply that the rod has a constant angular velocity? Guess your answer and write down your reasoning. No calculations.
(b)

Find an expression for the angular velocity of the rod in terms of $\theta$ , $\phi$ , $\ell$ , and $v_{0}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\frac{v_{0}\sin\phi}{\ell% \cos(\phi-\theta)}\hat{\boldsymbol{k}}$
(c)

Find the velocity of end B when $\theta=30^{\circ}$ .

Filename:bikefork1-ang-accel

Figure 17.27:

17.1.9 The disk shown in the figure drives the slider AB. The disk rotates clockwise at a constant angular speed $\omega_{D}=60\,{\rm rpm}$ . The radius of the disk is $R=0.25\,\mathrm{m}$ .

(a)

Find the velocity of collar A when $\theta=30^{\circ}$ .
(b)

Find the acceleration of collar A when $\theta=30^{\circ}$ .
(c)

Find the maximum and minimum velocity of the collar and the corresponding angle $\theta$ of the slider.

Filename:bikefork-ang-accel

Figure 17.28:

17.1.10 A rod AC of length $\ell=2\,\mathrm{m}$ is mounted on a cart at point C. The cart slides down under gravity on a frictionless surface inclined at $\phi=45^{\circ}$ from the horizontal. Assume that bar has negligible mass compared to that of the cart. The cart is released from rest and it reaches the position shown in $t=2\,\mathrm{s}$ . At the instant shown, $\theta=60^{\circ}$ , and rod AC has an angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=0.5\,\mathrm{rad}/\mathrm{s% }\hat{\boldsymbol{k}}$ and angular acceleration ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=1\,\mathrm{rad}/\mathrm{s}^% {2}\hat{\boldsymbol{k}}$ . Find the velocity and acceleration of the center of mass of the rod at the instant shown.

Filename:bikefork1-alt

Figure 17.29:

17.1.11 A vertical bar pendulum AB of length $\ell=4\,\mathrm{ft}$ is mounted on a cart at point A. The cart has a motor that drives the bar pendulum back and forth about the vertical position such that $\theta(t)=\theta_{0}\sin\omega t$ where $\theta_{0}=\pi/6$ and $\omega=\pi\,\mathrm{rad}/\mathrm{s}$ . The cart moves at a constant speed $v_{0}=2\,{\mathrm{ft/s}}$ on a sinusoidal track represented by $y(x)=y_{0}\sin\lambda x$ where $y_{0}=0.5\,\mathrm{ft}$ and $\lambda=\pi/4/\,\mathrm{ft}$ .

(a)

Find the velocity of point B when $\theta=0$ .
(b)

Find and plot the position of point B for one full oscillation of the bar, assuming the bar starts from its leftmost angular position (that is, $\theta=-\theta_{0}$ ) when the cart is on one of the crests of the track.

Filename:bikefork-alt

Figure 17.30:

17.1.12 The center of mass of a javelin travels on a more or less parabolic path while the javelin rotates during its flight. In a particular throw, the velocity of the center of mass of a javelin is measured to be ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=10\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$ when the center of mass is at its highest point $h=6\,\mathrm{m}$ . As the javelin lands on the ground, its nose hits the ground at G such that the javelin is almost tangent to the path of the center of mass at G. Neglect the air drag and lift on the javelin.

(a)

Given that the javelin is at an angle $\theta=45^{\circ}$ at the highest point, find the angular velocity of the javelin. Assume the angular velocity is constant during the flight and that the javelin makes less than a full revolution.

Filename:pfigure-blue-128-1

Figure 17.31:

17.2 General planar mechanics of a rigid-object

We now apply the kinematics ideas of the last section to the general mechanics principles in Table I in the inside cover. The goal is to understand the relation between forces and motion for a planar object in general 2-D motion. The simple measures^†^†margin: Advanced aside: What we call “simple measures” are examples of “generalized coordinates” in more advanced books. The idea sounds intimidating, but is simply this: If something can only move in a few ways, you should only keep track of the motion with that many variables. The kinematics of a rigid object (Sect. 17.1) allow us to “evaluate” the motion quantities, namely linear momentum, angular momentum, kinetic energy, and their rates of change in terms of these “simple measures”. By “evaluate” we mean express the motion quantities in terms of these measures. The alternative is as sums over Avogadro’s number of particles (There are on the order of $10^{23}$ atoms in a typical engineering part.). Even neglecting atoms and viewing matter as continuous we would still be stuck with integrals over complicated regions if we did not describe the motion with as few variables as possible. In the case of 2-D rigid object motion, the position of a reference point ( $x$ and $y$ ) with the rotation $\theta$ is called a set of minimal coordinates. These, and their time derivatives are the minimal information needed to describe all important mechanics motion quantities. of motion will be the displacement of one reference point $0^{\prime}$ on the object and the rotation of the object. We also need the first and second time derivatives of the displacement and rotation, altogether we use ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0^{\prime}}},{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{0^{\prime}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{0^{\prime}}},\theta,{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}$ .

We will treat all bodies as if they are squished into the plane (planar) and moving in the plane (planar motion). But the analysis is sensible for an object that is symmetric with respect to the plane containing the velocities (see Box 17.2 on page 17.2).

The balance laws for a rigid object

As always, once you have defined the system and the forces acting on it by drawing a free-body diagram, the basic momentum balance equations are applicable (and exact for engineering purposes). Namely,

		Linear momentum balance:	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{L}}}\qquad\mbox{and}$
		Angular momentum balance:	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{i/O}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.% 0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/O}$}}}.$

The same point 0, any point, is used on both sides of the angular momentum balance equation.

We also have power balance which, for a system with no internal energy or dissipation, is

\mbox{Power balance:}\qquad P={\dot{E}_{\mathrm{K}}}.

The left hand sides of the momentum balance equations are evaluated the same way, whether the system is composed of one object or many, whether the bodies are deformable or not, and whether the points move in straight lines, circles, hither and thither, or not at all. It is the right hand sides of the momentum equations that involve the motion.

Similarly, in the energy balance equations the applied power $P$ only depends on the position of the forces and the motions of the material points at those positions. But the kinetic energy ${E_{\mathrm{K}}}$ and its rate of change depend on the motion of the whole system.

You already know how to evaluate the momenta and energy, and their rates of change, for a variety of special cases, namely

•

Systems composed of particles where all the positions and accelerations are known (chapter 13);
•

Systems where all points have the same acceleration. That is, the system moves like a rigid object that does not rotate (chapter 14); and
•

Systems where all points move in circles about the same fixed axis in 2D (chapter 15).

Now we go on to consider the general 2-D motions of a planar rigid object. It’s now a little harder to evaluate ${\accentset{\rightharpoonup}{\boldsymbol{L}}},\dot{\accentset{\rightharpoonup}% {\boldsymbol{L}}},{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}},\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle% \mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}% },{E_{\mathrm{K}}}$ and ${\dot{E}_{\mathrm{K}}}$ . But not much.

Summary of the motion quantities

Table I in the back of the book describes the motion quantities for various special cases, including the planar motions we consider in this chapter. Most relevant is row (7).

The basic idea is that the momenta for general motion, which involves translation and rotation, are the sum of the momenta (both linear and angular, and their rates of change too) from those two effects.

Namely, the linear momentum is described, as for any system with any motion, by the motion of the center-of-mass

{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\hbox to36.135pt{\hfil and\hfil% }\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset% {\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}},

(17.32)

and the angular momentum has two contributions, one from the motion of the center-of-mass and one from rotation of the object about the center of mass,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{cm/O}}\times(m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{\mathrm{cm}})}^{\text{Angular momentum due to motion of the center-of-% mass}}+\overbrace{I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}}^{\text{Angular momentum due to motion relative to the center-of-% mass}}$		(17.33)
	$\displaystyle\hbox{and\quad}\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_% {\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}% $}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/O}}% \times(m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{% cm}})+I^{\mathrm{cm}}_{zz}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}.$		(17.34)

These simplifications, for 2D objects moving in the plane, are discussed in box 17.2 on page 17.2.

An important special case for the angular momentum evaluation is when the reference point is coincident with the center-of-mass. Then the angular momentum and its rate of change simplify to

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}=I^{\mathrm{cm}}_{% zz}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\qquad\qquad\hbox to36.13% 5pt{\hfil and\hfil}\qquad\qquad\dot{\accentset{\rightharpoonup}{\boldsymbol{H}% }}_{\mathrm{cm}}=I^{\mathrm{cm}}_{zz}\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}.

(17.35)

The kinetic energy and its rate of change are given by

	$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\overbrace{{1\over 2}m_{\mathrm{tot}}\underbrace{v_{cm}^{2}}_{{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{cm}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{cm}}}}^{% \text{kinetic energy from\hfil\break center-of-mass motion }}\par+\overbrace{{% 1\over 2}I^{\mathrm{cm}}_{zz}\omega^{2}}^{\text{kinetic energy relative to the center-of-mass}}$		(17.36)
	$\displaystyle\hbox{and\quad}{\dot{E}_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle m_{\mathrm{tot}}\underbrace{v_{cm}\dot{v}_{cm}}_{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}}\quad+\ I^{\mathrm{cm}}_{zz}% \omega\dot{\omega}$		(17.37)

The relations above are easily derived from the general center of mass theorems (see box 17.2 on page 17.2 for some of these derivations).

Equations of motion

Putting together the general balance equations and the expressions for the motion quantities we can now write linear momentum balance, angular momentum balance and power balance as:

\begin{array}[]{crclr}{\rm LMB:}&\displaystyle\sum{\accentset{\rightharpoonup}% {\boldsymbol{F}}}_{i}&=&\displaystyle m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}},&\hbox{(a)}\\ \\ {\rm AMB:}&\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.% 0pt\hbox{$\scriptscriptstyle\mathrm{O}$}}}&=&\displaystyle{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/O}}\times(m_{\mathrm{tot}}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}})+I^{\mathrm{cm}}_{% zz}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}&\hbox{(b)}\\ \\ {\rm or}&\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathrm{cm}}&=&I^{\mathrm{cm}}_{zz}\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}},\qquad\mbox{and}\\ \\ {\rm Power:}&\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{tot}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{cm}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}&=&m_{\mathrm{tot}}v% \dot{v}+I^{\mathrm{cm}}_{zz}\omega\dot{\omega}.\displaystyle&\hbox{(c)}\end{array}

(17.38)

Independent equations?

Equations are only independent if no one of them can be derived from the others. When counting equations and unknowns one needs to make sure one is writing independent equations. How many independent equations are in the set eqns. (17.38)abc applied to one free object diagram? The short answer is 3.

The linear momentum balance equation 17.38a yields two independent equations by dotting with any two non-parallel vectors (say, $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ ). Dotting with a third vector yields a dependent equation.

For any one reference point the angular momentum equation 17.38b yields one scalar equation. It is a vector equation but always has zero components in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions. But angular momentum equation can yield up to three independent equations by being applied to any set of three non-collinear points.

The power balance equation is one scalar equation.

In total, however, the full set of equations above only makes up a set of three independent equations.

To avoid thinking about what is or is not an independent set of equations some people prefer to stick with one of the canonical sets of independent equations:

•
The coordinate based set (“old standard”)
- –
  
  {LMB} $\cdot\hat{\boldsymbol{\imath}}$ or, equivalently, $\sum F_{x}=m_{\mathrm{tot}}a_{{\rm cm}_{x}}$ ,
- –
  
  {LMB} $\cdot\hat{\boldsymbol{\jmath}}$ or, equivalently, $\sum F_{y}=m_{\mathrm{tot}}a_{{\rm cm}_{y}}$ , and
- –
  
  {AMB} $\cdot\hat{\boldsymbol{k}}$ or, equivalently, $\sum M_{\rm cm}=I^{\mathrm{cm}}_{zz}\dot{\omega}$ .
•
Moment only (good for eliminating unknown reaction forces)
- –
  
  {AMB about pt A} $\cdot\hat{\boldsymbol{k}}$ (A is any point, on or off the object)
- –
  
  {AMB about pt B} $\cdot\hat{\boldsymbol{k}}$ (B is any other point)
- –
  
  {AMB about pt C} $\cdot\hat{\boldsymbol{k}}$ (C is a third point not on the line AB)
•
Two moments and a force component
- –
  
  {AMB about pt A} $\cdot\hat{\boldsymbol{k}}$ (A is any point, on or off the object)
- –
  
  {AMB about pt B} $\cdot\hat{\boldsymbol{k}}$ (B is any other point)
- –
  
  {LMB} $\cdot\hat{\boldsymbol{\lambda}}$ (where $\hat{\boldsymbol{\lambda}}$ is not perpendicular to the line AB)
•
Two force components and a moment (also good for eliminating unknown forces)
- –
  
  {LMB} $\cdot\hat{\boldsymbol{\lambda}}_{1}$ (where $\hat{\boldsymbol{\lambda}}_{1}$ is any unit vector)
- –
  
  {LMB} $\cdot\hat{\boldsymbol{\lambda}}_{2}$ (where $\hat{\boldsymbol{\lambda}}_{2}$ is any other unit vector)
- –
  
  {AMB about pt A} $\cdot\hat{\boldsymbol{k}}$ (A is any point, on or off the object)

Any of these will always do the job. The power balance equation is often used as a consistency check rather than an independent equation.

From a theoretical point of view one might ask the related question of which of the equations of motion can be derived from the others. There are many answers. Here are some of them:

•

Power balance follows from LMB and AMB,
•

AMB about three non-collinear points implies LMB, and
•

LMB and power balance yield AMB

Interestingly, there is no way to derive angular momentum balance from linear momentum balance without the questionable microscopic assumptions discussed in box LABEL:threeten on page LABEL:threeten.

Some simple examples

Here we consider some simple examples of unconstrained motion of a rigid object.

Example: The simplest case: no force and no moment.

If the net force and moment applied to an object are zero we have:

	$\displaystyle\hbox{LMB}\quad\Rightarrow{\accentset{\rightharpoonup}{% \boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{cm}}\quad\hbox{and}$
	$\displaystyle\hbox{AMB}\quad\Rightarrow{\accentset{\rightharpoonup}{% \boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}$

so ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ and $\dot{\omega}=0$ and the object moves at constant speed in a constant direction with a constant rate of rotation, all determined by the initial conditions. Throw an object in space and its center-of-mass goes in a straight line and it spins at constant rate (subject to the 2-D restrictions of this chapter).

Filename:tfigure8-syst-bods

Figure 17.32: The X marked at the center-of-mass of a thrown spinning clipboard follows a parabolic trajectory.

Example: Constant force applied to the center-of-mass.

In this case angular momentum balance about the center-of-mass again yields that the rotation rate is constant. Linear momentum balance is now the same as for a particle at the center-of-mass, i.e., the center-of-mass has a parabolic trajectory.

Near-earth (constant) gravity provides a simple example. An ‘X’ marked at the center-of-mass of a clipboard tossed across a room follows a parabolic trajectory (see fig. 17.32).

Example: Constant force not at the center-of-mass.

Assume the only force applied to an object is a constant force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\imath}}$ at A (see fig. 17.33). Then linear momentum balance gives us that

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{L}}}\quad\Rightarrow\quad F\hat{\boldsymbol{% \imath}}=m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}}\quad% \Rightarrow\quad{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}}=F/m% \hat{\boldsymbol{\imath}}=\hbox{constant}.

So if the object starts at rest, the point G will move in a straight line in the $\hat{\boldsymbol{\imath}}$ direction (The common intuition that point G will be pulled up is incorrect). Angular momentum about the center-of-mass gives

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm\,i}% }=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}\quad$	$\displaystyle\Rightarrow$	$\displaystyle\quad\left\{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{A/G}}\times F\hat{\boldsymbol{\imath}}=I^{\mathrm{cm}}_{zz}\ddot{% \theta}\hat{\boldsymbol{k}}\right\}$
	$\displaystyle\{\}\cdot\hat{\boldsymbol{k}}$	$\displaystyle\Rightarrow$	$\displaystyle\quad\ddot{\theta}+{F\ell\over I^{\mathrm{cm}}_{zz}}\sin\theta=0,$

with $\ell=|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/G}}|$ , which is the pendulum equation. That is, the object can swing back and forth about $\theta=0$ just like a pendulum, approximately sinusoidally if the angle $\theta$ starts small and with $\dot{\theta}$ initially also small. [One might wonder how to do this experiment. One way would be with a jet on a space craft that keeps re-orienting itself to keep in a constant spatial direction as the object changes orientation. Another would be with a string tied to A and pulled from a great distance.]

Filename:sfig8-7-2

Figure 17.33: The only force applied to the object is the constant force

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\imath}}

applied at point A. The resulting motion is a constant acceleration of the center-of-mass

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}}=(F/m)\hat{% \boldsymbol{\imath}}

and an oscillatory motion of

\theta

identical to that of a pendulum hinged at G. Note that the center of mass goes in a straight line.

Filename:sfig8-7-2a

Figure 17.34: A rocket is pointed in the direction

\hat{\boldsymbol{\lambda}}

which makes an angle

\theta

with the positive

x

axis. The position and velocity of the center-of-mass at G are called

{\accentset{\rightharpoonup}{\boldsymbol{r}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{v}}}

. The velocity of the tail is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}

Example: The flight of an arrow or rocket.

As a primitive model of an arrow or rocket assume that the only force is from drag on the fins at C and that this force opposes motion according to

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-c{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}

where c is a drag coefficient (see fig. 17.34). From linear momentum balance we have:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{L}}}\quad\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\quad{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$
$\displaystyle-c{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}$	$\displaystyle=$	$\displaystyle m\dot{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}$
$\displaystyle m\dot{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}$	$\displaystyle=$	$\displaystyle-c\left({\accentset{\rightharpoonup}{\boldsymbol{v}}}+{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{C/G}}\right)$
	$\displaystyle=$	$\displaystyle-c\left({\accentset{\rightharpoonup}{\boldsymbol{v}}}+\dot{\theta% }\hat{\boldsymbol{k}}\times(-\ell\hat{\boldsymbol{\lambda}})\right)$
$\displaystyle(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\lambda}}=\hat{% \boldsymbol{n}})\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\dot{{% \accentset{\rightharpoonup}{\boldsymbol{v}}}}$	$\displaystyle=$	$\displaystyle{c\over m}\left(\dot{\theta}\ell\hat{\boldsymbol{n}}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}\right).$

So if ${\accentset{\rightharpoonup}{\boldsymbol{v}}},\theta$ and $\dot{\theta}$ are known the acceleration $\dot{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}$ is calculated by the formula above.

Similarly angular momentum balance about G gives

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{G}}=% \dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm G}\quad\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}\quad\{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{C/G}}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}\}$
	$\displaystyle\{\}\cdot\hat{\boldsymbol{k}}\quad\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}\quad I^{\mathrm{cm}}_{zz}\dot{\omega}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/G}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\hat{\boldsymbol{k}}.$

Then, making the same substitutions as before for ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/G}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ we get

\dot{\omega}={c\ell\over I^{\mathrm{cm}}_{zz}}\left(\hat{\boldsymbol{\lambda}}% \times{\accentset{\rightharpoonup}{\boldsymbol{v}}}\cdot\hat{\boldsymbol{k}}-% \dot{\theta}\ell\right)

which determines the rate of change of $\omega$ if the present values of ${\accentset{\rightharpoonup}{\boldsymbol{v}}},\theta$ and $\dot{\theta}$ are known.

Setting up differential equations for solution

If one knows the forces and torques on an object in terms of its position, velocity, orientation and angular velocity one then has a ‘closed’ set of differential equations. That is, one has enough information to define the equations for a mathematician or a computer to solve them.

The full set of differential equations is gathered from linear and angular momentum balance and also from simple kinematics. Namely, one has the following set of 6 first order differential equations:

\begin{array}[]{ccl}\dot{x}&=&v_{x},\\ \dot{v}_{x}&=&F_{x}/m,\\ \dot{y}&=&v_{y},\\ \dot{v}_{y}&=&F_{y}/m,\\ \dot{\theta}&=&\omega,\hbox{ and}\\ \dot{\omega}&=&M_{\rm cm}/I^{\mathrm{cm}}_{zz},\end{array}

where the positions and velocities are the positions and velocities of the center-of-mass. The expressions for $F_{x},F_{y},$ and $M_{\rm cm}$ may well be complicated, as in the rocket example above.

Box 17.1 2-D mechanics makes sense in a 3-D world

The math for two-dimensional mechanics analysis is simpler than the math for three-dimensional analysis. And thus easier to learn first. But we do actually live in a three-dimensional world. You might wonder at the utility of learning something that is not right. There are three answers.

1.

Two-dimensional analysis can give partial information about the three-dimensional system that is exactly the same as the three-dimensional analysis would give by projection, no matter what the motion;
2.

if the motion is planar the 2-D kinematics can be used; and
3.

if the object is planar or symmetric about the motion plane, and any constraints that hold the object are also symmetric about the motion plane, the 2-D analysis is not only correct, but complete.

Of course no machine is exactly planar or exactly symmetric, but if the approximation seems reasonable most engineers will accept a small loss in accuracy for great gain in simplicity.

a) Projection

First let’s relax our assumption of 2-D motion. Consider arbitrary 3-D motions of an arbitrarily complex system. If we take the dot product of the linear momentum equations with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ and the angular momentum balance equation with $\hat{\boldsymbol{k}}$ we get

\begin{array}[]{lrcr}\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_% {i}=\sum m_{i}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{i}\right\}\cdot% \smash{\hat{\boldsymbol{\imath}}}\phantom{\hat{\boldsymbol{\jmath}}}% \Rightarrow\sum{F_{i}}_{x}=\sum m_{i}{a_{i}}_{x},&&\hbox{(a)}\\ \left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}=\sum m_{i}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{i}\right\}\cdot\hat{\boldsymbol{% \jmath}}\Rightarrow\sum{F_{i}}_{y}=\sum m_{i}{a_{i}}_{y},&\hbox{and}&\hbox{(b)% }\\ \left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i}=\sum{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{i}\times m_{i}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_% {i}\right\}\cdot\hat{\boldsymbol{k}}\\ \lx@intercol\hfil\qquad\Rightarrow\sum{r_{i}}_{x}{F_{i}}_{y}-{r_{i}}_{y}{F_{i}% }_{x}=\sum m_{i}({r_{i}}_{x}{a_{i}}_{y}-{r_{i}}_{y}{a_{i}}_{x}).\hfil% \lx@intercol&\hbox{(c)}\\ \end{array}

(17.39)

These are exactly the equations of 2-D mechanics. That is, if we only consider the planar components of the forces, the planar components of the positions, and the planar components of the motions, we get a correct but partial set of the 3-D equations. In this sense 2-D analysis is correct but incomplete.

b) Planar motion

If all the velocities of the parts of a 3-D system have no $z$ component the motion is planar (in the $x y$ plane). Thus the right-hand sides of eqns. (17.39) are not just projections, but the whole story. Further, in the case of rigid-object motion, the 2-D kinematics equation

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{P}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{G}}+\omega% \hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{P/G}}={\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{G}}+\omega\hat{\boldsymbol{k}}\times\left({r_{\rm P/G}}_{x}\hat{% \boldsymbol{\imath}}+{r_{\rm P/G}}_{y}\hat{\boldsymbol{\jmath}}\right)\\

(17.40)

also applies (the $z$ component of the position drops out of the cross product) and the expression for, say, the $z$ component of the angular momentum of an object about its center-of-mass is

{H_{\rm cm}}_{z}=I^{\mathrm{cm}}_{zz}\omega.

Differentiating, or adding up the $m_{i}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{i}$ terms we get,

{\dot{H}_{\rm cm}}_{z}=I^{\mathrm{cm}}_{zz}\dot{\omega}.

Similarly, the $z$ component of the full angular momentum balance equation for a 3-D rigid object in planar motion is the same as the $z$ component of eqn. (17.38)b.

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{O}$}}}\cdot\hat{\boldsymbol{k}}=\left({\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/O}}\times(m_{\mathrm{tot}}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}})\right)\cdot\hat{% \boldsymbol{k}}+I^{\mathrm{cm}}_{zz}\dot{\omega}

So for planar motion of 3-D rigid bodies one can do a correct 2-D analysis with the full ease of analyzing a planar object.

But this result is deceptively simple. The free object diagram in 3-D most likely shows forces in the $z$ direction, pairs of forces in the $x$ or $y$ directions that are applied at points with the same $x$ and $y$ coordinates but different $z$ values, or moments with components in the $x$ or $y$ directions. Full information about these force and moment components can’t be found from 2-D analysis. That is,

the nature of the forces that it takes to keep a system in planar motion can’t be found from a planar analysis.

For example, a system rotating about the $z$ axis which is statically balanced but is dynamically imbalanced

has no net $x$ or $y$ reaction force, as a planar analysis would reveal, yet the bearing reaction forces are not zero.

Another example would be a plan view of a car in a turn (assuming a stiff suspension). A 2-D analysis could be accurate, but would not be complete enough to describe the tire reaction forces needed to keep the car flat.

c) Symmetric bodies and planar bodies

If the rigid object has all its mass in the $x y$ plane, or its mass is symmetrically distributed about the $x y$ plane, and it is in planar motion in the $x y$ plane then

\begin{array}[]{ccc}\sum m_{i}{a_{i}}_{z}=0\hbox to36.135pt{\hfil and\hfil}\\ \left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}\times m_{i}{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{i}\right\}\cdot\hat{\boldsymbol{% \imath}}=0\hbox to36.135pt{\hfil and\hfil}\left\{\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i}\times m_{i}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{i}\right\}\cdot\hat{\boldsymbol{\jmath}}=0\end{array}

where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}$ is measured relative to any point in the plane. Thus, by linear and angular momentum balance,

\begin{array}[]{ccc}\sum F_{z}=0\hbox to36.135pt{\hfil and\hfil}\\ \left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i}\right\}\cdot\hat{\boldsymbol{\imath}}=0% \hbox to36.135pt{\hfil and\hfil}\left\{\sum{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{i}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}% \right\}\cdot\hat{\boldsymbol{\jmath}}=0\end{array}

A planar object or a symmetric object in planar motion needs no force in the $z$ direction and no moment in the $x$ or $y$ direction to keep it in the plane.

Systems that are symmetric or flat and moving in an approximately planar manner, are thus both accurately and completely modeled with a 2-D analysis. A slight generalization of the result is to any object or collection of objects whose centers of mass are on the plane and each of which is dynamically balanced for rotation about a $\hat{\boldsymbol{k}}$ axis through its center-of-mass.

Box 17.2 The center-of-mass theorems for 2-D rigid bodies

That all the particles in a system are part of one planar object in planar motion (in that plane) allows highly useful simplification of the expressions for the motion quantities, namely Eqns. 17.32 to 17.36. We can derive these expressions from the center-of-mass theorems from chapter LABEL:sec:KonigSchmonig. For completeness, we repeat some of those derivations here. To save space, we only use the integral ( $\int$ ) forms for the general expressions; the derivations with sums ( $\sum$ ) are similar. In all cases position, velocity, and acceleration are relative to a fixed point in space (that is ${\accentset{\rightharpoonup}{\boldsymbol{r}}},{\accentset{\rightharpoonup}{% \boldsymbol{v}}},$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}$ mean ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/0}},{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/0}},$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{/0}}$ respectively).

Linear momentum.

Here we show that to evaluate linear momentum and its rate of change you only need to know the motion of the center of mass.

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle\equiv$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{v}}}\,dm=\int{d\over dt% }{\accentset{\rightharpoonup}{\boldsymbol{r}}}\,dm={d\over dt}\int{\accentset{% \rightharpoonup}{\boldsymbol{r}}}\,dm={d\over dt}(m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}})$
		$\displaystyle=$	$\displaystyle m_{\mathrm{tot}}{d\over dt}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\mathrm{cm}}=m_{\mathrm{tot}}{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\mathrm{cm}}$

By identical reasoning, or by differentiating the expression above with respect to time,

\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}

Thus for linear momentum balance one need not pay heed to rotation, only the center-of-mass motion matters.

Angular momentum.

Here we attempt a derivation like the one above but get slightly more complicated results. For simplicity we evaluate angular momentum and its rate of change relative to the origin, but a very similar derivation would hold relative to any fixed point C.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}$	$\displaystyle\equiv$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}\,dm$
	$\displaystyle=$	$\displaystyle\int\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}-{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\right)\times\left({\accentset{% \rightharpoonup}{\boldsymbol{v}}}-{\accentset{\rightharpoonup}{\boldsymbol{v}}% }_{\mathrm{cm}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}% \right)\,dm$
	$\displaystyle=$	$\displaystyle\int\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{/cm}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}% \right)\times\left({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{/cm}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}% \right)\,dm$
	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/cm}% }\,dm+\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\,dm$
		$\displaystyle+\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm% }}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\,dm+\int{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/cm}}\,dm$
	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/cm}% }\,dm+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\int\,dm$
		$\displaystyle+\underbrace{\left(\int{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{\!\mathrm{/cm}}\,dm\right)}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}% }}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times\underbrace{% \left(\int{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/% cm}}\,dm\right)}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}$
	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/cm}% }\,dm+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}m_{\mathrm{tot}}.$

This much is true for any system in any motion. For a rigid object we know about the motions of the parts. Using the center-of-mass as a reference point we know that for all points on the object ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{/cm}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm}}$ . Thus we can continue the derivation above, following the same reasoning as was used in chapter 7 for circular motion of rigid bodies:

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm}% }\times\left({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/cm}}\right)\,dm+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}m_{\mathrm{tot}}.

Using the identity for the triple cross product (see box 17.2 on page 17.2) or using the geometry of the cross product directly with ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ as in chapters 7 and 8 we get

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle\omega\hat{\boldsymbol{k}}\int r_{\rm/cm}^{2}\,dm+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}m_{\mathrm{tot}}.

Then defining $I^{\mathrm{cm}}_{zz}\equiv\int r_{\rm/cm}^{2}\,dm$ we get the desired result:

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times% {\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}m_{\mathrm{tot}}+I^% {\mathrm{cm}}_{zz}\omega\hat{\boldsymbol{k}}.

A similar derivation, or differentiation of the result above (and using that $({d\over dt}{\accentset{\rightharpoonup}{\boldsymbol{r}}})\times{\accentset{% \rightharpoonup}{\boldsymbol{v}}}={\accentset{\rightharpoonup}{\boldsymbol{v}}% }\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ ) gives

\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}\times% {\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}m_{\mathrm{tot}}+I^% {\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}.

The results above hold for any reference point, not just the origin of the fixed coordinate system. Thus, relative to a point instantaneously coinciding with the center-of-mass

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}

\displaystyle=

\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{cm/cm}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}m_{\mathrm{tot}}+I^{% \mathrm{cm}}_{zz}\omega\hat{\boldsymbol{k}}}=I^{\mathrm{cm}}_{zz}\omega\hat{% \boldsymbol{k}}.

and similarly

\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}

\displaystyle=

\displaystyle I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}.

Kinetic energy.

Unsurprisingly the expression for kinetic energy and its rate of change are also simplified by derivations very similar to those above. Skipping the details (or leaving them as an exercise for the peppy reader):

	$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle\equiv$	$\displaystyle\int{1\over 2}{\accentset{\rightharpoonup}{\boldsymbol{v}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}\,dm$
		$\displaystyle=$	$\displaystyle{1\over 2}m_{\mathrm{tot}}v_{cm}^{2}+{1\over 2}I^{\mathrm{cm}}_{% zz}\omega^{2}$

and

	$\displaystyle{\dot{E}_{\mathrm{K}}}$	$\displaystyle\equiv$	$\displaystyle{d\over dt}{E_{\mathrm{K}}}$
		$\displaystyle=$	$\displaystyle m_{\mathrm{tot}}v\dot{v}+I^{\mathrm{cm}}_{zz}\omega\dot{\omega}.$

Box 17.3 The work of a moving force and of a couple

The work of a force acting on an object from state one to state two is

W_{12}=\int_{t_{1}}^{t_{2}}Pdt.

But sometimes we like to think not of the time integral of the power, but of the path integral of the moving force. So we rearrange this integral as follows.

$\displaystyle W_{12}$	$\displaystyle=$	$\displaystyle\int_{t_{1}}^{t_{2}}Pdt$	(17.41)
	$\displaystyle=$	$\displaystyle\int_{t_{1}}^{t_{2}}{\accentset{\rightharpoonup}{\boldsymbol{F}}}% \cdot\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}dt}_{d{% \accentset{\rightharpoonup}{\boldsymbol{r}}}}$
	$\displaystyle=$	$\displaystyle\int_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}}^{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}\cdot d{\accentset{\rightharpoonup}{\boldsymbol{r}}}$

The validity of equation 17.41 depends on the force acting on the same material point of the moving object as it moves from position 1 to position 2; i.e., the force moves with the object. If the material point of force application changes with time, eqn. (17.41) is senseless and should be replaced with the following more generally applicable equation:

W_{12}\ \equiv\ \int_{t_{1}}^{t_{2}}P\ dt\ =\ \int_{t_{1}}^{t_{2}}{{\accentset% {\rightharpoonup}{\boldsymbol{F}}}}\cdot{{\accentset{\rightharpoonup}{% \boldsymbol{v}}}}dt\

(17.42)

where ${{\accentset{\rightharpoonup}{\boldsymbol{v}}}}$ is the velocity of the material point at the instantaneous location of the applied force.

Hand drags on a passing train: a subtlety

There is a subtle distinction between eqn. (17.41) and eqn. (17.42). As an example think of standing still and dragging your hand on a passing train. Your hand slows down the train with the force

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{hand\ on\ train}}.

It might seem that the work of the hand on the train is zero because your hand doesn’t move; work is force times distance and the distance is zero and eqn. (17.41) superficially evaluates to

\int_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}}^{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{2}}{\accentset{\rightharpoonup}{\boldsymbol% {F}}}\cdot d{\accentset{\rightharpoonup}{\boldsymbol{r}}}=0.

But we have violated the condition for the validity of eqn. (17.41): the force be applied to a fixed material point as time progresses. Whereas on the train your hand smears a whole line of material points.

Clearly your hand does slow the train, so it must do (negative) work on the train, as eqn. (17.42) correctly shows because

P_{\rm force\ on\ train}={\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{hand\ on\ train}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{train}}\neq 0.

The power of the hand force on the train is the force on the train dotted with the velocity of the train (not with the velocity of your hand). Thus, your hand does negative work on the train. eqn. (17.42) applies to the train and eqn. (17.41) does not.

On the other hand (so to speak) if one looks at the power of the force on the hand we have:

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{train\ on\ hand}}=-% {\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{hand\ on\ train}}

while the velocity of the hand is zero so

P_{\rm force\ on\ hand}={\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{train\ on\ hand}}\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mkern 0.5mu\mathrm{hand}}=0.

So the train does no work on your hand while your hand does (negative) work on the train. The difference, of course, is mechanical energy lost to heat.

Work of an applied torque

By thinking of an applied torque as really a distribution of forces, the work of an applied torque is the sum of the contributions of the applied forces. If a collection of forces equivalent to a torque is applied to one rigid object the power of these forces turns out to be ${\accentset{\rightharpoonup}{\boldsymbol{M}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}$ . At a given angular velocity a bigger torque applies more power. And a given torque applies more power to a faster spinning object.

Here’s a quick derivation for a collection of forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}$ that add to zero acting at points with positions ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i}}$ relative to a reference point on the object o^′.

$\displaystyle P$	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}% }\cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{i}}$	(17.43)
	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}% }\cdot\left({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm% {o^{\prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/o^{\prime}}}\right)$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{o^{\prime}}}\cdot\underbrace{\sum{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{i}}}_{{\accentset{\rightharpoonup}{\boldsymbol{0% }}}}+\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}\cdot% \left({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/o^{\prime}}}\right)$
	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot\left(% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/o^{\prime}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}\right)$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot\sum{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/o^{\prime}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{o^{\prime}}}\par$

Work of a general force distribution

A general force distribution has, by reasoning close to that above, a power of:

\displaystyle P

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{tot}}% \cdot{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{o^{% \prime}}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{o^{\prime}}}.\par

(17.44)

For a given force system applied to a given object in a given motion any point o^′ can be used. The terms in the formula above will depend on o^′, but the sum does not. Besides the center-of-mass, another convenient locations for o^′ is a fixed hinge, in which case the applied force has no power.

Box 17.4 The vector triple product ${\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})$

The formula

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})=({\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{C}}}){\accentset{\rightharpoonup}{% \boldsymbol{B}}}-({\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{B}}}){\accentset{\rightharpoonup}{% \boldsymbol{C}}}.

(17.45)

can be verified by writing each of the vectors in terms of its orthogonal components (e.g., ${\accentset{\rightharpoonup}{\boldsymbol{A}}}=A_{x}\hat{\boldsymbol{\imath}}+A% _{y}\hat{\boldsymbol{\jmath}}+A_{z}\hat{\boldsymbol{k}}$ ) and checking equality of the $27$ terms on the two sides of the equations (only 12 are non-zero). If this 20 minute proof seems tedious it can be replaced by a more abstract geometric argument partly presented below that surely takes more than 20 minutes to grasp.

Geometry of the vector triple product

Filename:tfigure1-14

Because

{\accentset{\rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup% }{\boldsymbol{C}}}

is perpendicular to both

{\accentset{\rightharpoonup}{\boldsymbol{B}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{C}}}

it is perpendicular to the plane of

{\accentset{\rightharpoonup}{\boldsymbol{B}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{C}}}

, that is, it is ‘normal’ to the plane

B C

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})

is perpendicular to both

{\accentset{\rightharpoonup}{\boldsymbol{A}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup% }{\boldsymbol{C}}}

, so it is perpendicular to the normal to the plane of

B C

. That is, it must be in the plane of

{\accentset{\rightharpoonup}{\boldsymbol{B}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{C}}}

. But any vector in the plane of

{\accentset{\rightharpoonup}{\boldsymbol{B}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{C}}}

must be a combination of

{\accentset{\rightharpoonup}{\boldsymbol{B}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{C}}}

. Also, the vector triple product must be proportional in magnitude to each of

{\accentset{\rightharpoonup}{\boldsymbol{A}}}

{\accentset{\rightharpoonup}{\boldsymbol{B}}}

and

{\accentset{\rightharpoonup}{\boldsymbol{C}}}

. Finally, the triple cross product of

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})

must be the negative of

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{C}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{B}}})

because

{\accentset{\rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup% }{\boldsymbol{C}}}=-{\accentset{\rightharpoonup}{\boldsymbol{C}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{B}}}

. So the identity

{\accentset{\rightharpoonup}{\boldsymbol{A}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{B}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{C}}})=({\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{C}}}){\accentset{\rightharpoonup}{% \boldsymbol{B}}}-({\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{B}}}){\accentset{\rightharpoonup}{% \boldsymbol{C}}}

is almost natural: The expression above is almost the only expression that is a linear combination of ${\accentset{\rightharpoonup}{\boldsymbol{A}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{B}}}$ that is linear in both, also linear in ${\accentset{\rightharpoonup}{\boldsymbol{C}}}$ and switches sign if ${\accentset{\rightharpoonup}{\boldsymbol{B}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{C}}}$ are interchanged. These properties would be true if the whole expression were multiplied by any constant scalar. But a test of the equation with three unit vectors shows that such a multiplicative constant must be one. This reasoning constitutes an informal derivation of the identity 17.45.

Using the triple cross product in dynamics equations

We will use identity 17.45 for two purposes in the development of dynamics equations:

1.

In the 2D expression for acceleration, the centripetal acceleration is given by ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{R}}})$ simplifies to $-\omega^{2}{\accentset{\rightharpoonup}{\boldsymbol{R}}}\ {\rm\ if\ }{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\perp{\accentset{% \rightharpoonup}{\boldsymbol{R}}}$ . This equation follows by setting ${\accentset{\rightharpoonup}{\boldsymbol{A}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{B}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{C}}}={\accentset{\rightharpoonup}{% \boldsymbol{R}}}$ in equation 17.45 and using ${\accentset{\rightharpoonup}{\boldsymbol{R}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}=0$ if ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\perp{\accentset{% \rightharpoonup}{\boldsymbol{R}}}$ . In 3D ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})$ gives the vector shown in the lower figure below.
Filename:tfigure1-f

Filename:tfigure1-15
2.

The term ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})$ will appear in the calculation of the angular momentum of a rigid body. By setting ${\accentset{\rightharpoonup}{\boldsymbol{A}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{B}}}={\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{C}}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}$ , in equation 17.45 and use ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{r}}}=r^{2}$ because ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\parallel{\accentset{% \rightharpoonup}{\boldsymbol{r}}}$ we get the useful result that ${\accentset{\rightharpoonup}{\boldsymbol{r}}}\times({\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}})=r^{2}{{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}}-({% \accentset{\rightharpoonup}{\boldsymbol{r}}}\cdot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}){\accentset{\rightharpoonup}{\boldsymbol{r}}}$ .

SAMPLE 17.6

Filename:sfig8-7-2again

Figure 17.35:

Free planar motion. A rigid rod of length $\ell=1\,\mathrm{m}$ and mass $m_{\rm r}=1\,\mathrm{kg}$ , and a rigid square plate of side $1\,\mathrm{m}$ and mass $m_{\rm p}=10\,\mathrm{kg}$ are launched in motion on a frictionless plane (e.g., an ice hockey rink) with exactly the same initial velocities ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}(0)=10\,\mathrm{m}/% \mathrm{s}\hat{\boldsymbol{\imath}}+1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{% \jmath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}(0)=1\,\mathrm{rad}/\mathrm{% s}\hat{\boldsymbol{k}}$ . Both the rod and the plate have their center-of-mass at the baseline at $t=0$ .

1.

Which of the two is farther from the base line in 3 seconds and which one has undergone a greater number of revolutions?
2.

Find and draw the position of the bar at $t=1\sec$ and at $t=3\sec$ .

Solution

The free-body diagram of the rod is shown in fig. 17.36. There are no forces acting on the rod in the $x y$ -plane. Although there is force of gravity and the normal reaction of the surface acting on the rod, these forces are inconsequential since they act normal to the $x y$ -plane. Therefore, we do not include these forces in our free-body diagram .

Filename:sfig8-7-2disks

Figure 17.36:

The linear momentum balance for the rod gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m_{\rm r}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{% cm}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle m_{\rm r}{\dot{\accentset{\rightharpoonup}{\boldsymbol{v}}}}_{% \rm cm}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{0}}}\,dt=\mbox{% constant}={{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}_{0}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\int{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}% _{0}\,dt={{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}}_{0}+{{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}_{0}t$	(17.46)

It is clear from the analysis above that in the absence of any applied forces, the position of the body depends only on the initial position and the initial velocity. Since both the plate and the rod start from the same base line with the same initial velocity, they travel the same distance from the base line during any given time period; mass or its geometric distribution plays no role in the motion. Thus the center-of-mass of the rod and the plate will be exactly the same distance ( $|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}(t)-{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}}_{0}|=|{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}_{0}t|$ ) at time $t$ . Similarly, the angular momentum balance about the center-of-mass of the rod gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}{\dot{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{0}}}\,dt=\mbox{% constant}={\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{0}=\dot{\theta}_% {0}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta$	$\displaystyle=$	$\displaystyle\int\dot{\theta}_{0}\,dt=\theta_{0}+\dot{\theta}_{0}t$	(17.47)

Thus the angular position of the body is also, as expected, independent of the mass and mass distribution of the body, and depends entirely on the initial position and the initial angular velocity. Therefore, both the rod and the plate undergo exactly the same amount of rotation $\bigl{(}\theta(t)-\theta_{0}=\dot{\theta}_{0}t\bigr{)}$ during any given time.

We can find the position of the rod at $t=1\,\mathrm{s}$ and $t=3\,\mathrm{s}$ by substituting these values of $t$ in eqns. (17.46) and 17.47. For convenience, let us assume that ${{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}}_{0}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ . From the initial configuration of the rod, we also know that $\theta_{0}=0$ .

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}(t=1\,% \mathrm{s})$	$\displaystyle=$	$\displaystyle{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}_{0}% \cdot(1\,\mathrm{s})=(10\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+1\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}})\cdot(1\,\mathrm{s})10\,\mathrm% {m}\hat{\boldsymbol{\imath}}+1\,\mathrm{m}\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm}}(t=3\,% \mathrm{s})$	$\displaystyle=$	$\displaystyle{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}_{0}% \cdot(3\,\mathrm{s})=30\,\mathrm{m}\hat{\boldsymbol{\imath}}+3\,\mathrm{m}\hat% {\boldsymbol{\jmath}}.$
$\displaystyle\theta(t=1\,\mathrm{s})$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}\cdot(1\,\mathrm{s})=(1\,\mathrm{rad}/\mathrm{s})% \cdot(1\,\mathrm{s})=1\,{\rm rad}$
$\displaystyle\theta(t=3\,\mathrm{s})$	$\displaystyle=$	$\displaystyle\dot{\theta}_{0}\cdot(3\,\mathrm{s})=3\,{\rm rad}.$

Accordingly, we show the position of the rod in fig. 17.37.

Filename:sfig8-4-4

Figure 17.37:

SAMPLE 17.7

Filename:sfig8-4-4a

Figure 17.38:

A passive rigid diver. An experimental model of a rigid diver is to be launched from a diving board that is $3\,\mathrm{m}$ above the water level. Say that the initial velocity of the center-of-mass and the initial angular velocity of the diver can be controlled at launch. The diver is launched into the dive in almost vertical position, and it is required to be as vertical as possible at the very end of the dive (which is marked by the position of the diver’s center-of-mass at $1\,\mathrm{m}$ above the water level). If the initial vertical velocity of the diver’s center-of-mass is $3\,\mathrm{m}/\mathrm{s}$ , find the required initial angular velocity for the vertical entry of the diver into the water.

Solution

Filename:sfig8-4-4b

Figure 17.39:

Once the diver leaves the diving board, it is in free flight under gravity, i.e., the only force acting on it is the force due to gravity. The free-body diagram of the diver is shown in fig. 17.39. The linear momentum balance for the diver gives

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}$
$\displaystyle-mg\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle m\ddot{y}\hat{\boldsymbol{\jmath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{y}$	$\displaystyle=$	$\displaystyle-g$
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}\ddot{\theta}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}$	$\displaystyle=$	$\displaystyle 0.$

From these equations of motion, it is clear that the linear and the angular motions of the diver are uncoupled. We can easily solve the equations of motion to get

	$\displaystyle y(t)$	$\displaystyle=$	$\displaystyle y_{0}+\dot{y}_{0}t-\frac{1}{2}gt^{2}$
	$\displaystyle\theta(t)$	$\displaystyle=$	$\displaystyle\theta_{0}+\dot{\theta}_{0}t.$

We need to find the initial angular speed $\dot{\theta}_{0}$ such that $\theta=\pi$ when $y=1\,\mathrm{m}$ (the center-of-mass position at the water entry). From the expression for $\theta(t)$ , we get, $\dot{\theta}_{0}=\pi/t$ . Thus we need to find the value of $t$ at the instant of water entry. We can find $t$ from the expression for $y(t)$ since we know that $y=1\,\mathrm{m}$ at that instant, and that $y_{0}=3\,\mathrm{m}$ and $\dot{y}_{0}=3\,\mathrm{m}/\mathrm{s}$ . We have,

$\displaystyle y$	$\displaystyle=$	$\displaystyle y_{0}+\dot{y}_{0}-\frac{1}{2}gt^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t$	$\displaystyle=$	$\displaystyle\frac{\dot{y}_{0}\pm\sqrt{\dot{y}_{0}^{2}+2g(y_{0}-y)}}{g}$
	$\displaystyle=$	$\displaystyle\frac{3\,\mathrm{m}/\mathrm{s}\pm\sqrt{(3\,\mathrm{m}/\mathrm{s})% ^{2}+2\cdot 9.8\,\mathrm{m}/\mathrm{s}^{2}\cdot(3\,\mathrm{m}-1\,\mathrm{m})}}% {9.8\,\mathrm{m}/\mathrm{s}^{2}}$
	$\displaystyle=$	$\displaystyle 1.15\mbox{ or }-0.53\,\mathrm{s}.$

We reject the negative value of time as meaningless in this context. Thus it takes the diver $1.15\,\mathrm{s}$ to complete the dive. Since, the diver must rotate by $\pi$ during this time, we have

\dot{\theta}_{0}=\pi/t=\pi/(1.15\,\mathrm{s})=2.73\,\mathrm{rad}/\mathrm{s}.

Answer: $\dot{\theta}_{0}=2.73\,\mathrm{rad}/\mathrm{s}$

SAMPLE 17.8

Filename:sfig8-4-4c

Figure 17.40:

A plate tumbling in space. A rectangular plate of mass $m=0.5\,\mathrm{kg},I^{\mathrm{cm}}_{zz}=2.08\times 10^{-3}\,\mathrm{kg}\cdot\,% \mathrm{m}^{2}$ , and dimensions $a=200\,{\rm mm}$ and $b=100\,{\rm mm}$ is pushed by a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=0.5\,{\mbox{N}}\hat{\boldsymbol{% \imath}}$ , acting at $d=30\,{\rm mm}$ away from the mass-center, as shown in the figure. Assume that the force remains constant in magnitude and direction but remains attached to the material point P of the plate. There is no gravity.

1.

Find the initial acceleration of the mass-center.
2.

Find the initial angular acceleration of the plate.
3.

Write the equations of motion of the plate (for both linear and angular motion).

Solution

The only force acting on the plate is the applied force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ . Thus, fig. 17.40 is also the free-body diagram of the plate at the start of motion.

1.

From the linear momentum balance we get,

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$ $\displaystyle=$ $\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}$ $\displaystyle=$ $\displaystyle\frac{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}}{m}=\frac% {0.5\,{\mbox{N}}\hat{\boldsymbol{\imath}}}{0.5\,\mathrm{kg}}=1\,\mathrm{m}/% \mathrm{s}^{2}\hat{\boldsymbol{\imath}}.$

which is the initial acceleration of the mass-center.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}=1\,\mathrm{m}/% \mathrm{s}^{2}\hat{\boldsymbol{\imath}}$
2.

From the angular momentum balance about the mass-center, we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}$ $\displaystyle=$ $\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$

$\displaystyle Fd\hat{\boldsymbol{k}}$ $\displaystyle=$ $\displaystyle I^{\mathrm{cm}}_{zz}\dot{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}$ $\displaystyle=$ $\displaystyle\frac{Fd}{I^{\mathrm{cm}}_{zz}}\hat{\boldsymbol{k}}=\frac{0.5\,{% \mbox{N}}\cdot 0.03\,\mathrm{m}}{2.08\,\mathrm{kg}\cdot\,\mathrm{m}^{2}}=7.2\,% \mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}$

which is the initial angular acceleration of the plate.

Answer: $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=7.2\,\mathrm{rad}/% \mathrm{s}^{2}\hat{\boldsymbol{k}}$

To find the equations of motion, we can use the linear momentum balance and the angular momentum balance as we have done above. So, why aren’t the equations obtained above for the linear acceleration, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}=F/m\hat{% \boldsymbol{\imath}}$ , and the angular acceleration, $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=Fd/I^{\mathrm{cm}}_{zz}% \hat{\boldsymbol{k}}$ , qualified to be called equations of motion? Because, they are not valid for a general configuration of the plate during its motion. The expressions for the accelerations are valid only in the initial configuration (and hence those are initial accelerations).

Let us first draw a free-body diagram of the plate in a general configuration during its motion (see fig. 17.41). Assume the center-of-mass to be displaced by $x\hat{\boldsymbol{\imath}}$ and $y\hat{\boldsymbol{\jmath}}$ , and the longitudinal axis of the plate to be rotated by $\theta\hat{\boldsymbol{k}}$ with respect to the vertical (inertial $y$ -axis). The applied force remains horizontal and attached to the material point P, as stated in the problem.

Filename:pfigure-s94h13p2

Figure 17.41: Free-body diagram of the plate at some instant

t

when the longitudinal axis of the plate makes an angle

\theta(t)

with the fixed vertical axis.

The linear momentum balance gives

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}$	$\Rightarrow$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}=\frac% {\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}}{m}$
	$\displaystyle\hbox to36.135pt{or\hfil}\ddot{x}\hat{\boldsymbol{\imath}}+\ddot{% y}\hat{\boldsymbol{\jmath}}=\frac{F}{m}\hat{\boldsymbol{\imath}}$	$\Rightarrow$	$\displaystyle\ddot{x}=\frac{F}{m},\quad\ddot{y}=0.$

Since $F/m$ is constant, the equations of motion of the center-of-mass indicate that the acceleration is constant and that the mass-center moves in the $x$ -direction.

Similarly, we now use angular momentum balance to determine the rotation (angular motion) of the plate. The angular momentum balance about the mass-center give

Filename:pfigure-f93f5

Figure 17.42: Geometry of the plate at the instant

t

when the longitudinal axis of the plate makes an angle

\theta(t)

with the fixed

y

-axis. The position of point P is

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/cm}

which makes a fixed angle

\alpha(=\tan^{-1}\frac{d}{b/2})

with the transverse axis of the plate. This angle is shown here merely for ease of calculation.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/cm}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}\ddot{\theta}\hat{\boldsymbol{k}}.$
Now,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/cm}}$	$\displaystyle=$	$\displaystyle-r[\cos(\theta+\alpha)\hat{\boldsymbol{\imath}}+\sin(\theta+% \alpha)\hat{\boldsymbol{\jmath}}]$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle F\hat{\boldsymbol{\imath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/cm}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle Fr\sin(\theta+\alpha)\hat{\boldsymbol{k}}.$
Thus,
$\displaystyle\ddot{\theta}$	$\displaystyle=$	$\displaystyle\frac{Fr}{I^{\mathrm{cm}}_{zz}}\sin(\theta+\alpha)$

where $r=\sqrt{d^{2}+(b/2)^{2}}$ and $\alpha=\tan^{-1}(2d/b).$

Thus, we have got the equations of motion for both the linear and the angular motion.

Answer: $\ddot{x}=\frac{F}{m},\quad\ddot{y}=0,\quad\ddot{\theta}=\frac{Fr}{I^{\mathrm{% cm}}_{zz}}\sin(\theta+\alpha)$

4.

The equations of linear motion of the plate are very simple and we can solve them at once to get

$\displaystyle x(t)$ $\displaystyle=$ $\displaystyle x_{0}+\dot{x}_{0}t+\frac{1}{2}\frac{F}{m}t^{2}$

$\displaystyle y(t)$ $\displaystyle=$ $\displaystyle y_{0}+\dot{y}_{0}t.$

If the plate starts from rest ( $\dot{x}_{0}=0,\,\dot{y}_{0}=0$ ) with the center-of-mass at the origin ( $x_{0}=0,\,y_{0}=0$ ), then we have

$x(t)=\frac{F}{2m}t^{2},\quad\mbox{and}\quad y(t)=0.$

Thus the center-of-mass moves along the $x$ -axis with acceleration $F/m$ .

The equation of angular motion of the plate is not so simple. In fact, it is a nonlinear ODE. It is difficult to get an analytical solution of this equation. However, we can solve it numerically using, say, a Runge-Kutta ODE solver:

Filename:pfigure-s94h13p3

Figure 17.43: Variation of plate orientation $\theta$ with time.
```
ODEs = {thetadot = w, wdot = (F*r/Icm)*sin(theta+a)}
IC   = {theta(0) = 0, w(0) = 0}
Set  F=.5, d=0.03; b=0.1; Icm=2.08e-03
compute r = sqrt(d^2+.25*b^2),  a = atan(2*d/b)
Solve ODEs with IC for t=0 to t=10
Plot theta(t)
```
The plot obtained from this calculation is shown in fig. 17.43 which shows that the plate oscillates in its plane as the center of mass races along the $x$ -axis.

The trajectory of point P can now be obtained from ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}-r\cos(\theta+\alpha)\hat{% \boldsymbol{\imath}}-r\sin(\theta+\alpha)\hat{\boldsymbol{\jmath}}$ . Using this relationship, the trajectory of point P, along with the trajectory of the center of mass, is shown in fig. 17.44 for the first 3 seconds.

Filename:p-s96-p3-3

Figure 17.44: Motion of the plate during the first 3 seconds: Position of the plate at various time instants along with the trajectories of the center of mass and point P.

SAMPLE 17.9 Impulse-momentum. Consider the plate problem of Sample 17.39 (page 17.39) again. Assume that the plate is at rest at $t=0$ in the vertical upright position and that the force acts on the plate for 2 seconds.

1.

Find the velocity of the center-of-mass of the plate at the end of 2 seconds.
2.

Can you also find the angular velocity of the plate at the end of 2 seconds?

Solution

Since we are interested in finding the velocity at a particular instant $t$ , given the velocity at another instant $t=0$ , we can use the impulse-momentum equations to find the desired velocity.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{2}-{\accentset{% \rightharpoonup}{\boldsymbol{L}}}_{1}$	$\displaystyle=$	$\displaystyle\int_{t_{1}}^{t_{2}}\sum{\accentset{\rightharpoonup}{\boldsymbol{% F}}}\,dt$
$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}(t)-% m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}(0)$	$\displaystyle=$	$\displaystyle\int_{0}^{t}{\accentset{\rightharpoonup}{\boldsymbol{F}}}\,dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}(t)$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}(0)+% \frac{1}{m}\int_{0}^{2}{\accentset{\rightharpoonup}{\boldsymbol{F}}}\,dt$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}+\frac{1}{0.5\,% \mathrm{kg}}\int_{0}^{2}(0.5\,{\mbox{N}}\hat{\boldsymbol{\imath}})\,dt$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}(2\,\mathrm{s})=2\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$

Now, let us try to find the angular velocity the same way, using angular impulse-momentum relation. We have,

$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}})_{2}% -({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}})_{1}$	$\displaystyle=$	$\displaystyle\int_{t_{1}}^{t_{2}}\sum{\accentset{\rightharpoonup}{\boldsymbol{% M}}}_{\mathrm{cm}}\,dt$
$\displaystyle I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}(t)-I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}(0)$	$\displaystyle=$	$\displaystyle\int_{0}^{t}\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathrm{cm}}\,dt$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}(t)$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}(0)+\frac{1}{I^% {\mathrm{cm}}_{zz}}\int_{0}^{t}\sum{\accentset{\rightharpoonup}{\boldsymbol{M}% }}_{\mathrm{cm}}\,dt$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}(0)+\frac{1}{I^% {\mathrm{cm}}_{zz}}\int_{0}^{t}({\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{P/cm}}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}})\,dt$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}+\frac{1}{I^{\mathrm% {cm}}_{zz}}\int_{0}^{t}(Fr\sin(\theta+\alpha)\hat{\boldsymbol{k}})\,dt$
	$\displaystyle=$	$\displaystyle\frac{Fr}{I^{\mathrm{cm}}_{zz}}\left(\int_{0}^{t}\sin(\theta+% \alpha)\,dt\right)\hat{\boldsymbol{k}}.$

Now, we are in trouble; how do we evaluate the integral? In the integrand, we have $\theta$ which is an implicit function of $t$ . Unless we know how $\theta$ depends on $t$ we cannot evaluate the integral. To find $\theta(t)$ we have to solve the equation of angular motion we derived in the previous sample. However, we were not able to solve for $\theta(t)$ analytically; we had to resort to a numerical solution. Thus, it is not possible to evaluate the integral above and, therefore, we cannot find the angular velocity of the plate at the end of 2 seconds using impulse-momentum equations. We could, however, find the desired velocity easily from the numerical solution.

Problems for 17.2 Dynamics of a rigid object

17.2.1 The uniform rectangle of width $a=1\,\mathrm{m}$ , length $b=2\,\mathrm{m}$ , and mass $m=1\,\mathrm{kg}$ in the figure is sliding on the $x y$ -plane with no friction. At the moment in question, point C is at $x_{C}=3\,\mathrm{m}$ and $y_{C}=2\,\mathrm{m}$ . The linear momentum is ${\accentset{\rightharpoonup}{\boldsymbol{L}}}=4\hat{\boldsymbol{\imath}}+3\hat% {\boldsymbol{\jmath}}\ (\,{\mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{\mathrm{s}})$ and the angular momentum about the center of mass is ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{cm}=5\hat{\boldsymbol{k}}(\,{% \rm kg}\mathord{\cdot}{\rm m}/{\rm s}^{2})$ . Find the acceleration of any point on the body that you choose. (Mark it.) [Hint: You have been given some redundant information.]

Filename:bikefork1-ang-mom

Figure 17.45

17.2.2 The vertical pole AB of mass $m$ and length $\ell$ is initially, at rest on a frictionless surface. A tension $T$ is suddenly applied at A. What is $\ddot{x}_{cm}$ ? What is $\ddot{\theta}_{AB}$ ? What is $\ddot{x}_{B}$ ? Gravity may be ignored.

Filename:bikefork-ang-mom

Figure 17.46

17.2.3 Force on a stick in space. 2-D . No gravity. A uniform thin stick with length $\ell$ and mass $m$ is, at the instant of interest, parallel to the $y$ axis and has no velocity and no angular velocity. The force ${{\accentset{\rightharpoonup}{\boldsymbol{F}}}}=F{\hat{\boldsymbol{\imath}}}$ with $F>0$ is suddenly applied at point $A$ . The questions below concern the instant after the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is applied.

(a)

What is the acceleration of point $C$ , the center of mass? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}=\frac{F}{m}\hat{\boldsymbol% {\imath}}$ .
(b)

What is the angular acceleration of the stick? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=-\frac{6F}{m\ell}\hat{% \boldsymbol{k}}$ .
(c)

What is the acceleration of the point $A$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{4F}{m}\hat{\boldsymbol% {\imath}}$ .
(d)

(relatively harder) What additional force would have to be applied to point $B$ to make point $B$ ’s acceleration zero? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}=\frac{F}{2}\hat{\boldsymbol{% \imath}}$

Filename:summer95f-5-a

Figure 17.47

17.2.4 A uniform thin rod of length $\ell$ and mass $m$ stands vertically, with one end resting on a frictionless surface and the other held by someone’s hand. The rod is released from rest, displaced slightly from the vertical. No forces are applied during the release. There is gravity.

(a)

Find the path of the center of mass.
(b)

Find the force of the floor on the end of the rod just before the rod is horizontal.

17.2.5 A uniform disk, with mass center labeled as point G, is sitting motionless on the frictionless $x y$ plane. A massless peg is attached to a point on its perimeter. This disk has radius of $1\,\mathrm{m}$ and mass of $10\,\mathrm{kg}$ . A constant force of $F=1000\,{\mbox{N}}\boldsymbol{\imath}$ is applied to the peg for $.0001\,\mathrm{s}$ (one ten-thousandth of a second).

(a)

What is the velocity of the center of mass of the center of the disk after the force is applied?
(b)

Assuming that the idealizations named in the problem statement are exact is your answer to (a) exact or approximate?
(c)

What is the angular velocity of the disk after the force is applied?
(d)

Assuming that the idealizations named in the problem statement are exact is your answer to (c) exact or approximate?

17.2.6 A uniform thin flat disc is floating in space. It has radius $R$ and mass $m$ . A force $F$ is applied to it at a distance $d$ from the center in the $y$ direction. Treat this problem as two-dimensional.

(a)

What is the acceleration of the center of the disc? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\frac{F}{m}\hat{\boldsymbol{% \jmath}}$ .
(b)

What is the angular acceleration of the disk? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=\frac{Fd}{m\frac{R^{2}}{2}}% \hat{\boldsymbol{k}}$ .

Filename:pfigure4-2-rp10

Figure 17.48

17.2.7 A uniform 1kg plate that is one meter on a side is initially at rest in the position shown. A constant force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=1\,{\mbox{N}}\hat{\boldsymbol{% \imath}}$ is applied at $t=0$ and maintained henceforth. If you need to calculate any quantity that you don’t know, but can’t do the calculation to find it, assume that the value is given.

(a)

Find the position of G as a function of time (the answer should have numbers and units).
(b)

Find a differential equation, and initial conditions, that when solved would give $\theta$ as a function of time. $\theta$ is the counterclockwise rotation of the plate from the configuration shown.
(c)

Write computer commands that would generate a drawing of the outline of the plate at $t=1\,\mathrm{s}$ . You can use hand calculations or the computer for as many of the intermediate commands as you like. Hand work and sketches should be provided as needed to justify or explain the computer work.
(d)

Run your code and show clear output with labeled plots. Mark output by hand to clarify any points.

Filename:pfigure-blue-125-2

Figure 17.49

17.2.8 A uniform rectangular metal beam of mass $m$ hangs symmetrically by two strings as shown in the figure.

(a)

Draw a free-body diagram of the beam and evaluate $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$ .
(b)

Repeat (a) immediately after the left string is cut.

Filename:pfigure-blue-68-1

Figure 17.50

17.2.9 A uniform slender bar AB of mass $m$ is suspended from two springs (each of spring constant $K$ ) as shown. Immediately after spring 2 breaks, determine

(a)

the angular acceleration of the bar,
(b)

the acceleration of point $A$ , and
(c)

the acceleration of point $B$ .

Filename:pfigure-blue-58-1

Figure 17.51

17.2.10 Two small spheres A and B are connected by a rigid rod of length $\ell=1.0\,\mathrm{ft}$ and negligible mass. The assembly is hung from a hook, as shown. Sphere A is struck, suddenly breaking its contact with the hook and giving it a horizontal velocity $v_{0}=3.0\,{\mathrm{ft/s}}$ which sends the assembly into free fall. Determine the angular momentum of the assembly about its mass center at point G immediately after A is hit. After the center of mass has fallen two feet, determine:

(a)

the angle $\theta$ through which the rod has rotated,
(b)

the velocity of sphere A,
(c)

the total kinetic energy of the assembly of spheres A and B and the rod, and
(d)

the acceleration of sphere A.

Filename:pfigure-blue-157-1

Figure 17.52

The next several problems concern Work, power and energy

17.2.11 Verify that the expressions for work done by a force $F$ , $W=F\Delta S$ , and by a moment $M$ , $W=M\Delta\theta$ , are dimensionally correct if $\Delta S$ and $\Delta\theta$ are linear and angular displacements respectively.

17.2.12 A uniform disc of mass $m$ and radius $r$ rotates with angular velocity $\omega\hat{\boldsymbol{k}}$ . Its center of mass translates with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=\dot{x}\hat{\boldsymbol{\imath}}% +{\dot{y}}\hat{\boldsymbol{\jmath}}$ in the $x y$ -plane. What is the total kinetic energy of the disk?

17.2.13

Calculate the energy stored in a spring using the expression ${E_{\mathrm{P}}}=\frac{1}{2}k\delta^{2}$ if the spring is compressed by 100 mm and the spring constant is 100 N/m.

17.2.14 In a rack and pinion system, the rack is acted upon by a constant force $F=50\,{\mbox{N}}$ and has speed $v=2\,\mathrm{m}/\mathrm{s}$ in the direction of the force. Find the input power to the system.

Filename:summer95f-5

Figure 17.53

17.2.15 The driving gear in a compound gear train rotates at constant speed $\omega_{0}$ . The driving torque is $M_{in}$ . If the driven gear rotates at a constant speed $\omega_{out}$ , find:

(a)

the input power to the system, and
(b)

the output torque of the system assuming there is no power loss in the system; i.e., power in = power out.

Filename:pfigure-blue-127-2

Figure 17.54

17.2.16 An elaborate frictionless gear box has an input and output roller with $V_{in}=const$ . Assuming that $V_{out}=7V_{in}$ and the force between the left belt and roller is $F_{in}=3\,{\rm lb}$ :

(a)

What is $F_{out}$ (draw a picture defining the signs of $F_{in}$ and $F_{out}$ )?. Answer: $F_{out}=\frac{3}{7}\,{\rm lb}$ .
(b)

Is $F_{out}$ greater or less than the $F_{in}$ ? (Assume $F_{in}>0$ .) Why? Answer: $F_{out}$ is always less than the $F_{in}$ .

Filename:pfigure-s94h13p4

Figure 17.55

17.3 Kinematics of rolling and sliding

Pure rolling in 2-D

In this section, we would like to add to the vocabulary of special motions by considering pure rolling. Most commonly, one discusses pure rolling of round objects on flat ground, like wheels and balls, and rolling of round things on other round things like gears and cams.

2-D rolling of a round wheel on level ground

The simplest case, the no-slip rolling of a round wheel, is an instructive starting point. First, we define the geometric and kinematic variables as shown in fig. 17.56.

Filename:pfigure-blue-90-2

Figure 17.56: Pure rolling of a round wheel on a level support.

For convenience, we pick a point $D$ which was at $x_{D}=0$ at the start of rolling, when $x_{C}=0$ . The key to the kinematics is that:

The arc length traversed on the wheel is the distance traveled by the wheel center.

That is,

$\displaystyle x_{C}$	$\displaystyle=$	$\displaystyle s_{D}$
	$\displaystyle=$	$\displaystyle R\phi$
$\displaystyle\Rightarrow v_{C}=\dot{x}_{C}$	$\displaystyle=$	$\displaystyle R\dot{\phi}$
$\displaystyle\Rightarrow a_{C}=\dot{v}_{C}=\ddot{x}_{C}$	$\displaystyle=$	$\displaystyle R\ddot{\phi}$

So the rolling condition amounts to the following set of restrictions on the position of $C$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ , and the rotations of the wheel $\phi$ :

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}=R\phi\hat{\boldsymbol{\imath% }}+R\hat{\boldsymbol{\jmath}},\quad{\accentset{\rightharpoonup}{\boldsymbol{v}% }}_{C}=R\dot{\phi}\hat{\boldsymbol{\imath}},\quad{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{C}=R\ddot{\phi}\hat{\boldsymbol{\imath}},\quad{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}=-\dot{\phi}\hat{\boldsymbol{k}},\mbox{ % and}\quad{\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}=-\ddot{\phi}\hat{\boldsymbol{k}}.

If we want to track the motion of a particular point, say $D$ , we could do so by using the following parametric formula:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{D}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{D/C}$
	$\displaystyle=$	$\displaystyle R(\phi\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})+R(-% \sin\phi\hat{\boldsymbol{\imath}}-\cos\phi\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle R\left[(\phi-\sin\phi)\hat{\boldsymbol{\imath}}+(1-\cos\phi)\hat% {\boldsymbol{\jmath}})\right]$
$\displaystyle\Rightarrow{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}$	$\displaystyle=$	$\displaystyle R\left[(\dot{\phi}(1-\cos\phi)\hat{\boldsymbol{\imath}}+\dot{% \phi}\sin\phi\hat{\boldsymbol{\jmath}})\right]$	(17.48)
$\displaystyle\Rightarrow{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{D}$	$\displaystyle=\text{assuming $\dot{\phi}$= constant }$	$\displaystyle R\dot{\phi}^{2}(\sin\phi\hat{\boldsymbol{\imath}}+\cos\phi\hat{% \boldsymbol{\jmath}}).$

Note that if $\phi=0$ or $2\pi$ or $4\pi$ , etc., then the point $D$ is on the ground and eqn. (17.48) correctly gives that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}=R\left[\dot{\phi}(% \underbrace{1-\overbrace{\cos(2n\pi}^{1})}_{0}\right]\hat{\boldsymbol{\imath}}% +\dot{\phi}\underbrace{\sin(2n\pi)}_{0}\hat{\boldsymbol{\jmath}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}.

Instantaneous kinematics

Instead of tracking the wheel from its start, we could analyze the kinematics at the instant of interest. Here, we make the observation that the wheel rolls without slip. Therefore, the point on the wheel touching the ground has no velocity relative to the ground.

\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}}^{\text{Velocity % of point on the wheel touching the ground}}=\overbrace{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}}^{\text{Velocity of ground = {\accentset% {\rightharpoonup}{\boldsymbol{0}}}}}

(17.49)

Now, we know how to calculate the velocity of points on a rigid body. So,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{C}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A/C},

where, since $A$ and $C$ are on the same rigid body (fig. 17.56), we have from eqn. (16.2) that

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A/C}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{A/C}.

Putting this equation together with eqn. (17.49), we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$
$\displaystyle\Rightarrow\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v% }}}_{C}}_{v_{C}\hat{\boldsymbol{\imath}}}+\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}}_{\omega\hat{\boldsymbol{k}}}\times% \underbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/C}}_{-R\hat{% \boldsymbol{\jmath}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\Rightarrow v_{C}\hat{\boldsymbol{\imath}}+\omega R\hat{% \boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$

\Rightarrow\qquad v_{C}=-\omega R.

(17.50)

We use ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}=v\hat{\boldsymbol{\imath}}$ since the center of the wheel goes neither up nor down.

Note that if you measure the angle by $\phi$ , like we did before, then ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\dot{\phi}\hat{\boldsymbol% {k}}$ so that positive rotation rate is in the counter-clockwise direction. Thus, $v_{C}=-\omega R=-(-\dot{\phi})R=\dot{\phi}R$ .

Since there is always some point of the wheel touching the ground, we know that $v_{C}=-\omega R$ for all time. Therefore,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}=\dot{v}_{C}\hat{\boldsymbol{% \imath}}=-\dot{\omega}R\hat{\boldsymbol{\imath}}.

Rolling of round objects on round surfaces

For round objects rolling on or in another round object, the analysis is similar to that for rolling on a flat surface. Common applications are the so-called epicyclic, hypo-cyclic, or planetary gears (See Box 17.3 on planetary gears on page 17.3).

Filename:s92f1p7

Figure 17.57: One round object rolling on another round object.

Referring to fig. 17.57, we can calculate the velocity of C with respect to a fixed frame two ways and compare:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{C/B}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$	$\displaystyle=$	$\displaystyle\overbrace{\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v% }}}_{A}}_{{\accentset{\rightharpoonup}{\boldsymbol{0}}}}+\underbrace{{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{B/A}}_{{\accentset{% \rightharpoonup}{\boldsymbol{0}}}}}^{{\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{B}}+{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C/B}.$
$\displaystyle\dot{\theta}(R_{1}+R_{2})\hat{\boldsymbol{e}}_{\scriptscriptstyle% \!\theta}$	$\displaystyle=$	$\displaystyle\omega_{\cal B}R_{2}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$\displaystyle\Rightarrow\omega_{\cal B}$	$\displaystyle=$	$\displaystyle\frac{\dot{\theta}(R_{1}+R_{2})}{R_{2}}=\dot{\theta}(1+\frac{R_{1% }}{R_{2}}).$

Example: Two quarters.

The formula above can be tested in the case of $R_{1}=R_{2}$ by using two quarters or two dimes on a table. Roll one quarter, call it $\mathcal{B}$ , around another quarter pressed fast to the table. You will see that as the rolling quarter $\mathcal{B}$ travels around the stationary quarter one time, it makes two full revolutions. That is, the orientation of $\mathcal{B}$ changes twice as fast as the angle of the line from the center of the stationary quarter to its center. Or, in the language of the calculation above, $\omega_{\mathcal{B}}=2\dot{\theta}$ .

Sliding

Although wheels and balls are known for rolling, they do sometimes slide such as when a car screeches at fast acceleration or sudden braking or when a bowling ball is released on the lane.

The sliding velocity is the velocity of the material point on the wheel (or ball) relative to its contacting substrate. In the case of pure rolling, the sliding velocity is zero. In the case of a ball or wheel moving against a stationary support surface, whether flat or curved, the sliding velocity is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{sliding}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{circle\ % center}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{contact/center}}

(17.51)

Example: Bowling ball

Filename:twodisks-ang-mom

Figure 17.58: The bowling ball is sliding so long as

v_{G}\neq-\omega R

The velocity of that point on the bowling ball which is instantaneously in contact with the alley (ground) is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}=v_{\rm G% }\hat{\boldsymbol{\imath}}+\omega\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/G}}=(v_{\rm G}+\omega R)\hat{% \boldsymbol{\imath}}$ . So unless $\omega=-v_{\rm G}/R$ the ball is sliding.

Note that, if sliding, the friction force on the ball opposes the slip of the ball and tends to accelerate the ball’s rotation towards rolling. That is, for example, if the ball is not rotating the sliding velocity is $v_{G}\hat{\boldsymbol{\imath}}$ , the friction force is in the $-\hat{\boldsymbol{\imath}}$ direction and angular momentum balance about the center-of-mass implies $\dot{\omega}<0$ and a clockwise rotational acceleration. No matter what the initial velocity or rotational rate the ball will eventually roll.

Box 17.5 The Sturmey-Archer hub

In 1903, the year the Wright Brothers first flew powered airplanes, the Sturmey-Archer company patented the internal-hub three-speed bicycle transmission. This marvel of engineering was sold on the best bikes until finicky but fast racing bicycles using derailleurs started to push them out of the market in the 1960’s. Now, a hundred years later, internal bicycle hubs (now made by Shimano and Sachs) are having something of a revival, particularly in Europe. These internal-hub transmissions utilize a system called planetary gears, gears which roll around other gears. See the figure below.

Filename:tfigure7-2D-planetary

In order to understand this gear system, we need to understand its kinematics—the motion of its parts. The central ‘sun’ gear $\cal F$ is stationary, at least we treat it as stationary in this discussion since it is fixed to the bike frame, so it is fixed in body $\cal F$ . The ‘planet’ gears roll around the sun gear. Let’s call one of these planets $\cal P$ . The spider $\cal S$ connects the centers of the rolling planets. Finally, the ring gear $\cal R$ rotates around the sun.

Filename:tfigure7-2D-planetary-b

The gear transmission steps up the angular velocity when the spider $\cal S$ is driven and ring $\cal R$ , which moves faster, is connected to the wheel. The transmission steps down the angular velocity when the ring gear is driven and the slower spider is connected to the wheel. The third ‘speed’ in the three-speed gear transmission is direct drive (the wheel is driven directly).

What are the ‘gear ratios’ in the planetary gear system? The ‘trick’ is to recognize that for rolling contact that the contacting points have the same velocity, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{B}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{E}$ . Let’s define some terms.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal S}$	$\displaystyle=$	$\displaystyle\omega_{\cal S}\hat{\boldsymbol{k}}$	angular velocity of the spider
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal P}$	$\displaystyle=$	$\displaystyle\omega_{\cal P}\hat{\boldsymbol{k}}$	angular velocity of the planet
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal R}$	$\displaystyle=$	$\displaystyle\omega_{\cal R}\hat{\boldsymbol{k}}$	angular velocity of the ring

Now, we can find the relation of these angular velocities as follows. Look at the velocity of point $C$ in two ways. First,

$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}}^{% \text{A point on the spider}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}}^{% \text{A point on the planetary gear}}$
$\displaystyle\Rightarrow{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{% \cal S}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$	$\displaystyle=$	$\displaystyle\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}}_{{% \accentset{\rightharpoonup}{\boldsymbol{0}}}}+{\accentset{\rightharpoonup}{% \boldsymbol{\omega}}}_{\cal P}\times{\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{C/B}$
$\displaystyle\Rightarrow\omega_{\cal S}r_{C}$	$\displaystyle=$	$\displaystyle\omega_{\cal P}R_{P}$
$\displaystyle\Rightarrow\omega_{\cal P}$	$\displaystyle=$	$\displaystyle\frac{r_{C}}{R_{P}}\omega_{\cal S}$	(17.52)

Next, let’s look at point $D$ and $E$ :

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{E}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{D/A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal R}\times% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{R}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}_{\cal P}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{D/A}$	$\displaystyle=$	$\displaystyle\omega_{\cal R}\hat{\boldsymbol{k}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{R}$
$\displaystyle\omega_{\cal P}(2R_{P})\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$	$\displaystyle=$	$\displaystyle\omega_{\cal R}r_{R}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}$
$\displaystyle\Rightarrow\omega_{\cal R}$	$\displaystyle=$	$\displaystyle\frac{2R_{P}}{r_{R}}\underbrace{\omega_{\cal P}}_{\text{$\omega_{% \cal P}=\frac{r_{C}}{R_{P}}\omega_{\cal S}$}}$
$\displaystyle\omega_{\cal R}$	$\displaystyle=$	$\displaystyle\frac{2R_{P}}{r_{R}}\frac{\overbrace{r_{C}}^{\text{$r_{C}=r_{S}+R% _{P}$}}}{R_{P}}\omega_{\cal S}$
	$\displaystyle=$	$\displaystyle\frac{2(r_{S}+R_{P})}{\underbrace{r_{R}}_{\text{$r_{R}=r_{S}+2R_{% P}$}}}\omega_{\cal S}$
$\displaystyle\Rightarrow\frac{\omega_{\cal R}}{\omega_{\cal S}}$	$\displaystyle=$	$\displaystyle 2\frac{1+\frac{R_{P}}{r_{S}}}{1+\frac{2R_{P}}{r_{S}}}=\mbox{ % angular velocity step-up}.$

Filename:tfigure7-2D-planetary-a

Typically, the gears have radius ratio of $\frac{R_{P}}{r_{S}}=\frac{3}{2}$ which gives a gear ratio of $\frac{5}{4}$ . Thus, the ratio of the highest gear to the lowest gear on a Sturmey-Archer hub is $\frac{5}{4}/\frac{4}{5}=\frac{25}{16}=1.5625$ . You might compare this ratio to that of a modern mountain bike, with eighteen or twenty-one gears, where the ratio of the highest gear to the lowest is about 4:1.

SAMPLE 17.10

Filename:pfigure-blue-90-1

Figure 17.59:

Falling ladder: The ends of a ladder of length $L=3\,\mathrm{m}$ slip along the frictionless wall and floor shown in Figure 17.59. At the instant shown, when $\theta=60^{{}^{\circ}}$ , the angular speed $\dot{\theta}=1.15\,\mathrm{rad}/\mathrm{s}$ and the angular acceleration $\ddot{\theta}=2.5\,\mathrm{rad}/\mathrm{s}^{2}$ . Find the absolute velocity and acceleration of end B of the ladder.

Solution

Since the ladder is falling, it is rotating clockwise. From the given information:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\dot{\theta}\hat{\boldsymbol{k}}=-1.15\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$	$\displaystyle=$	$\displaystyle\ddot{\theta}\hat{\boldsymbol{k}}=-2.5\,\mathrm{rad}/\mathrm{s}^{% 2}\hat{\boldsymbol{k}}.$

We need to find ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ , the absolute velocity of end B, and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$ , the absolute acceleration of end B.

Since the end A slides along the wall and end B slides along the floor, we know the directions of ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A},\;{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B},\;{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{A}\;\mbox{and}\;{\accentset{\rightharpoonup}{\boldsymbol{a}}% }_{B}$ .

Let ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=v_{A}\hat{\boldsymbol{\jmath% }},\;{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=a_{A}\hat{\boldsymbol{% \jmath}},\;{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=v_{B}\hat{% \boldsymbol{\imath}}\;\mbox{and}\;{\accentset{\rightharpoonup}{\boldsymbol{a}}% }_{B}=a_{B}\hat{\boldsymbol{\imath}}$ where the scalar quantities $v_{A},\;a_{A},\;v_{B}\;\mbox{and}\;a_{B}$ are unknown.

$\displaystyle\mbox{Now,}\quad{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}+{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A/B}={\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{B}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/B}$
$\displaystyle\quad\mbox{or}\quad v_{A}\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle v_{B}\hat{\boldsymbol{\imath}}+\dot{\theta}\hat{\boldsymbol{k}}% \times\underbrace{L(-\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}})}_{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/B}}$
	$\displaystyle=$	$\displaystyle(v_{B}+\dot{\theta}L\sin\theta)\hat{\boldsymbol{\imath}}-\dot{% \theta}L\cos\theta\hat{\boldsymbol{\jmath}}.$

Filename:pfigure-blue-110-1

Figure 17.60:

Dotting both sides of the equation with $\hat{\boldsymbol{\imath}}$ , we get:

$\displaystyle v_{A}\underbrace{\hat{\boldsymbol{\jmath}}\mathord{\cdot}\hat{% \boldsymbol{\imath}}}_{0}$	$\displaystyle=$	$\displaystyle(v_{B}+\dot{\theta}L\sin\theta)\underbrace{\hat{\boldsymbol{% \imath}}\mathord{\cdot}\hat{\boldsymbol{\imath}}}_{1}+\dot{\theta}L\cos\theta% \underbrace{\hat{\boldsymbol{\jmath}}\mathord{\cdot}\hat{\boldsymbol{\imath}}}% _{0}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=$	$\displaystyle v_{B}+\dot{\theta}L\sin\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{B}$	$\displaystyle=$	$\displaystyle-\dot{\theta}L\sin\theta=-(-1.15\,\mathrm{rad}/\mathrm{s})% \mathord{\cdot}3\,\mathrm{m}\mathord{\cdot}\frac{\sqrt{3}}{2}$
	$\displaystyle=$	$\displaystyle 2.99\,\mathrm{m}/\mathrm{s}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}=2.9\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}$

Similarly,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}+\dot{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{A/B}+\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}\times({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{A/B})}^{-\omega^{2}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{A/B}}$
$\displaystyle a_{A}\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle a_{B}\hat{\boldsymbol{\imath}}+\ddot{\theta}\hat{\boldsymbol{k}}% \times L(-\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{% \jmath}})-\dot{\theta}^{2}L(-\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta% \hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(a_{B}+\ddot{\theta}L\sin\theta+\dot{\theta}^{2}L\cos\theta)\hat{% \boldsymbol{\imath}}+(-\ddot{\theta}L\cos\theta+\dot{\theta}^{2}L\sin\theta)% \hat{\boldsymbol{\jmath}}.$

Dotting both sides of this equation with $\hat{\boldsymbol{\imath}}$ (as we did for velocity) we get:

$\displaystyle 0$	$\displaystyle=$	$\displaystyle a_{B}+\ddot{\theta}L\sin\theta+\dot{\theta}^{2}L\cos\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a_{B}$	$\displaystyle=$	$\displaystyle-\ddot{\theta}L\sin\theta-\dot{\theta}^{2}L\cos\theta$
	$\displaystyle=$	$\displaystyle-(-2.5\,\mathrm{rad}/\mathrm{s}^{2}\mathord{\cdot}3\,\mathrm{m}% \mathord{\cdot}\frac{\sqrt{3}}{2})-(-1.15\,\mathrm{rad}/\mathrm{s})^{2}% \mathord{\cdot}3\,\mathrm{m}\mathord{\cdot}\frac{1}{2}$
	$\displaystyle=$	$\displaystyle 4.51\,\mathrm{m}/\mathrm{s}^{2}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=4.51\,\mathrm{m}/\mathrm{s}^% {2}$

SAMPLE 17.11

Filename:pfigure-blue-107-1

Figure 17.61:

A cylinder of diameter $500\,{\rm mm}$ rolls down an inclined plane with uniform acceleration (of the center-of-mass) $a=0.1\,\mathrm{m}/\mathrm{s}^{2}$ . At an instant $t_{0}$ , the mass-center has speed $v_{0}=0.5\,\mathrm{m}/\mathrm{s}$ .

1.

Find the angular speed $\omega$ and the angular acceleration $\dot{\omega}$ at $t_{0}$ .
2.

How many revolutions does the cylinder make in the next 2 seconds?
3.

What is the distance travelled by the center-of-mass in those 2 seconds?

Solution

This problem is about simple kinematic calculations. We are given the velocity, $\dot{x}$ , and the acceleration, $\ddot{x}$ , of the center-of-mass. We are supposed to find angular velocity $\omega$ , angular acceleration $\dot{\omega}$ , angular displacement $\theta$ in 2 seconds, and the corresponding linear distance $x$ along the incline. The radius of the cylinder $R=\mbox{diameter}/2=0.25\,\mathrm{m}$ .

1.

From the kinematics of pure rolling,

$\displaystyle\omega$ $\displaystyle=$ $\displaystyle\frac{\dot{x}}{R}=\frac{0.5\,\mathrm{m}/\mathrm{s}}{0.25\,\mathrm% {m}}=2\,\mathrm{rad}/\mathrm{s},$

$\displaystyle\dot{\omega}$ $\displaystyle=$ $\displaystyle\frac{\ddot{x}}{R}=\frac{0.1\,\mathrm{m}/\mathrm{s}^{2}}{0.25\,% \mathrm{m}}=0.4\,\mathrm{rad}/\mathrm{s}^{2}.$

Answer: $\omega=2\,\mathrm{rad}/\mathrm{s},\quad\dot{\omega}=0.4\,\mathrm{rad}/\mathrm{% s}^{2}$
2.

We can find the number of revolutions the cylinder makes in 2 seconds by solving for the angular displacement $\theta$ in this time period. Since,

$\ddot{\theta}\equiv\dot{\omega}=\mbox{constant},$

we integrate this equation twice and substitute the initial conditions, $\dot{\theta}(t=0)=\omega=2\,\mathrm{rad}/\mathrm{s}$ and $\theta(t=0)=0$ , to get

$\displaystyle\theta(t)$ $\displaystyle=$ $\displaystyle\omega t+\frac{1}{2}\dot{\omega}t^{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta(t=2\,\mathrm{s})$ $\displaystyle=$ $\displaystyle(2\,\mathrm{rad}/\mathrm{s})\cdot(2\,\mathrm{s})+\frac{1}{2}(0.4% \,\mathrm{rad}/\mathrm{s})\cdot(4\,\mathrm{s}^{2})$

$\displaystyle=$ $\displaystyle 4.8\,{\rm rad}=\frac{4.8}{2\pi}\,\mbox{rev}=0.76\,\mbox{rev}.$

Answer: $\theta=0.76\,\mbox{rev}$
3.

Now that we know the angular displacement $\theta$ , the distance travelled by the mass-center is the arc-length corresponding to $\theta$ , i.e.,

$x=R\theta=(0.25\,\mathrm{m})\cdot(4.8)=1.2\,\mathrm{m}.$

Answer: $x=1.2\,\mathrm{m}$

Note that we could have found the distance travelled by the mass-center by integrating the equation $\ddot{x}=0.1\,\mathrm{m}/\mathrm{s}^{2}$ twice.

SAMPLE 17.12 Condition of pure rolling. A cylinder of radius $R=20\,\mathrm{cm}$ rolls on a flat surface with absolute angular speed $\omega=12\,\mathrm{rad}/\mathrm{s}$ under the conditions shown in the figure (In cases (ii) and (iii), you may think of the ‘flat surface’ as a conveyor belt). In each case,

1.

Write the condition for pure rolling.
2.

Find the velocity of the center C of the cylinder.

Filename:pfigure-s94h14p4

Figure 17.62:

Solution

Filename:pfigure-blue-112-1

Figure 17.63: The cylinder rolls on the flat surface. Instantaneously, point P on the cylinder is in contact with point Q on the flat surface. For pure rolling, points P and Q must have the same velocity.

At any instant during rolling, the cylinder makes a point-contact with the flat surface. Let the point of instantaneous contact on the cylinder be P, and let the corresponding point on the flat surface be Q. The condition of pure rolling, in each case, is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{Q}$ , that is, there is no relative motion between the two contacting points (a relative motion will imply slip). Now, we analyze each case.

Case(i)

In this case, the bottom surface is fixed. Therefore,

1.

The condition of pure rolling is: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{Q}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.$
2.

Velocity of the center:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{C/P}={\accentset{\rightharpoonup}{\boldsymbol{0}}}+(-\omega% \hat{\boldsymbol{k}})\times R\hat{\boldsymbol{\jmath}}$

$\displaystyle=$ $\displaystyle\omega R\hat{\boldsymbol{\imath}}=(12\,\mathrm{rad}/\mathrm{s})% \cdot(0.2\,\mathrm{m})\hat{\boldsymbol{\imath}}=2.4\,\mathrm{m}/\mathrm{s}\hat% {\boldsymbol{\imath}}.$

Case(ii)

In this case, the bottom surface moves with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=1\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}$ . Therefore, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{Q}=1\,\mathrm{m}/\mathrm{s}\hat% {\boldsymbol{\imath}}$ . Thus,

1.

The condition of pure rolling is: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{Q}=1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}.$
2.

Velocity of the center:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{C/P}=v_{0}\hat{\boldsymbol{\imath}}+\omega R\hat{\boldsymbol% {\imath}}$

$\displaystyle=$ $\displaystyle 1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+2.4\,\mathrm{m% }/\mathrm{s}\hat{\boldsymbol{\imath}}=3.4\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}.$

Case(iii)

In this case, the bottom surface moves with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=-1\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}$ . Therefore, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{Q}=-1\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}}$ . Thus,

1.

The condition of pure rolling is: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{Q}=-1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}.$
2.

Velocity of the center:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}+{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{C/P}=-v_{0}\hat{\boldsymbol{\imath}}+\omega R\hat{% \boldsymbol{\imath}}$

$\displaystyle=$ $\displaystyle-1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}+2.4\,\mathrm{m% }/\mathrm{s}\hat{\boldsymbol{\imath}}=1.4\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}.$

Answer: $\par\begin{array}[]{llll}\mbox{(a)}:&\mbox{(i)}{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}{\boldsymbol{0}}},&\mbox{(ii)% }{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=1\,\mathrm{m}/\mathrm{s}% \hat{\boldsymbol{\imath}},&\mbox{(iii)}{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{P}=-1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}},\\ \mbox{(b)}:&\mbox{(i)}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}=2.4\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}},&\mbox{(ii)}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{C}=3.4\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}},&\mbox{(iii)}{\accentset{\rightharpoonup}{\boldsymbol{v}}% }_{C}=1.4\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}\end{array}\par$

SAMPLE 17.13

Filename:tfigure8-alt-app2c

Figure 17.64: A uniform disk of radius

r

rolls without slipping inside a fixed cylinder.

Motion of a point on a disk rolling inside a cylinder. A uniform disk of radius $r$ rolls without slipping with constant angular speed $\omega$ inside a fixed cylinder of radius $R$ . A point P is marked on the disk at a distance $\ell$ ( $\ell<r$ ) from the center of the disk. At a general time $t$ during rolling, find

1.

the position of point P,
2.

the velocity of point P, and
3.

the acceleration of point P

Solution

Let the disk be vertically below the center of the cylinder at $t=0\,\mathrm{s}$ such that point P is vertically above the center of the disk (Fig. 17.65). At this instant, $Q$ is the point of contact between the disk and the cylinder. Let the disk roll for time $t$ such that at instant $t$ the line joining the two centers (line OC) makes an angle $\phi$ with its vertical position at $t=0\,\mathrm{s}$ . Since the disk has rolled for time $t$ at a constant angular speed $\omega$ , point P has rotated counter-clockwise by an angle $\theta=\omega t$ from its original vertical position P’.

Filename:sfig4-6-4a

Figure 17.65: Geometry of motion: keeping track of point P while the disk rolls for time

t

, rotating by angle

\theta=\omega t

inside the cylinder.

Position of point P: From Fig. 17.65(b) we can write

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{C}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/C}=(R-r% )\hat{\boldsymbol{\lambda}}_{OC}+\ell\hat{\boldsymbol{\lambda}}_{CP}

where

	$\displaystyle\hat{\boldsymbol{\lambda}}_{OC}$	$\displaystyle=$	$\displaystyle\mbox{a unit vector along OC }=-\sin\phi\hat{\boldsymbol{\imath}}% -\cos\phi\hat{\boldsymbol{\jmath}},$
	$\displaystyle\hat{\boldsymbol{\lambda}}_{CP}$	$\displaystyle=$	$\displaystyle\mbox{a unit vector along CP }=-\sin\theta\hat{\boldsymbol{\imath% }}+\cos\theta\hat{\boldsymbol{\jmath}}.$

Thus,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}=[-(R-r)\sin\phi-\ell\sin% \theta]\hat{\boldsymbol{\imath}}+[-(R-r)\cos\phi+\ell\cos\theta]\hat{% \boldsymbol{\jmath}}.

We have thus obtained an expression for the position vector of point P as a function of $\phi$ and $\theta$ . Since we also want to find velocity and acceleration of point P, it will be nice to express ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$ as a function of $t$ . As noted above, $\theta=\omega t$ ; but how do we find $\phi$ as a function of $t$ ? Note that the center of the disk $C$ is going around point O in circles with angular velocity $-\dot{\phi}\hat{\boldsymbol{k}}$ . The disk, however, is rotating with angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ about the instantaneous center of rotation, point D. Therefore, we can calculate the velocity of point C in two ways:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$
$\displaystyle\hbox to36.135pt{or\hfil}{\accentset{\rightharpoonup}{\boldsymbol% {\omega}}}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/D}$	$\displaystyle=$	$\displaystyle-\dot{\phi}\hat{\boldsymbol{k}}\times{\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{C/O}$
$\displaystyle\hbox to36.135pt{or\hfil}\omega\hat{\boldsymbol{k}}\times r(-\hat% {\boldsymbol{\lambda}}_{OC})$	$\displaystyle=$	$\displaystyle-\dot{\phi}\hat{\boldsymbol{k}}\times(R-r)\hat{\boldsymbol{% \lambda}}_{OC}$
$\displaystyle\hbox to36.135pt{or\hfil}-\omega r(\hat{\boldsymbol{k}}\times\hat% {\boldsymbol{\lambda}}_{OC})$	$\displaystyle=$	$\displaystyle-\dot{\phi}(R-r)(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{% \lambda}}_{OC})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\frac{r}{R-r}\omega$	$\displaystyle=$	$\displaystyle\dot{\phi}.$

Integrating the last expression with respect to time, we obtain

\phi=\frac{r}{R-r}\omega t.

Let

q=\frac{r}{R-r},

then, the position vector of point P may now be written as

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}=[-(R-r)\sin(q\omega t)-\ell% \sin(\omega t)]\hat{\boldsymbol{\imath}}+[-(R-r)\cos(q\omega t)+\ell\cos(% \omega t)]\hat{\boldsymbol{\jmath}}.

(17.53)

2.

Velocity of point P: Differentiating Eqn. (17.53) once with respect to time we get

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=-\omega[(R-r)q\cos(q\omega t% )+\ell\cos(\omega t)]\hat{\boldsymbol{\imath}}+\omega[(R-r)q\sin(q\omega t)-% \ell\sin(\omega t)]\hat{\boldsymbol{\jmath}}.$

Substituting $(R-r)q=r$ in ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ we get

${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}=-\omega r[\{\cos(q\omega t)+% \frac{\ell}{r}\cos(\omega t)\}\hat{\boldsymbol{\imath}}-\{\sin(q\omega t)-% \frac{\ell}{r}\sin(\omega t)\}\hat{\boldsymbol{\jmath}}].$ (17.54)
3.

Acceleration of point P: Differentiating Eqn. (17.54) once with respect to time we get

${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=-\omega^{2}r[-\{q\sin(q% \omega t)+\frac{\ell}{r}\sin(\omega t)\}\hat{\boldsymbol{\imath}}-\{q\cos(q% \omega t)-\frac{\ell}{r}\cos(\omega t)\}\hat{\boldsymbol{\jmath}}].$ (17.55)

Answer: $\begin{array}[]{lll}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}&=&[-(R-r% )\sin(q\omega t)-\ell\sin(\omega t)]\hat{\boldsymbol{\imath}}+[-(R-r)\cos(q% \omega t)+\ell\cos(\omega t)]\hat{\boldsymbol{\jmath}}\\ {\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}&=&-\omega r[\{\cos(q\omega t% )+\frac{\ell}{r}\cos(\omega t)\}\hat{\boldsymbol{\imath}}-\{\sin(q\omega t)-% \frac{\ell}{r}\sin(\omega t)\}\hat{\boldsymbol{\jmath}}]\\ {\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}&=&-\omega^{2}r[-\{q\sin(q% \omega t)+\frac{\ell}{r}\sin(\omega t)\}\hat{\boldsymbol{\imath}}-\{q\cos(q% \omega t)-\frac{\ell}{r}\cos(\omega t)\}\hat{\boldsymbol{\jmath}}]\end{array}$

SAMPLE 17.14 The rolling disk: instantaneous kinematics. For the rolling disk in Sample 17.63, let $R=4\,\mathrm{ft},\;r=1\,\mathrm{ft}$ and point P be on the rim of the disk. Assume that at $t=0$ , the center of the disk is vertically below the center of the cylinder and point P is on the vertical line joining the two centers. If the disk is rolling at a constant speed $\omega=\pi\,\mathrm{rad}/\mathrm{s}$ , find

1.

the position of point P and center C at $t=1\,\mathrm{s},\;3\,\mathrm{s},$ and $5.25\,\mathrm{s}$ ,
2.

the velocity of point P and center C at those instants, and
3.

the acceleration of point P and center C at the same instants as above.

Draw the position of the disk at the three instants and show the velocities and accelerations found above.

Solution

The general expressions for position, velocity, and acceleration of point P obtained in Sample 17.63 can be used to find the position, velocity, and acceleration of any point on the disk by substituting an appropriate value of $\ell$ in equations (17.53), (17.54), and (17.55). Since $R=4r$ ,

q=\frac{r}{R-r}=\frac{1}{3}.

Now, point P is on the rim of the disk and point C is the center of the disk. Therefore,

	$\displaystyle\mbox{ for point P: \hskip 21.68121pt}\ell$	$\displaystyle=$	$\displaystyle r,$
	$\displaystyle\mbox{ for point C: \hskip 21.68121pt}\ell$	$\displaystyle=$	$\displaystyle 0.$

Substituting these values for $\ell$ , and $q=1/3$ in equations (17.53), (17.54), and (17.55) we get the following.

1.

Position:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ $\displaystyle=$ $\displaystyle-3r\left[\sin\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{% \imath}}+\cos\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{\jmath}}\right],$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}+r\left[-sin% \left(\omega t\right)\hat{\boldsymbol{\imath}}+\cos\left(\omega t\right)\hat{% \boldsymbol{\jmath}}\right].$
2.

Velocity:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$ $\displaystyle=$ $\displaystyle-\omega r\left[\cos\left(\frac{\omega t}{3}\right)\hat{% \boldsymbol{\imath}}-\sin\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{% \jmath}}\right],$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ $\displaystyle=$ $\displaystyle-\omega r\left[\left\{\cos\left(\frac{\omega t}{3}\right)+\cos% \left(\omega t\right)\right\}\hat{\boldsymbol{\imath}}-\left\{\sin\left(\frac{% \omega t}{3}\right)-\sin\left(\omega t\right)\right\}\hat{\boldsymbol{\jmath}}% \right].$

Acceleration:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}$	$\displaystyle=$	$\displaystyle\frac{\omega^{2}r}{3}\left[\sin\left(\frac{\omega t}{3}\right)% \hat{\boldsymbol{\imath}}+\cos\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{% \jmath}}\right],$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$	$\displaystyle=$	$\displaystyle\omega^{2}r\left[\left\{\frac{1}{3}\sin\left(\frac{\omega t}{3}% \right)+\sin\left(\omega t\right)\right\}\hat{\boldsymbol{\imath}}+\left\{% \frac{1}{3}\cos\left(\frac{\omega t}{3}\right)-\cos\left(\omega t\right)\right% \}\hat{\boldsymbol{\jmath}}\right].$

We can now use these expressions to find the position, velocity, and acceleration of the two points at the instants of interest by substituting $r=1\,\mathrm{ft}$ , $\omega=\pi\,\mathrm{rad}/\mathrm{s}$ , and appropriate values of $t$ . These values are shown in Table 17.1.

$t$	1 $\,\mathrm{s}$	3 $\,\mathrm{s}$	5.25 $\,\mathrm{s}$
${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$ ( $\,\mathrm{ft}$ )	$3(-\frac{\sqrt{3}}{2}\hat{\boldsymbol{\imath}}-\frac{1}{2}\hat{\boldsymbol{% \jmath}})$	$3\hat{\boldsymbol{\jmath}}$	$3(\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}}-\frac{1}{\sqrt{2}}\hat{% \boldsymbol{\jmath}})$
${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$ ( $\,\mathrm{ft}$ )	${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}-\hat{\boldsymbol{\jmath}}$	${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}-\hat{\boldsymbol{\jmath}}$	$4(\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}}-\frac{1}{\sqrt{2}}\hat{% \boldsymbol{\jmath}})$
${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$ ( ft/s)	$\pi(-\frac{1}{2}\hat{\boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{% \jmath}})$	$\pi\hat{\boldsymbol{\imath}}$	$\pi(-\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}}-\frac{1}{\sqrt{2}}\hat{% \boldsymbol{\jmath}})$
${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$ ( ft/s)	$\pi(\frac{1}{2}\hat{\boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{% \jmath}})$	$2\pi\hat{\boldsymbol{\imath}}$	⇀ 0
${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}\;(\,{\mathrm{ft}}/{\mathrm{s% }}^{2})$	$\frac{\pi^{2}}{3}(\frac{\sqrt{3}}{2}\hat{\boldsymbol{\imath}}+\frac{1}{2}\hat{% \boldsymbol{\jmath}})$	$-\frac{\pi^{2}}{3}\hat{\boldsymbol{\jmath}}$	$\frac{\pi^{2}}{3}(-\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}}+\frac{1}{\sqrt{% 2}}\hat{\boldsymbol{\jmath}})$
${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}\;(\,{\mathrm{ft}}/{\mathrm{s% }}^{2})$	$11.86(.24\hat{\boldsymbol{\imath}}+.97\hat{\boldsymbol{\jmath}})$	$\frac{2\pi^{2}}{3}\hat{\boldsymbol{\jmath}}$	$13.16(-\frac{1}{\sqrt{2}}\hat{\boldsymbol{\imath}}+\frac{1}{\sqrt{2}}\hat{% \boldsymbol{\jmath}})$

Table 17.1: Position, velocity, and acceleration of point P and point C

The velocity and acceleration of the two points are shown in Figures 17.66(a) and (b) respectively.

Filename:sfig4-6-3

Figure 17.66: (a) Velocity and (b) Acceleration of points P and C at

t=1\,\mathrm{s},\;3\,\mathrm{s},

and

5.25\,\mathrm{s}

It is worthwhile to check the directions of velocities and the accelerations by thinking about the velocity and acceleration of point P as a vector sum of the velocity (same for acceleration) of the center of the disk and the velocity (same for acceleration) of point P with respect to the center of the disk. Since the motions involved are circular motions at constant rate, a visual inspection of the velocities and the accelerations is not very difficult. Try it.

SAMPLE 17.15 The rolling disk: path of a point on the disk. For the rolling disk in Sample 17.63, take $\omega=\pi\,\mathrm{rad}/\mathrm{s}$ . Draw the path of a point on the rim of the disk for one complete revolution of the center of the disk around the cylinder for the following conditions:

1.

$R=8r$ ,
2.

$R=4r$ , and
3.

$R=2r$ .

Solution In Sample 17.63, we obtained a general expression for the position of a point on the disk as a function of time. By computing the position of the point for various values of time $t$ up to the time required to go around the cylinder for one complete cycle, we can draw the path of the point. For the various given conditions, the variable that changes in Eqn. (17.53) is $q$ . We can write a computer program to generate the path of any point on the disk for a given set of $R$ and $r$ . Here is a pseudocode to generate the required path on a computer according to Eqn. (17.53).

A pseudocode to plot the path of a point on the disk:

(pseudo-code) program rollingdisk
%-------------------------------------------------------------
% This code plots the path of any point on a disk of radius
% ’r’ rolling with speed ’w’ inside a cylinder of radius ’R’.
% The point of interest is distance ’l’ away from the center of
% the disk. The coordinates x and y of the specified point P are
% calculated according to the relation mentioned above.
%--------------------------------------------------------------
phi = pi/50*[1,2,3,...,100]   % make a vector phi from 0 to 2*pi
x = R*cos(phi)                % create points on the outer cylinder
y = R*sin(phi)
plot y vs x                   % plot the outer cylinder
hold this plot                % hold to overlay plots of paths
q = r/(R-r)                   % calculate q.
T = 2*pi/(q*w)                % calculate time T for going around-
                              % the cylinder once at speed ’w’.
t = T/100*[1,2,3, ..., 100]   % make a time vector t from 0 to T-
                              % taking 101 points.

rcx = -(R-r)*(sin(q*w*t))     % find the x coordinates of pt. C.
rcy = -(R-r)*(cos(q*w*t))     % find the y coordinates of pt. C.
rpx = rcx-l*sin(w*t)          % find the x coordinates of pt. P.
rpy = rcy + l*cos(q*t)        % find the y coordinates of pt. P.
plot rpy vs rpx               % plot the path of P and the path
plot rcy vs rcx               % of C. For path of C

Once coded, we can use this program to plot the paths of both the center and the point P on the rim of the disk for the three given situations. Note that for any point on the rim of the disk $l=r$ (see Fig 17.65).

1.

Let $R=4$ units. Then $r=0.5$ for $R=8r$ . To plot the required path, we run our program rollingdisk with desired input,

R = 4

r = 0.5

w = pi

l = 0.5

execute rollingdisk

The plot generated is shown in Fig.17.67 with a few graphic elements added for illustrative purposes.

Filename:sfig4-6-3a

Figure 17.67: Path of point P and the center C of the disk for $R=8r$ .
2.
Filename:pfigure4-4-rp12

Figure 17.68: Path of point P and the center C of the disk for $R=4r$ .

Similarly, for $R=4r$ we type:

R = 4

r = 1

w = pi

l = 1

execute rollingdisk

to plot the desired paths. The plot generated in this case is shown in Fig.17.68
3.
Filename:pfigure-blue-38-2

Figure 17.69: Path of point P and the center C of the disk for $R=2r$ .

The last one is the most interesting case. The plot obtained in this case by typing:

R = 4

r = 2

w = pi

l = 2

execute rollingdisk

is shown in Fig.17.69. Point P just travels on a straight line! In fact, every point on the rim of the disk goes back and forth on a straight line. Most people find this motion odd at first sight. You can roughly verify the result by cutting a hole twice the diameter of a coin (say a US quarter or dime) in a piece of cardboard and rolling the coin around inside while watching a marked point on the perimeter.

A curiosity. We just discovered something simple about the path of a point on the edge of a circle rolling in another circle that is twice as big. The edge point moves in a straight line. In contrast one might think about the motion of the center G of a straight line segment that slides against two straight walls as in sample 17.89. A problem couldn’t be more different. Naturally the path of point G is a circle (as you can check physically by looking at the middle of a ruler as you hold it sliding against a wall-floor corner).

Problems for 17.3 Kinematics of rolling and sliding

17.3.1 A stone in a wheel. A round wheel rolls to the right. At time $t=0$ it picks up a stone from the road. The stone is stuck in the edge of the wheel. You want to know the direction of the rock’s motion just before and after it next hits the ground. Here are some candidate answers:

•

When the stone approaches the ground its motion is tangent to the ground.
•

The stone approaches the ground at angle $x$ (you name it).
•

When the stone approaches the ground its motion is perpendicular to the ground.
•

The stone approaches the ground at various angles depending on the following conditions(…you list the conditions.)

Although you could address this question analytically, you are to try to get a clear answer by looking at computer generated plots. In particular, you are to plot the pebble’s path for a small interval of time near when the stone next touches the ground. You should pick the parameters that make your case for an answer the strongest. You may make more than one plot.

Here are some steps to follow:

(a)

Assuming the wheel has radius $R_{w}$ and the pebble is a distance $R_{p}$ from the center (not necessarily equal to $R_{w}$ ). The pebble is directly below the center of the wheel at time $t=0$ . The wheel spins at constant clockwise rate $\omega$ . The $x$ -axis is on the ground and $x(t=0)=0$ . The wheel rolls without slipping. Using a clear well labeled drawing (use a compass and ruler or a computer drawing program), show that

$x(t)=\omega tR_{w}-R_{p}\sin(\omega t)$

$y(t)=R_{w}-R_{p}\cos(\omega t)$
(b)

Using this relation, write a program to make a plot of the path of the pebble as the wheel makes a little more than one revolution. Also show the outline of the wheel and the pebble itself at some intermediate time of interest. [Use any software and computer that pleases you.]
(c)

Change whatever you need to change to make a good plot of the pebble’s path for a small amount of time as the pebble approaches and leaves the road. Also show the wheel and the pebble at some time in this interval.
(d)

In this configuration the pebble moves a very small distance in a small time so your axes need to be scaled. But make sure your $x$ - and $y$ - axes have the same scale so that the path of the pebble and the outline of the wheel will not be distorted.
(e)

How does your computer output buttress your claim that the pebble approaches and leaves the ground at the angles you claim?
(f)

Think of something about the pebble in the wheel that was not explicitly asked in this problem and explain it using the computer, and/or hand calculation and/or a drawing.

17.3.2 A uniform disk of radius $r$ rolls at a constant rate without slip. A small ball of mass $m$ is attached to the outside edge of the disk. What is the force required to hold the disk in place when the mass is above the center of the disk?

17.3.3 Rolling at constant rate. A round disk rolls on the ground at constant rate. It rolls $1{1\over 4}$ revolutions over the time of interest.

(a)

Particle paths. Accurately plot the paths of three points: the center of the disk C, a point on the outer edge that is initially on the ground, and a point that is initially half way between the former two points. [Hint: Write a parametric equation for the position of the points. First find a relation between $\omega$ and $v_{C}$ . Then note that the position of a point is the position of the center plus the position of the point relative to the center.] Draw the paths on the computer, make sure $x$ and $y$ scales are the same.
(b)

Velocity of points. Find the velocity of the points at a few instants in the motion: after $1\over 4$ , $1\over 2$ , $3\over 4$ , and $1$ revolution. Draw the velocity vector (by hand) on your plot. Draw the direction accurately and draw the lengths of the vectors in proportion to their magnitude. You can find the velocity by differentiating the position vector or by using relative motion formulas appropriately. Draw the disk at its position after one quarter revolution. Note that the velocity of the points is perpendicular to the line connecting the points to the ground contact.
(c)

Acceleration of points. Do the same as above but for acceleration. Note that the acceleration of the points is parallel to the line connecting the points to the center of the disk.

Filename:pfigure-blue-49-2

Figure 17.70

17.3.4 The concentric wheels are welded to each other and roll without slip on the rack and stationary support. The rack moves to the right at $v_{r}=1\,\mathrm{m}/\mathrm{s}$ . What is the velocity of point A in the middle of the wheels shown?

Filename:pfigure4-4-rp13

Figure 17.71

17.3.5 Questions (a) - (e) refer to the cylinders in the configuration shown in the figure. Question (f) is closely related. Answer the questions in terms of the given quantities (and any other quantities you define if needed).

(a)

What is the speed (magnitude of velocity) of point c?
(b)

What is the speed of point P?
(c)

What is the magnitude of the acceleration of point c?
(d)

What is the magnitude of the acceleration of point P?
(e)

What is the radius of curvature of the path of the particle P at the point of interest?
(f)

In the special case of $A=2b$ what is the curve which particle P traces (for all time)? Sketch the path.

Filename:f92h7p1

Figure 17.72

A little cylinder (with outer radius $b$ and center at point c) rolls without slipping inside a bigger fixed cylinder (with inner radius A and center at point O). The absolute angular velocity of the little cylinder $\omega$ is constant. P is attached to the outside edge of the little cylinder. At the instant of interest, P is on the line between O and c.

17.4 Mechanics of contacting bodies: rolling and sliding

A typical machine part has forces that come from contact with other parts. In fact, with the major exceptions of

•

Gravity,
•

Electromagnetic forces inside motors, and
•

Magnetic attraction/repulsion

most of the forces that act on bodies of engineering interest come from contact. Many of the forces you have drawn in free-body diagrams have been contact forces: The force of the ground on a wheel, of an axle on a bearing, of any part on any other part it touches.

We’d now like to consider some mechanics problems that involve sliding or rolling contact. Once you understand the kinematics from the previous section, there is nothing new in the mechanics. As always, the mechanics is linear momentum balance, angular momentum balance and energy balance. Because we are considering single rigid objects in 2D the expressions for the motion quantities are especially simple (as you can look up in Table I at the back of the book):

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$	$\displaystyle=m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{cm}},$		(17.56)
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/C}}% \times(m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{% cm}})+I\dot{\omega}\hat{\boldsymbol{k}}$	$\displaystyle(\text{where}I=I^{\mathrm{cm}}_{zz})$	(17.57)
$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=m_{\mathrm{tot}}v_{cm}^{2}/2+I\omega^{2}/2$		(17.58)

The key to success, as usual, is the drawing of appropriate free-body diagrams (see Chapter 3). For rolling and sliding the two cases one needs to consider as possible are, self-evidently, rolling, where the contact point has no relative velocity and the tangential reaction force is unknown but less than $\mu N$ , and sliding where the relative velocity could be anything and the tangential reaction force is usually assumed to have a magnitude of $\mu N$ but oppose the relative motion.

Work-energy relations and impulse-momentum relations are useful to solve some problems both with and without slip. In pure rolling contact the contact force does no work because the material point of contact has no velocity. However, when there is sliding mechanical energy is dissipated. The rate of loss of kinetic and potential energy is

\hbox{Rate of frictional dissipation}=P_{diss}=F_{\rm friction}\cdot v_{\rm slip}

(17.59)

where $v_{\rm slip}$ is the relative velocity of the contacting slipping points. If either the friction force (ideal lubrication) or sliding velocity (no slip) is zero there is no dissipation.

As for various problems throughout the text, it is often a savings of calculation to use angular momentum balance (or moment balance in statics) relative to a point where there are unknown reaction forces. For rolling and slipping problems this often means making use of contact points.

Filename:pfigure-blue-52-2

Figure 17.73: A ball rolls or slides on level ground.

Example: Pure rolling on level ground

A ball or wheel rolling on level ground, with no air friction etc, rolls at constant speed (see fig. 17.73). This is most directly deduced from angular momentum balance about the contact point C:

\begin{array}[]{lccccc}{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.% 0pt\hbox{$\scriptscriptstyle\mathrm{C}$}}}=\dot{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\rm C}&\Rightarrow&{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{G/C}}\times-mg\hat{\boldsymbol{\jmath}}&=&{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}\times m{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}}+\dot{\omega}I^{\mathrm{cm}}_{zz% }\hat{\boldsymbol{k}}\\ &\Rightarrow&{\accentset{\rightharpoonup}{\boldsymbol{0}}}&=&R\hat{\boldsymbol% {\jmath}}\times(-m\dot{\omega}R\hat{\boldsymbol{\imath}})+\dot{\omega}I^{% \mathrm{cm}}_{zz}\hat{\boldsymbol{k}}\\ \hbox{dotting with $\hat{\boldsymbol{k}}$}&\Rightarrow&\dot{\omega}=0&% \Rightarrow&\omega=\hbox{constant}.\\ \end{array}

Because for rolling $v_{G}=-\omega R$ we thus have that $v_{G}$ is a constant. [The result can also be obtained by combining angular momentum balance about the center-of-mass with linear momentum balance.]

Finally, linear momentum balance gives the reaction force at C to be ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{tot}=F\hat{\boldsymbol{\imath}}% +n\hat{\boldsymbol{\jmath}}=mg\hat{\boldsymbol{\jmath}}$ . So,

assuming point contact, there is no rolling resistance.

Example: Bowling ball with initial sliding

A bowling ball is released with an initial speed of $v_{0}$ and no rotation rate. What is its subsequent motion? To start with, the motion is incompatible with rolling, the bottom of the ball is sliding to the right. So there is a frictional force which opposes motion and $F=-\mu N$ (see fig. 17.73). Linear and angular momentum balance give:

\begin{array}[]{lccl}\hbox{LMB:}&&\Rightarrow&\left\{-F\hat{\boldsymbol{\imath% }}+N\hat{\boldsymbol{\jmath}}-mg\hat{\boldsymbol{\jmath}}=ma\hat{\boldsymbol{% \imath}}\right\}\\ &\{\}\cdot\hat{\boldsymbol{\jmath}}&\Rightarrow&N=mg\\ &\{\}\cdot\hat{\boldsymbol{\imath}}&\Rightarrow&a=-\mu g\\ \hbox{AMB${}_{\rm/G}$:}&&\Rightarrow&-R\mu mg=I^{\mathrm{cm}}_{zz}\dot{\omega}% \\ \\ \Rightarrow&v=v_{0}-\mu gt&\hbox to36.135pt{\hfil and\hfil}&\omega=-\mu Rmgt/I% ^{\mathrm{cm}}_{zz}\end{array}

Thus the forward speed of the ball decreases linearly with time while the counter-clockwise angular velocity decreases linearly with time.

This solution is only appropriate so long as there is rightward slip, $v_{G}>-\omega R$ . Just like for a sliding block, there is no impetus for reversal, and the ball switches to pure rolling when

v=-\omega R\Rightarrow v_{0}-\mu gt=-\left(-\mu Rmgt/I^{\mathrm{cm}}_{zz}% \right)R\Rightarrow t={v_{0}\over\mu g\left(1+{mR^{2}\over I^{\mathrm{cm}}_{zz% }}\right)}.

Note that the energy lost during sliding is less than $\mu mg$ times the distance the center of the ball moves during slip.

Filename:pfigure4-4-rp16

Figure 17.74: A ball rolls or slides down a slope.

Example: Ball rolling down hill.

Assuming rolling we can find the acceleration of a ball as it rolls downhill (see fig. 17.74). We start out with the kinematic observations that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm G}=a_{\rm G}\hat{% \boldsymbol{\lambda}}$ , that $R\omega=-v_{\rm G}$ and that $R\dot{\omega}=-a_{\rm G}$ . Angular momentum balance about the stationary point on the ground instantaneously coinciding with the contact point gives

\begin{array}[]{lccl}\hbox{AMB${}_{\rm/C}$}&&\Rightarrow&{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}\times(-mg\hat{\boldsymbol{% \jmath}})={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}\times m% {\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}}+I^{\mathrm{cm}}_{zz% }\dot{\omega}\hat{\boldsymbol{k}}\\ &&&\left\{-R\sin\phi mg\hat{\boldsymbol{k}}=(R\hat{\boldsymbol{n}})\times(ma_{% \rm G}\hat{\boldsymbol{\lambda}})+I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{% \boldsymbol{k}}\right\}\\ \left\{\right\}\cdot\hat{\boldsymbol{k}}&&\Rightarrow&-Rmg\sin\phi=-Rma_{\rm G% }-I^{\mathrm{cm}}_{zz}a_{\rm G}/R\\ &&\Rightarrow&a_{\rm G}={g\sin\phi\over 1+I^{\mathrm{cm}}_{zz}/(mR^{2})}.\par% \end{array}

Which is less than the acceleration of a block sliding on a ramp without friction: $a=g\sin\phi$ (unless the mass of the rolling ball is concentrated at the center with $I^{\mathrm{cm}}_{zz}=0$ ). Note that a very small ball rolls just as slowly. In the limit as the ball radius goes to zero the behavior does not approach that of a point mass that slides; the rolling remains significant.

Example: Ball rolling down hill: energy approach

We can find the acceleration of the rolling ball using power balance or conservation of energy. For example

\begin{array}[]{cclcl}0={d\over dt}E_{\mathrm{T}}\Rightarrow&0=&\dot{E_{% \mathrm{K}}}+\dot{E_{\mathrm{P}}}\\ &=&{d\over dt}\left(mv^{2}/2+I^{\mathrm{cm}}_{zz}\omega^{2}/2\right)+{d\over dt% }\left(mgy\right)\\ &=&mv\dot{v}+I^{\mathrm{cm}}_{zz}\omega\dot{\omega}+mg\dot{y}\\ &=&mv\dot{v}+I^{\mathrm{cm}}_{zz}(v/R)\dot{v}/R-mg(\sin\phi)v\\ \hbox{assuming $v\neq 0\Rightarrow$}&0=&(m+I^{\mathrm{cm}}_{zz}/R^{2})\dot{v}-% mg\sin\phi\\ \\ \Rightarrow&\dot{v}=&{g\sin\phi\over 1+I^{\mathrm{cm}}_{zz}/(mR^{2})}\\ \end{array}

as before.

Example: Does the ball slide?

How big is the coefficient of friction $\mu$ needed to prevent slip for a ball rolling down a hill? Use angular momentum balance to find the normal and frictional components of the contact force, using the rolling example above.

\begin{array}[]{lccl}\hbox{LMB (${\accentset{\rightharpoonup}{\boldsymbol{F}}}% _{\!\!\mathrm{tot}}=m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{G}% }$)}&&\Rightarrow&\left\{N\hat{\boldsymbol{n}}+F\hat{\boldsymbol{\lambda}}-mg% \hat{\boldsymbol{\jmath}}=ma_{\rm G}\hat{\boldsymbol{\lambda}}\right\}\\ &\left\{\right\}\cdot\hat{\boldsymbol{n}}&\Rightarrow&N=mg\cos\phi\\ &\left\{\right\}\cdot\hat{\boldsymbol{\lambda}}&\Rightarrow&F+mg\sin\phi=m{g% \sin\phi\over 1+I^{\mathrm{cm}}_{zz}/(mR^{2})}\\ &&&F={-mg\sin\phi\over 1+mR^{2}/I^{\mathrm{cm}}_{zz}}\\ \\ \hbox{Critical condition:}&&\Rightarrow&\mu={|F|\over N}={\tan\phi\over 1+mR^{% 2}/I^{\mathrm{cm}}_{zz}}\\ \end{array}

If $I^{\mathrm{cm}}_{zz}$ is very small (the mass concentrated near the center of the ball) then small friction is needed to prevent rolling. For a uniform rubber ball on pavement (with $\mu\approx 1$ and $I^{\mathrm{cm}}_{zz}\approx 2mR^{2}/5$ ) the steepest slope for rolling without slip is a steep $\phi=\tan^{-1}(7/2)\approx 74^{{}^{\circ}}$ . A metal hoop on the other hand (with $\mu\approx.3$ and $I^{\mathrm{cm}}_{zz}\approx mR^{2}$ ) will only roll without slip for slopes less than about $\phi=\tan^{-1}(.6)\approx 31^{{}^{\circ}}$ .

Filename:pfigure4-4-rp14

Figure 17.75: A ball rolls in a round cross-section bowl.

Example: Oscillations of a ball in a bowl.

A round ball can oscillate back and forth in the bottom of a circular cross section bowl or pipe (see fig. 17.75). Similarly, a cylindrical object can roll inside a pipe. What is the period of oscillation? Start with angular momentum balance about the contact point

\begin{array}[]{rcll}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{% G/C}}\times(-mg\hat{\boldsymbol{\jmath}})&=&{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{G/C}}\times m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{G}}+I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol% {k}}\\ rmg\sin\theta\hat{\boldsymbol{k}}&=&-r\hat{\boldsymbol{e}}_{r}\times\left(m% \left((R-r)\ddot{\theta}\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta}-(R-r% )\dot{\theta}^{2}\hat{\boldsymbol{e}}_{r}\right)\right)\\ &&\quad+I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}.\\ \end{array}

Evaluating the cross products (using that $\hat{\boldsymbol{e}}_{r}\times\hat{\boldsymbol{e}}_{\scriptscriptstyle\!\theta% }=\hat{\boldsymbol{k}}$ ) and using the kinematics from the previous section (that $(R-r)\dot{\theta}=-r\omega$ ) and dotting the left and right sides with $\hat{\boldsymbol{k}}$ gives

(R-r)\ddot{\theta}={-g\sin\theta\over 1+I^{\mathrm{cm}}_{zz}/mr^{2}},

the tangential acceleration is the same as would have been predicted by putting the ball on a constant slope of $-\theta$ . Using the small angle approximation that $\sin\theta=\theta$ the equation can be rearranged as a standard harmonic oscillator equation

\ddot{\theta}+\left({g\over(R-r)(1+I^{\mathrm{cm}}_{zz}/mr^{2})}\right)\theta=0.

If all the ball’s mass were concentrated in its middle, keeping $r$ a fixed non-zero size, (so $I^{\mathrm{cm}}_{zz}=0$ , like a lead pellet inside a styrofoam ball) this is the same as for a simple pendulum with length $R-r$ . For any parameter values the period of small oscillation is

T=2\pi\sqrt{(R-r)(1+I^{\mathrm{cm}}_{zz}/mr^{2})\over g}.

For a marble or ball bearing in a sideways glass (with $R-r\approx 4\,\mathrm{cm}=.04\,\mathrm{m}$ , $I^{\mathrm{cm}}_{zz}/mr^{2}\approx 2/5$ and $g\approx 10\,\mathrm{m}/\mathrm{s}^{2}$ ) this gives about one oscillation every half second. See page 19.2 for the energy approach to this problem.

SAMPLE 17.16

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Figure 17.76

Equation of motion of a driven wheel Consider a wheel of mass $m$ and radius $R$ driven by an axle force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ as shown in the figure. The wheel rolls to the left without slipping. Write the equation of motion of the wheel.

Solution

Filename:tfigure4-spherical-rotaxis

Figure 17.77: FBD of a wheel with mass

m

. Force

F

is applied by the axle.

The free-body diagram of the wheel is shown in figure 17.77. We can write the equation of motion of the wheel in terms of either the center-of-mass position $x$ or the angular displacement of the wheel $\theta$ . Since in pure rolling, these two variables share a simple relationship ( $x=R\theta$ ), we can easily get the equation of motion in terms of $x$ if we have the equation in terms of $\theta$ and vice versa. Let ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\omega}\hat{% \boldsymbol{k}}$ .

Since all the forces are shown in the free-body diagram, we can readily write the angular momentum balance for the wheel. We choose the point of contact C as our reference point for the angular momentum balance (because the gravity force, $-mg\hat{\boldsymbol{\jmath}}$ , the friction force $-F_{\rm friction}\hat{\boldsymbol{\imath}}$ , and the normal reaction of the ground $N\hat{\boldsymbol{\jmath}}$ , all pass through the contact point $C$ and therefore, produce no moment about this point). We have

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$
where
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{cm/C}}^% {R\hat{\boldsymbol{\jmath}}}\times(F\hat{\boldsymbol{\lambda}})$
	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{\jmath}}\times F(-\cos\phi\hat{\boldsymbol{% \imath}}-\sin\phi\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle FR\cos\phi\hat{\boldsymbol{k}}$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm/C}}% \times m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}+I^{\mathrm% {cm}}_{zz}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$
	$\displaystyle=$	$\displaystyle R\hat{\boldsymbol{\jmath}}\times m\underbrace{\ddot{x}}_{-\dot{% \omega}R}\hat{\boldsymbol{\imath}}+I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{% \boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle m\dot{\omega}R^{2}\hat{\boldsymbol{k}}+I^{\mathrm{cm}}_{zz}\dot{% \omega}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(I^{\mathrm{cm}}_{zz}+mR^{2})\dot{\omega}\hat{\boldsymbol{k}}.$
Thus,
$\displaystyle FR\cos\phi\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle(I^{\mathrm{cm}}_{zz}+mR^{2})\dot{\omega}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\omega}\equiv\ddot{\theta}$	$\displaystyle=$	$\displaystyle\frac{FR\cos\phi}{I^{\mathrm{cm}}_{zz}+mR^{2}}$

which is the equation of motion we are looking for.

Note that we can easily substitute $\ddot{\theta}=-\ddot{x}/R$ in the equation of motion above to get the equation of motion in terms of the center-of-mass displacement $x$ as

\ddot{x}=-\frac{FR^{2}\cos\phi}{I^{\mathrm{cm}}_{zz}+mR^{2}}.

Answer: $\ddot{\theta}=\frac{FR\cos\phi}{I^{\mathrm{cm}}_{zz}+mR^{2}}$

Comments: We could have, of course, used linear momentum balance with angular momentum balance about the center-of-mass to derive the equation of motion. Note, however, that the linear momentum balance will essentially give two scalar equations in the $x$ and $y$ directions involving all forces shown in the free-body diagram . The angular momentum balance , on the other hand, gets rid of some of them. Depending on which forces are known, we may or may not need to use all the three scalar equations. In the final equation of motion, we must have only one unknown.

SAMPLE 17.17

Filename:tfigure5-7

Figure 17.78

Energy and power of a rolling wheel. A wheel of diameter $2\,\mathrm{ft}$ and mass $20\,{\mathrm{lbm}}$ rolls without slipping on a horizontal surface. The kinetic energy of the wheel is $1700\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}$ . Assume the wheel to be a thin, uniform disk.

1.

Find the rate of rotation of the wheel.
2.

Find the average power required to bring the wheel to a complete stop in 5 $\,\mathrm{s}$ .

Solution

Let $\omega$ be the rate of rotation of the wheel. Since the wheel rotates without slip, its center-of-mass moves with speed $v_{\rm cm}=\omega r$ . The wheel has both translational and rotational kinetic energy. The total kinetic energy is

$\displaystyle{E_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}mv_{\rm cm}^{2}+\frac{1}{2}I^{\rm cm}\omega^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}m\omega^{2}r^{2}+\frac{1}{2}I^{\rm cm}\omega^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(mr^{2}+\underbrace{I^{\rm cm}}_{\frac{1}{2}mr^{2}})% \omega^{2}$
	$\displaystyle=$	$\displaystyle\frac{3}{4}mr^{2}\omega^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega^{2}$	$\displaystyle=$	$\displaystyle\frac{4{E_{\mathrm{K}}}}{3mr^{2}}$
	$\displaystyle=$	$\displaystyle\frac{4\times 1700\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}}{3% \times 20\,{\mathrm{lbm}}\mathord{\cdot}1\,\mathrm{ft}^{2}}$
	$\displaystyle=$	$\displaystyle\frac{4\times 1700\times 32.2\cancel{\,{\mathrm{lbm}}}\mathord{% \cdot}\cancel{\,\mathrm{ft}}/\,\mathrm{s}^{2}}{3\times 20\cancel{\,{\mathrm{% lbm}}}\mathord{\cdot}\cancel{\,\mathrm{ft}}}$
	$\displaystyle=$	$\displaystyle 3649.33\frac{1}{\,\mathrm{s}^{2}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega$	$\displaystyle=$	$\displaystyle 60.4\,\mathrm{rad}/\mathrm{s}.$

Answer: $\omega=60.4\,\mathrm{rad}/\mathrm{s}$

Note: This rotational speed, by the way, is extremely high. At this speed the center-of-mass moves at 60.4 ft/s!

2.

Power is the rate of work done on a body or the rate of change of kinetic energy. Here we are given the initial kinetic energy, the final kinetic energy (zero) and the time to achieve the final state. Therefore, the average power is,

$\displaystyle P$ $\displaystyle=$ $\displaystyle\frac{{E_{\mathrm{K}}}_{1}-{E_{\mathrm{K}}}_{2}}{\Delta t}$

$\displaystyle=$ $\displaystyle\frac{1700\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}-0}{5\,\mathrm{s% }}=340\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}/\,\mathrm{s}$

$\displaystyle=$ $\displaystyle 340\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}/\,\mathrm{s}\cdot% \frac{1\mbox{ hp}}{550\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}/\,\mathrm{s}}$

$\displaystyle=$ $\displaystyle 0.62\mbox{ hp}$

Answer: $P=0.62\mbox{ hp}$

SAMPLE 17.18

Filename:tfigure5-gen-rigid-body

Figure 17.79: FBD of a rolling wheel.

Equation of motion of a rolling wheel from energy balance. Consider the wheel with mass $m$ from figure 17.79. The free-body diagram of the wheel is shown here again. Derive the equation of motion of the wheel using energy balance. Assume CCW (counter clockwise) rotation is positive.

Solution

From energy balance, we have

$\displaystyle P$	$\displaystyle=$	$\displaystyle{\dot{E}_{\mathrm{K}}}$
where
$\displaystyle P$	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\cdot{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}$
	$\displaystyle=$	$\displaystyle-F_{\rm friction}\hat{\boldsymbol{\imath}}\cdot\overbrace{{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}}^{{\accentset{\rightharpoonup% }{\boldsymbol{0}}}}+N\hat{\boldsymbol{\jmath}}\cdot\overbrace{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{C}}^{{\accentset{\rightharpoonup}{% \boldsymbol{0}}}}-mg\hat{\boldsymbol{\jmath}}\cdot\overbrace{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{cm}}^{v\hat{\boldsymbol{\imath}}}+F\hat{% \boldsymbol{\lambda}}\cdot\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{cm}}^{v\hat{\boldsymbol{\imath}}}$
	$\displaystyle=$	$\displaystyle-mgv(\underbrace{\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{% \jmath}}}_{0})+Fv(\underbrace{\hat{\boldsymbol{\lambda}}\cdot\hat{\boldsymbol{% \imath}}}_{-\cos\phi})$
	$\displaystyle=$	$\displaystyle-Fv\cos\phi$
and
$\displaystyle{\dot{E}_{\mathrm{K}}}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}(\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}I^{\mathrm{cm}}_{% zz}\overbrace{\omega^{2}}^{(\dot{x}/R)^{2}})$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{d}{dt}\left[\left(m+\frac{I^{\mathrm{cm}}_{zz}}{% R^{2}}\right)\dot{x}^{2}\right]$
	$\displaystyle=$	$\displaystyle\left(m+\frac{I^{\mathrm{cm}}_{zz}}{R^{2}}\right)\dot{x}\ddot{x}.$
Thus,
$\displaystyle-Fv\cos\phi$	$\displaystyle=$	$\displaystyle\left(m+\frac{I^{\mathrm{cm}}_{zz}}{R^{2}}\right)\dot{x}\ddot{x}$
$\displaystyle\hbox to36.135pt{\hfil or\hfil}-F\cancel{\dot{x}}\cos\phi$	$\displaystyle=$	$\displaystyle\left(m+\frac{I^{\mathrm{cm}}_{zz}}{R^{2}}\right)\cancel{\dot{x}}% \ddot{x}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{x}$	$\displaystyle=$	$\displaystyle-\frac{F\cos\phi}{m+\frac{I^{\mathrm{cm}}_{zz}}{R^{2}}}.$

We can also write the equation of motion in terms of $\theta$ by replacing $\ddot{x}$ with $-\ddot{\theta}R$ giving,

\ddot{\theta}=\frac{(F/R)\cos\phi}{m+\frac{I^{\mathrm{cm}}_{zz}}{R^{2}}}.

Answer: $\ddot{x}=-\frac{F\cos\phi}{m+I^{\mathrm{cm}}_{zz}/R^{2}}$

Comments: In the equations above (for calculating $P$ ), we have set ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ because in pure rolling, the instantaneous velocity of the contact point is zero. Note that the force due to gravity is normal to the direction of the velocity of the center-of-mass. So, the only power supplied to the wheel is due to the force $F\hat{\boldsymbol{\lambda}}$ acting at the center-of-mass.

SAMPLE 17.19

Filename:tfigure5-term1-a

Figure 17.80

Equation of motion of a rolling disk on an incline. A uniform circular disk of mass $m=1\,\mathrm{kg}$ and radius $R=0.4\,\mathrm{m}$ rolls down an inclined shown in the figure. Write the equation of motion of the disk assuming pure rolling, and find the distance travelled by the center-of-mass in 2 $\,\mathrm{s}$ .

Solution

The free-body diagram of the disk is shown in fig. 17.81. In addition to the base unit vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , let us use unit vectors $\hat{\boldsymbol{\lambda}}$ and $\hat{\boldsymbol{n}}$ along the plane and perpendicular to the plane, respectively, to express various vectors.

Filename:tfigure5-term1-b

Figure 17.81

We can write the equation of motion using linear momentum balance or angular momentum balance. However, note that if we use linear momentum balance we have two unknown forces in the equation. On the other hand, if we use angular momentum balance about the contact point C, these forces do not show up in the equation. So, let us use angular momentum balance about point C:

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$
where
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}% \times m{\accentset{\rightharpoonup}{\boldsymbol{g}}}=R\hat{\boldsymbol{n}}% \times(-mg\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-Rmg\sin\alpha\hat{\boldsymbol{k}}$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle-I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}+\overbrace{% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/C}}}^{R\hat{% \boldsymbol{n}}}\times m\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{a}% }}_{\mathrm{cm}}}^{R\dot{\omega}\hat{\boldsymbol{\lambda}}}$
	$\displaystyle=$	$\displaystyle-I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}+mR^{2}\dot{% \omega}(\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\lambda}})$
	$\displaystyle=$	$\displaystyle-(I^{\mathrm{cm}}_{zz}+mR^{2})\dot{\omega}\hat{\boldsymbol{k}}.$
Thus,
$\displaystyle-Rmg\sin\alpha\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle-(I^{\mathrm{cm}}_{zz}+mR^{2})\dot{\omega}\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\omega}$	$\displaystyle=$	$\displaystyle\frac{g\sin\alpha}{R[1+I^{\mathrm{cm}}_{zz}/(mR^{2})]}.$

Answer: $\dot{\omega}=\frac{g\sin\alpha}{R[1+I^{\mathrm{cm}}_{zz}/(mR^{2})]}$

Note that in the above equation of motion, the right hand side is constant. So, we can solve the equation for $\omega$ and $\theta$ by simply integrating this equation and substituting the initial conditions $\omega(t=0)=0$ and $\theta(t=0)=0$ . Let us write the equation of motion as $\dot{\omega}=\beta$ where $\beta=g\sin\alpha/R(1+I^{\mathrm{cm}}_{zz}/mR^{2})$ . Then,

	$\displaystyle\omega$	$\displaystyle\equiv$	$\displaystyle\dot{\theta}=\beta t+C_{1}$
	$\displaystyle\theta$	$\displaystyle=$	$\displaystyle\frac{1}{2}\beta t^{2}+C_{1}t+C_{2}.$

Substituting the given initial conditions $\dot{\theta}(0)=0$ and $\theta(0)=0$ , we get $C_{1}=0$ and $C_{2}=0$ , which implies that $\theta=\frac{1}{2}\beta t^{2}$ . Now, in pure rolling, $x=R\theta$ . Therefore,

$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle R\theta(t)=\frac{1}{2}\beta t^{2}=\cancel{R}\cdot\frac{1}{2}% \frac{g\sin\alpha}{\cancel{R}(1+I^{\mathrm{cm}}_{zz}/mR^{2})}t^{2}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{g\sin\alpha}{1+\frac{\frac{1}{2}mR^{2}}{mR^{2}}}% t^{2}=\frac{1}{3}(g\sin\alpha)t^{2}$
$\displaystyle x(2\,\mathrm{s})$	$\displaystyle=$	$\displaystyle\frac{1}{3}\cdot{9.8\,\mathrm{m}/\mathrm{s}^{2}\cdot\sin(30^{{}^{% \circ}})}\cdot(2\,\mathrm{s})^{2}=6.53\,\mathrm{m}.$

Answer: $x(2\,\mathrm{s})=6.53\,\mathrm{m}$

SAMPLE 17.20

Filename:sfig1-2-12

Figure 17.82

Using Work and energy in pure rolling. Consider the disk of Sample 17.79 rolling down the incline again. Suppose the disk starts rolling from rest. Find the speed of the center-of-mass when the disk is 2 $\,\mathrm{m}$ down the inclined plane.

Solution We are given that the disk rolls down, starting with zero initial velocity. We are to find the speed of the center-of-mass after it has travelled 2 $\,\mathrm{m}$ along the incline. We can, of course, solve this problem using equation of motion, by first solving for the time $t$ the disk takes to travel the given distance and then evaluating the expression for speed $\omega(t)$ or $x(t)$ at that $t$ . However, it is usually easier to use work energy principle whenever positions are specified at two instants, speed is specified at one of those instants, and speed is to be found at the other instant. This is because we can, presumably, compute the work done on the system in travelling the specified distance and relate it to the change in kinetic energy of the system between the two instants.

Filename:sfig4-6-5

Figure 17.83

In the problem given here, let $\omega_{1}$ and $\omega_{2}$ be the initial and final (after rolling down by $d=2\,\mathrm{m}$ ) angular speeds of the disk, respectively. We know that in rolling, the kinetic energy is given by

{E_{\mathrm{K}}}=\frac{1}{2}m\overbrace{v_{\mathrm{cm}}^{2}}^{(\omega R)^{2}}+% \frac{1}{2}I^{\mathrm{cm}}_{zz}\omega^{2}=\frac{1}{2}(mR^{2}+I^{\mathrm{cm}}_{% zz})\omega^{2}.

Therefore,

\Delta{E_{\mathrm{K}}}={E_{\mathrm{K}}}_{2}-{E_{\mathrm{K}}}_{1}=\frac{1}{2}(% mR^{2}+I^{\mathrm{cm}}_{zz})(\omega_{2}^{2}-\omega_{1}^{2}).

(17.60)

Now, let us calculate the work done by all the forces acting on the disk during the displacement of the mass-center by $d$ along the plane. Note that in ideal rolling, the contact forces do no work. Therefore, the work done on the disk is only due to the gravitational force:

W=(-mg\hat{\boldsymbol{\jmath}})\cdot(d\hat{\boldsymbol{\lambda}})=-mgd(% \overbrace{\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{\lambda}}}^{-\sin% \alpha})=mgd\sin\alpha.

(17.61)

From work-energy principle (integral form of power balance, $P={\dot{E}_{\mathrm{K}}}$ ), we know that $W=\Delta{E_{\mathrm{K}}}$ . Therefore, from eqn. (17.60) and eqn. (17.61), we get

$\displaystyle mgd\sin\alpha$	$\displaystyle=$	$\displaystyle\frac{1}{2}(mR^{2}+I^{\mathrm{cm}}_{zz})(\omega_{2}^{2}-\omega_{1% }^{2})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{2}^{2}$	$\displaystyle=$	$\displaystyle\omega_{1}^{2}+\frac{2mgd\sin\alpha}{mR^{2}+I^{\mathrm{cm}}_{zz}}% =\omega_{1}^{2}+\frac{2gd\sin\alpha}{R^{2}\left(1+\frac{I^{\mathrm{cm}}_{zz}}{% mR^{2}}\right)}$
	$\displaystyle=$	$\displaystyle\omega_{1}^{2}+\frac{4gd\sin\alpha}{3R^{2}}.$
Substituting the values of $g,d,\alpha,R$ , etc., and setting $\omega_{1}=0$ , we get
$\displaystyle\omega_{2}^{2}$	$\displaystyle=$	$\displaystyle\frac{4\cdot(9.8\,\mathrm{m}/\mathrm{s}^{2})\cdot(2\,\mathrm{m})% \cdot(\sin(30^{{}^{\circ}})}{3\cdot(0.4\,\mathrm{m})^{2}}=81.67/s^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{2}$	$\displaystyle=$	$\displaystyle 9.04\,\mathrm{rad}/\mathrm{s}.$

The corresponding speed of the center-of-mass is

v_{\mathrm{cm}}=\omega_{2}R=9.04\,\mathrm{rad}/\mathrm{s}\cdot 0.4\,\mathrm{m}% =3.61\,\mathrm{m}/\mathrm{s}.

Answer: $v_{\mathrm{cm}}=3.61\,\mathrm{m}/\mathrm{s}$

SAMPLE 17.21

Filename:sfig4-6-5a

Figure 17.84

Impulse and momentum calculations in pure rolling. Consider the disk of Sample 17.79 rolling down the incline again. Find an expression for the rolling speed ( $\omega$ ) of the disk after a finite time $\Delta t$ , given the rolling speed $\omega_{1}$ at some instant $t_{1}$ .

Solution

Once again, this problem can be solved by integrating the equation of motion (as done in Sample 17.79). However, we will solve this problem here using impulse-momentum relationship. Let the desired angular speed at $t_{2}=t_{1}+\Delta t$ be $\omega_{2}$ . We need to find $\omega_{2}$ , given the $\omega_{1}$ at $t=t_{1}$ . Since the forces acting on the disk do not change during this time (assuming pure rolling), it is easy to calculate impulse and then relate it to the change in the momenta of the disk between the two instants.

Filename:sfig4-6-5b

Figure 17.85

Now, from the linear impulse momentum relationship, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\Delta t={\accentset{% \rightharpoonup}{\boldsymbol{L}}}_{2}-{\accentset{\rightharpoonup}{\boldsymbol% {L}}}_{1}$ , we have

$\displaystyle(-F\hat{\boldsymbol{\lambda}}+N\hat{\boldsymbol{\imath}}-mg\hat{% \boldsymbol{\jmath}})\Delta t$	$\displaystyle=$	$\displaystyle m(v_{2}-v_{1})\hat{\boldsymbol{\lambda}}.$	(17.62)
Dotting eqn. (17.62) with $\hat{\boldsymbol{\lambda}}$ gives
$\displaystyle(-F-mg(\underbrace{\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol% {\lambda}}}_{-\sin\alpha}))\Delta t$	$\displaystyle=$	$\displaystyle m(v_{2}-v_{1})$
$\displaystyle(-F+mg\sin\alpha)\Delta t$	$\displaystyle=$	$\displaystyle mR(\omega_{2}-\omega_{1}).$	(17.63)

Similarly, the angular impulse-momentum relationship about the mass-center, ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{O}$}}}\Delta t=({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2% .0pt\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/O}$}}{\lower 1.0pt% \hbox{$\textstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/O}$}}{% \lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{/O}$}}})_{2}-({\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/O}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{/O}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{/O}$}}})_{1}$ , gives

	$\displaystyle(-FR\hat{\boldsymbol{k}})\Delta t$	$\displaystyle=$	$\displaystyle-I^{\mathrm{cm}}_{zz}(\omega_{2}-\omega_{1})\hat{\boldsymbol{k}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}FR\Delta t$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}(\omega_{2}-\omega_{1}).$		(17.64)

Note that the other forces ( $N$ and $m g$ ) do not produce any moment about the mass-center as they pass through this point. We can now eliminate the unknown force $F$ from eqn. (17.63) and eqn. (17.64) by multiplying eqn. (17.63) with $R$ and adding to eqn. (17.64):

$\displaystyle(-F+mg\sin\alpha)\Delta t\cdot R+FR\Delta t$	$\displaystyle=$	$\displaystyle mR(\omega_{2}-\omega_{1})\cdot R+I^{\mathrm{cm}}_{zz}(\omega_{2}% -\omega_{1})$
$\displaystyle\hbox to36.135pt{or\hfil}mgR\sin\alpha\Delta t$	$\displaystyle=$	$\displaystyle(I^{\mathrm{cm}}_{zz}+mR^{2})(\omega_{2}-\omega_{1})$
$\displaystyle\hbox to36.135pt{or\hfil}g\sin\alpha\Delta t$	$\displaystyle=$	$\displaystyle R\left(1+\frac{I^{\mathrm{cm}}_{zz}}{mR^{2}}\right)(\omega_{2}-% \omega_{1})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{2}$	$\displaystyle=$	$\displaystyle\omega_{1}+\frac{g\sin\alpha}{R\left(1+\frac{I^{\mathrm{cm}}_{zz}% }{mR^{2}}\right)}\Delta t.$

Answer: $\omega_{2}=\omega_{1}+\frac{g\sin\alpha}{R\left(1+\frac{I^{\mathrm{cm}}_{zz}}{% mR^{2}}\right)}\Delta t$

Note: From the answer obtained, we clearly see that $\omega_{2}=\omega_{1}+\dot{\omega}\Delta t$ where $\dot{\omega}$ is the angular acceleration given by the expression $\dot{\omega}=\frac{g\sin\alpha}{R(1+I^{\mathrm{cm}}_{zz}/mR^{2})}$ . This is the same expression we obtained for the angular acceleration in Sample 17.79.

SAMPLE 17.22

Filename:sfig4-6-5c

Figure 17.86: A ladder, modeled as a uniform rod of mass

m

and length

\ell

, falls from a rest position at

\theta=\theta_{o}\;(<\pi/2)

such that its ends slide along frictionless vertical and horizontal surfaces.

Falling ladder. A ladder AB, modeled as a uniform rigid rod of mass $m$ and length $\ell$ , rests against frictionless horizontal and vertical surfaces. The ladder is released from rest at $\theta=\theta_{o}\;(\theta_{o}<\pi/2)$ . Assume the motion to be planar (in the vertical plane).

1.

As the ladder falls, what is the path of the center-of-mass of the ladder?
2.

Find the equation of motion (e.g., a differential equation in terms of $\theta$ and its time derivatives) for the ladder.
3.

How does the angular speed $\omega\;(=\dot{\theta})$ depend on $\theta$ ?

Solution Since the ladder is modeled by a uniform rod AB, its center-of-mass is at G, half way between the two ends. As the ladder slides down, the end A moves down along the vertical wall and the end B moves out along the floor. Note that it is a single degree of freedom system as angle $\theta$ (a single variable) is sufficient to determine the position of every point on the ladder at any instant of time.

1.

Path of the center-of-mass:

Filename:sfig4-6-5d

Figure 17.87

Let the origin of our $x$ - $y$ coordinate system be the intersection of the two surfaces on which the ends of the ladder slide (see Fig. 17.87). The position vector of the center-of-mass G may be written as

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{G/B}$ (17.65)

$\displaystyle=$ $\displaystyle\ell\cos\theta\hat{\boldsymbol{\imath}}+\frac{\ell}{2}(-\cos% \theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$

$\displaystyle=$ $\displaystyle\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat% {\boldsymbol{\jmath}}).$

Thus the coordinates of the center-of-mass are

$x_{G}=\frac{\ell}{2}\cos\theta\hbox to36.135pt{\hfil and\hfil}y_{G}=\frac{\ell% }{2}\sin\theta,$

from which we get

$x_{G}^{2}+y_{G}^{2}=\frac{\ell^{2}}{4}$

which is the equation of a circle of radius $\frac{\ell}{2}$ . Therefore, the center-of-mass of the ladder follows a circular path of radius $\frac{\ell}{2}$ centered at the origin. Of course, the center-of-mass traverses only that part of the circle which lies between its initial position at $\theta=\theta_{o}$ and the final position at $\theta=0$ .

Equation of motion:

Filename:sfig4-5-6

Figure 17.88: The free-body diagram of the ladder.

The free-body diagram of the ladder is shown in Fig. 17.88. Since there is no friction, the only forces acting at the end points A and B are the normal reactions from the contacting surfaces.

Now, writing the linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ ) for the ladder we get

N_{1}\hat{\boldsymbol{\imath}}+(N_{2}-mg)\hat{\boldsymbol{\jmath}}=m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=m\ddot{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}}_{G}.

Differentiating eqn. (17.65) twice we get $\ddot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{G}$ as

\ddot{{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{G}=\frac{\ell}{2}[(-% \ddot{\theta}\,\sin\theta-\dot{\theta}^{2}\cos\theta)\hat{\boldsymbol{\imath}}% +(\ddot{\theta}\cos\theta-\dot{\theta}^{2}\sin\theta)\hat{\boldsymbol{\jmath}}].

Substituting this expression in the linear momentum balance equation above and dotting both sides of the equation by $\hat{\boldsymbol{\imath}}$ and then by $\hat{\boldsymbol{\jmath}}$ we get

	$\displaystyle N_{1}$	$\displaystyle=$	$\displaystyle-\frac{1}{2}m\ell(\ddot{\theta}\,\sin\theta+\dot{\theta}^{2}\cos\theta)$
	$\displaystyle N_{2}$	$\displaystyle=$	$\displaystyle\frac{1}{2}m\ell(\ddot{\theta}\,\cos\theta-\dot{\theta}^{2}\sin% \theta)+mg.$

Next, we write the angular momentum balance for the ladder about its center-of-mass, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/G}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/G}$ , where

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/G}$	$\displaystyle=$	$\displaystyle\left(-N_{1}\frac{\ell}{2}\sin\theta+N_{2}\frac{\ell}{2}\cos% \theta\right)\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}m\ell(\ddot{\theta}\,\sin\theta+\dot{\theta}^{2}\cos% \theta)\frac{\ell}{2}\sin\theta\hat{\boldsymbol{k}}$
		$\displaystyle+\left[\frac{1}{2}m\ell(\ddot{\theta}\,\cos\theta-\dot{\theta}^{2% }\sin\theta)+mg\right]\frac{\ell}{2}\cos\theta\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\left(\frac{1}{4}m\ell^{2}\ddot{\theta}+\frac{1}{2}mg\ell\cos% \theta\right)\hat{\boldsymbol{k}}$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/G}$	$\displaystyle=$	$\displaystyle I_{zz/G}\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=% \frac{1}{12}m\ell^{2}\ddot{\theta}(-\hat{\boldsymbol{k}}),$

where $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\ddot{\theta}(-\hat{% \boldsymbol{k}})$ because $\theta$ is measured positive in the clockwise direction $(-\hat{\boldsymbol{k}})$ . Now, equating the two quantities $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/G}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{/G}$ and dotting both sides with $\hat{\boldsymbol{k}}$ we get

$\displaystyle\frac{1}{4}\cancel{m}\ell^{2}\ddot{\theta}+\frac{1}{2}\cancel{m}g% \ell\cos\theta$	$\displaystyle=$	$\displaystyle-\frac{1}{12}\cancel{m}\ell^{2}\ddot{\theta}$
$\displaystyle\hbox to36.135pt{or\hfil}(\frac{1}{12}+\frac{1}{4})\ell^{2}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{1}{2}g\ell\cos\theta$
$\displaystyle\hbox to36.135pt{or\hfil}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{3g}{2\ell}\cos\theta$	(17.66)

which is the required equation of motion. Unfortunately, it is a nonlinear equation which does not have a nice closed form solution for $\theta(t)$ .

3.

Angular Speed of the ladder: To solve for the angular speed $\omega\;(=\dot{\theta})$ as a function of $\theta$ we need to express eqn. (17.66) in terms of $\omega,\;\theta$ , and derivatives of $\omega$ with respect to $\theta$ . Now,

$\ddot{\theta}=\dot{\omega}=\frac{d\omega}{dt}=\frac{d\omega}{d\theta}\cdot% \frac{d\theta}{dt}=\omega\,\frac{d\omega}{d\theta}.$

Substituting in eqn. (17.66) and integrating both sides from the initial rest position to an arbitrary position $\theta$ we get

$\displaystyle\int_{0}^{\omega}\omega\,d\omega$ $\displaystyle=$ $\displaystyle-\int_{\theta_{0}}^{\theta}\frac{3g}{2\ell}\cos\theta\,d\theta$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\frac{1}{2}\omega^{2}$ $\displaystyle=$ $\displaystyle-\frac{3g}{2\ell}(\sin\theta-\sin\theta_{0})$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega$ $\displaystyle=$ $\displaystyle\pm\sqrt{\frac{3g}{\ell}(\sin\theta_{0}-\sin\theta)}.$

Since end B is sliding to the right, $\theta$ is decreasing; hence it is the negative sign in front of the square root which gives the correct answer, i.e.,

${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\dot{\theta}(-\hat{% \boldsymbol{k}})=-\sqrt{\frac{3g}{\ell}(\sin\theta_{0}-\sin\theta)}\;\hat{% \boldsymbol{k}}.$

Note: To do this problem we have assumed that the upper end of the ladder stays in contact with the wall as it slides down. If a real ladder were sliding against a slippery wall and floor, would it not lose contact? The answer is yes, it would. One way of finding when the contact would be lost is to calculate the normal reaction $N_{1}$ and finding out at what value of $\theta$ it passes through zero. The answer depends on $\theta_{0}$ . For example, if $\theta_{0}=80^{{}^{\circ}}$ , then $N_{1}$ is zero at about $\theta=41^{\circ}$ . You can verify this result by substituting the expressions for $\ddot{\theta}$ and $\dot{\theta}$ in the expression for $N_{1}$ and solving for $\theta$ when $N_{1}=0$ .

SAMPLE 17.23

Filename:sfig4-5-6a

Figure 17.89

The falling ladder again. Consider the falling ladder of Sample 17.58 again. The mass of the ladder is $m$ and the length is $\ell$ . The ladder is released from rest at $\theta=80^{{}^{\circ}}$ .

1.

At the instant when $\theta=45^{{}^{\circ}}$ , find the speed of the center-of-mass of the ladder using energy.
2.

Derive the equation of motion of the ladder using work-energy balance.

Solution

Filename:sfig4-6-8

Figure 17.90

Since there is no friction, there is no loss of energy between the two states: $\theta_{0}=80^{{}^{\circ}}$ and $\theta_{f}=45^{{}^{\circ}}$ . The only external forces on the ladder are $N_{1},\;N_{2},\;\mbox{and}\;mg$ as shown in the free-body diagram. Since the displacements of points A and B are perpendicular to the normal reactions of the walls, $N_{1}$ and $N_{2}$ , respectively, no work is done by these forces on the ladder. The only force that does work is the force due to gravity. But this force is conservative. Therefore, the conservation of energy holds between any two states of the ladder during its fall.

Filename:sfig4-6-8a

Figure 17.91

Let $E_{1}$ and $E_{2}$ be the total energy of the ladder at $\theta_{0}$ and $\theta_{f}$ , respectively. Then

E_{1}=E_{2}\quad\mbox{(conservation of energy).}

$\displaystyle\mbox{Now}\quad E_{1}$	$\displaystyle=$	$\displaystyle\underbrace{E_{K_{1}}}_{\rm K.E.}+\underbrace{E_{P_{1}}}_{\rm P.E.}$
	$\displaystyle=$	$\displaystyle 0+mgh_{1}$
	$\displaystyle=$	$\displaystyle mg\frac{\ell}{2}\sin\theta_{0}$
$\displaystyle\mbox{and}\quad E_{2}$	$\displaystyle=$	$\displaystyle E_{K_{2}}+E_{P_{2}}=\underbrace{\frac{1}{2}mv_{G}^{2}+\frac{1}{2% }I_{zz}^{\rm G}\omega^{2}}_{E_{K_{2}}}+mgh_{2}.$

Equating $E_{1}$ and $E_{2}$ we get

	$\displaystyle\cancel{m}g\frac{\ell}{\cancel{2}}(\sin\theta_{0}-\sin\theta_{f})$	$\displaystyle=$	$\displaystyle\frac{1}{\cancel{2}}(\cancel{m}v_{G}^{2}+\underbrace{\frac{1}{12}% \cancel{m}\ell^{2}}_{I_{zz}^{\rm G}}\omega^{2})$
	$\displaystyle\mbox{or}\quad g\ell(\sin\theta_{0}-\sin\theta_{f})$	$\displaystyle=$	$\displaystyle v_{G}^{2}+\frac{1}{12}\ell^{2}\omega^{2}.$		(17.67)

Clearly, we cannot find $v_{G}$ from this equation alone because the equation contains another unknown, $\omega$ . So we need to find another equation which relates $v_{G}$ and $\omega$ . To find this equation we turn to kinematics. Note that

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}$	$\displaystyle=$	$\displaystyle\frac{\ell}{2}(\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat% {\boldsymbol{\jmath}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{G}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G}=\frac{\ell}% {2}(-\sin\theta\mathord{\cdot}\dot{\theta}\hat{\boldsymbol{\imath}}+\cos\theta% \mathord{\cdot}\dot{\theta}\hat{\boldsymbol{\jmath}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{G}$	$\displaystyle=$	$\displaystyle\|{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}\|=\sqrt{\frac{% \ell^{2}}{4}(\cos^{2}\theta+\sin^{2}\theta)\;\dot{\theta}^{2}}$
	$\displaystyle=$	$\displaystyle\frac{\ell}{2}\dot{\theta}=\frac{\ell}{2}\omega$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega$	$\displaystyle=$	$\displaystyle\frac{2v_{G}}{\ell}.$

Substituting the expression for $\omega$ in eqn. (17.67) we get

$\displaystyle g\ell(\sin\theta_{0}-\sin\theta_{f})$	$\displaystyle=$	$\displaystyle v_{G}^{2}+\frac{1}{12}\cancel{\ell^{2}}\mathord{\cdot}\frac{4v_{% G}^{2}}{\cancel{\ell^{2}}}$
	$\displaystyle=$	$\displaystyle\frac{4}{3}v_{G}^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{G}$	$\displaystyle=$	$\displaystyle\sqrt{\frac{3g\ell}{4}(\sin\theta_{0}-\sin\theta_{f})}$
	$\displaystyle=$	$\displaystyle 0.46\sqrt{g\ell}.$

Answer: $v_{G}=0.46\sqrt{g\ell}$

Equation of motion: Since the ladder is a single degree of freedom system, we can use the power equation to derive the equation of motion:

P=\dot{E}_{K}.

For the ladder, the only force that does work is $m g$ . This force acts at the center-of-mass G. Therefore,

$\displaystyle P$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}\mathord{\cdot}{% \accentset{\rightharpoonup}{\boldsymbol{v}}}=-mg\hat{\boldsymbol{\jmath}}\;% \mathord{\cdot}\;{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{G}$
	$\displaystyle=$	$\displaystyle-mg\hat{\boldsymbol{\jmath}}\mathord{\cdot}\left[\frac{\ell}{2}(-% \sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}})\;\dot% {\theta}\right]$
	$\displaystyle=$	$\displaystyle-mg\frac{\ell}{2}\dot{\theta}\cos\theta.$
Now, the rate of change of kinetic energy is
$\displaystyle\dot{E}_{K}$	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(\frac{1}{2}mv_{G}^{2}+\frac{1}{2}I_{zz}^{\rm G}% \omega^{2}\right)$
	$\displaystyle=$	$\displaystyle\frac{d}{dt}\left(\frac{1}{2}m\frac{\ell^{2}\omega^{2}}{4}+\frac{% 1}{2}\frac{m\ell^{2}}{12}\omega^{2}\right)$
	$\displaystyle=$	$\displaystyle\frac{m\ell^{2}}{4}\omega\dot{\omega}+\frac{m\ell^{2}}{12}\omega% \dot{\omega}$
	$\displaystyle=$	$\displaystyle\frac{m\ell^{2}}{3}\omega\dot{\omega}\equiv\frac{m\ell^{2}}{3}% \dot{\theta}\ddot{\theta}\quad\mbox{(since $\omega=\dot{\theta}$ and $\dot{\omega}=\ddot{\theta}$)}.$

Now equating $P$ and $\dot{E}_{K}$ we get

	$\displaystyle\frac{\cancel{m}\ell^{2}}{3}\cancel{\dot{\theta}}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\cancel{m}g\frac{\ell}{2}\cancel{\dot{\theta}}\cos\theta$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\ddot{\theta}$	$\displaystyle=$	$\displaystyle-\frac{3g}{2\ell}\cos\theta$

which is the same expression as obtained in Sample 17.85 (b).

Answer: $\ddot{\theta}=-\frac{3g}{2\ell}\cos\theta$

Note: Just like in the last problem, we have assumed that the upper end of the ladder stays in contact with the vertical wall. This assumption is used implicitly in the kinematics of the ladder. In energy calculations, the normal reactions of the two walls make no contribution as the work done by them is zero and hence the assumption of wall contact does not figure in the calculations above.

SAMPLE 17.24

Filename:sfig4-6-8b

Figure 17.92: A composite wheel made of three uniform disks rolls down an inclined wedge without slipping.

Rolling on an inclined plane. A wheel is made up of three uniform disks— the center disk of mass $m=1\,\mathrm{kg}$ , radius $r=10\,\mathrm{cm}$ and two identical outer disks of mass $M=2\,\mathrm{kg}$ each and radius $R$ . The wheel rolls down an inclined wedge without slipping. The angle of inclination of the wedge with horizontal is $\theta=30^{{}^{\circ}}$ . The radius of the bigger disks is to be selected such that the linear acceleration of the wheel center does not exceed 0.2g. Find the radius $R$ of the bigger disks.

Solution

Since a bound is prescribed on the linear acceleration of the wheel and the radius of the bigger disks is to be selected to satisfy this bound, we need to find an expression for the acceleration of the wheel (hopefully) in terms of the radius $R$ .

The free-body diagram of the wheel is shown in Fig. 17.93. In addition to the weight $(m+2M)g$ of the wheel and the normal reaction $N$ of the wedge surface there is an unknown force of friction $F_{f}$ acting on the wheel at point C. This friction force is necessary for the condition of rolling motion. You must realize, however, that $F_{f}\neq\mu N$ because there is neither slipping nor a condition of impending slipping. Thus the magnitude of $F_{f}$ is not known yet.

Filename:sfig5-5-2

Figure 17.93: Free-body diagram of the wheel.

Let the acceleration of the center-of-mass of the wheel be

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=a_{G}\hat{\boldsymbol{% \lambda}}

and the angular acceleration of the wheel be

\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\dot{\omega}\hat{% \boldsymbol{k}}.

We assumed $\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ to be in the negative $\hat{\boldsymbol{k}}$ direction. But, if this assumption is wrong, we will get a negative value for $\dot{\omega}$ .

Now we write the equation of linear momentum balance for the wheel:

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle m_{\rm total}\,{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \rm cm}$
	$\displaystyle-(m+2M)g\hat{\boldsymbol{\jmath}}+N\hat{\boldsymbol{n}}-F_{f}\hat% {\boldsymbol{\lambda}}$	$\displaystyle=$	$\displaystyle(m+2M)a_{G}\hat{\boldsymbol{\lambda}}$

This 2-D vector equation gives (at the most) two independent scalar equations. But we have three unknowns: $N,\;F_{f},$ and $a_{G}$ . Thus we do not have enough equations to solve for the unknowns including the quantity of interest, $a_{G}$ . So, we now write the equation of angular momentum balance for the wheel about the point of contact C (using ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/C}=r\hat{\boldsymbol{n}}$ ):

Filename:sfig5-5-2a

Figure 17.94: Geometry of unit vectors. This diagram can be used to find various dot and cross products between any two unit vectors. For example,

\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\jmath}}=\sin\theta\hat{\boldsymbol% {k}}.

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$
where
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{C}$% }}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/C}\times(m+2M)g(% -\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle r\hat{\boldsymbol{n}}\times(m+2M)g(-\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-(m+2M)gr\sin\theta\hat{\boldsymbol{k}}\quad\mbox{(see Fig.~{}% \ref{sfig7-3-3b})}$
and
$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle=$	$\displaystyle I^{G}_{zz}\,\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}% }}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/C}\times m_{\rm total}\,{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}$
	$\displaystyle=$	$\displaystyle I^{G}_{zz}\,(-\dot{\omega}\hat{\boldsymbol{k}})-m_{\rm total}\,% \dot{\omega}r^{2}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(I^{G}_{zz}+m_{\rm total}r^{2})(-\dot{\omega}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle\left[(\frac{1}{2}mr^{2}+2\cdot\frac{1}{2}MR^{2})+\overbrace{(m+2% M)}^{m_{\rm total}}r^{2}\right](-\dot{\omega}\hat{\boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle-\left[\frac{3}{2}mr^{2}+M(R^{2}+2r^{2})\right]\dot{\omega}\hat{% \boldsymbol{k}}.$

Thus,

	$\displaystyle-(m+2M)gr\sin\theta\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle-\left[\frac{3}{2}mr^{2}+M(R^{2}+2r^{2})\right]\dot{\omega}\hat{% \boldsymbol{k}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\omega}$	$\displaystyle=$	$\displaystyle\frac{(m+2M)gr\sin\theta}{\frac{3}{2}mr^{2}+M(R^{2}+2r^{2})}.$		(17.68)

Now we need to relate $\dot{\omega}$ to $a_{G}$ . From the kinematics of rolling,

a_{G}=\dot{\omega}r.

Therefore, from Eqn. (17.68) we get

\displaystyle a_{G}

\displaystyle=

\displaystyle\frac{(m+2M)gr^{2}\sin\theta}{\frac{3}{2}mr^{2}+M(R^{2}+2r^{2})}.

Now we can solve for $R$ in terms of $a_{G}$ :

$\displaystyle\frac{3}{2}mr^{2}+M(R^{2}+2r^{2})$	$\displaystyle=$	$\displaystyle\frac{(m+2M)gr^{2}\sin\theta}{a_{G}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M(R^{2}+2r^{2})$	$\displaystyle=$	$\displaystyle\frac{(m+2M)g}{a_{G}}r^{2}\sin\theta-\frac{3}{2}mr^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}R^{2}$	$\displaystyle=$	$\displaystyle\frac{(m+2M)g}{Ma_{G}}r^{2}\sin\theta-\frac{3m}{2M}r^{2}-2r^{2}.$

Since we require $a_{G}\leq 0.2g$ we get

$\displaystyle R^{2}$	$\displaystyle\geq$	$\displaystyle\left(\frac{(m+2M)g}{M\cdot 0.2g}\sin\theta-\frac{3m}{2M}-2\right% )r^{2}$
	$\displaystyle\geq$	$\displaystyle\left(\frac{5\,\mathrm{kg}}{0.4\,\mathrm{kg}}\cdot\frac{1}{2}-% \frac{3\,\mathrm{kg}}{4\,\mathrm{kg}}-2\right)(0.1\,\mathrm{m})^{2}$
	$\displaystyle\geq$	$\displaystyle 0.035{\,\mathrm{m}}^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}R$	$\displaystyle\geq$	$\displaystyle 0.187\,\mathrm{m}.$

Thus the outer disks of radius $20\,\mathrm{cm}$ will do the job.

Answer: $R\geq 18.7\,\mathrm{cm}$

SAMPLE 17.25

Filename:sfig5-5-2b

Figure 17.95

Which one starts rolling first — a marble or a bowling ball? A marble and a bowling ball, made of the same material, are launched on a horizontal platform with the same initial velocity, say $v_{0}$ . The initial velocity is large enough so that both start out sliding. Towards the end of their motion, both have pure rolling motion. If the radius of the bowling ball is 16 times that of the marble, find the instant, for each ball, when the sliding motion changes to rolling motion.

Solution

Let us consider one ball, say the bowling ball, first. Let the radius of the ball be $r$ and mass $m$ . The ball starts with center-of-mass velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{o}=v_{0}\hat{\boldsymbol{\imath}}$ . The ball starts out sliding. During the sliding motion, the force of friction acting on the ball must equal $\mu N$ (see the FBD). The friction force creates a torque about the mass-center which, in turn, starts the rolling motion of the ball. However, rolling and sliding coexist for a while, till the speed of the mass-center slows down enough to satisfy the pure rolling condition, $v=\omega r$ . Let the instant of transition from the mixed motion to pure rolling be $t^{*}$ .

Filename:sfig7-4-2

Figure 17.96: Free-body diagram of the ball during sliding.

From linear momentum balance , we have

$\displaystyle m\dot{v}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-\mu N\hat{\boldsymbol{\imath}}+(N-mg)\hat{\boldsymbol{\jmath}}$	(17.69)
$\displaystyle\mbox{eqn.~{}(\ref{eq:bowling1.1})}\cdot\hat{\boldsymbol{\jmath}}% \hbox to36.135pt{\hfil$\Rightarrow$\hfil}N$	$\displaystyle=$	$\displaystyle mg$
$\displaystyle\mbox{eqn.~{}(\ref{eq:bowling1.1})}\cdot\hat{\boldsymbol{\imath}}% \hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\dot{v}$	$\displaystyle=$	$\displaystyle-\mu N=-\mu mg$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{v}$	$\displaystyle=$	$\displaystyle-\mu g$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v$	$\displaystyle=$	$\displaystyle v_{0}-\mu gt.$	(17.70)

Similarly, from angular momentum balance about the mass-center, we get

$\displaystyle-I^{\mathrm{cm}}_{zz}\dot{\omega}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle-\mu Nr\hat{\boldsymbol{k}}=-\mu mgr\hat{\boldsymbol{k}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\dot{\omega}$	$\displaystyle=$	$\displaystyle\frac{\mu mgr}{I^{\mathrm{cm}}_{zz}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega$	$\displaystyle=$	$\displaystyle\underbrace{\omega_{0}}_{0}+\frac{\mu mgr}{I^{\mathrm{cm}}_{zz}}t.$	(17.71)

At the instant of transition from mixed rolling and sliding to pure rolling, i.e., at $t=t^{*}$ , $v=\omega r$ . Therefore, from eqn. (17.70) and eqn. (17.71), we get

$\displaystyle v_{0}-\mu gt^{*}$	$\displaystyle=$	$\displaystyle\frac{\mu mgr^{2}}{I^{\mathrm{cm}}_{zz}}t^{*}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{0}$	$\displaystyle=$	$\displaystyle\mu gt^{*}(1+\frac{mr^{2}}{I^{\mathrm{cm}}_{zz}})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t^{*}$	$\displaystyle=$	$\displaystyle\frac{v_{0}}{\mu g(1+\frac{mr^{2}}{I^{\mathrm{cm}}_{zz}})}.$

Now, for a sphere, $I^{\mathrm{cm}}_{zz}=\frac{2}{5}mr^{2}$ . Therefore,

t^{*}=\frac{v_{0}}{\mu g(1+\frac{mr^{2}}{\frac{2}{5}mr^{2}})}=\frac{2v_{0}}{7% \mu g}.

Note that the expression for $t^{*}$ is independent of mass and radius of the ball! Therefore, the bowling ball and the marble are going to change their mixed motion to pure rolling at exactly the same instant. This is not an intuitive result.

Answer: $t^{*}=\frac{2v_{0}}{7\mu g}\mbox{ for both.}$

SAMPLE 17.26

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Figure 17.97

Transition from a mix of sliding and rolling to pure rolling, using impulse-momentum. Consider the problem in Sample 17.94 again: A ball of radius $r=10\,\mathrm{cm}$ and mass $m=1\,\mathrm{kg}$ is launched horizontally with initial velocity $v_{0}=5\,\mathrm{m}/\mathrm{s}$ on a surface with coefficient of friction $\mu=0.2$ . The ball starts sliding, rolls and slides simultaneously for a while, and then rolls without sliding. Find the time it takes to start pure rolling.

Solution

Let us denote the time of transition from mixed motion (rolling and sliding) to pure rolling by $t^{*}$ . At $t=0$ , we know that $v_{\mathrm{cm}}=v_{0}=5\,\mathrm{m}/\mathrm{s}$ , and $\omega_{0}=0$ . We also know that at $t=t^{*}$ , $v_{\mathrm{cm}}=v_{t^{*}}=\omega_{t^{*}}r$ , where $r$ is the radius of the ball. We do not know $t^{*}$ and $v_{t^{*}}$ . However, we are considering a finite time event (during $t^{*}$ ) and the forces acting on the ball during this duration are known. Recall that impulse momentum equations involve the net force on the body, the time of impulse, and momenta of the body at the two instants. Momenta calculations involve velocities. Therefore, we should be able to use impulse-momentum equations here and find the desired unknowns.

Filename:pfigure-blue-47-2

Figure 17.98: Free-body diagram of the ball during sliding.

From linear impulse-momentum, we have

$\displaystyle\left(\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\right)t^{*}$	$\displaystyle=$	$\displaystyle mv_{t^{*}}\hat{\boldsymbol{\imath}}-mv_{0}\hat{\boldsymbol{% \imath}}$
$\displaystyle\left(-\mu N\hat{\boldsymbol{\imath}}+(N-mg)\hat{\boldsymbol{% \jmath}}\right)t^{*}$	$\displaystyle=$	$\displaystyle m(v_{t^{*}}-v_{0})\hat{\boldsymbol{\imath}}.$
Dotting the above equation with $\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{\imath}}$ , respectively, we get
$\displaystyle N$	$\displaystyle=$	$\displaystyle mg$
$\displaystyle-\mu\underbrace{N}_{mg}t^{*}$	$\displaystyle=$	$\displaystyle m(v_{t^{*}}-v_{0})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-\mu gt^{*}$	$\displaystyle=$	$\displaystyle v_{t^{*}}-v_{0}.$	(17.72)

Similarly, from angular impulse-momentum relation about the mass-center, we get

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}t^% {*}$	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}})_{t^% {*}}-({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}})_{0}$
$\displaystyle(-\mu Nr\hat{\boldsymbol{k}})t^{*}$	$\displaystyle=$	$\displaystyle(I^{\mathrm{cm}}_{zz}\omega_{t^{*}}-I^{\mathrm{cm}}_{zz}% \underbrace{\omega_{0}}_{0})(-\hat{\boldsymbol{k}})$
$\displaystyle\hbox to36.135pt{or\hfil}-\mu mgrt^{*}$	$\displaystyle=$	$\displaystyle-I^{\mathrm{cm}}_{zz}\omega_{t^{*}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega_{t^{*}}$	$\displaystyle=$	$\displaystyle\mu mgrt^{*}/I^{\mathrm{cm}}_{zz}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}v_{t^{*}}$	$\displaystyle\equiv$	$\displaystyle\omega_{t^{}}r=\mu mgr^{2}t^{}/I^{\mathrm{cm}}_{zz}.$

Substituting this expression for $v_{t^{*}}$ in eqn. (17.72), we get

	$\displaystyle-\mu gt^{*}$	$\displaystyle=$	$\displaystyle\mu mgr^{2}t^{*}/I^{\mathrm{cm}}_{zz}-v_{0}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}t^{*}$	$\displaystyle=$	$\displaystyle\frac{v_{0}}{\mu g(1+\frac{mr^{2}}{I^{\mathrm{cm}}_{zz}})}$

which is, of course, the same expression we obtained for $t^{*}$ in Sample 17.94. Again, noting that $I^{\mathrm{cm}}_{zz}=\frac{2}{5}mr^{2}$ for a sphere, we calculate the time of transition as

t^{*}=\frac{2v_{0}}{7\mu g}=\frac{2\cdot(5\,\mathrm{m}/\mathrm{s})}{7\cdot(0.2% )\cdot(9.8\,\mathrm{m}/\mathrm{s}^{2})}=0.73\,\mathrm{s}.

Answer: $t^{*}=0.73\,\mathrm{s}$

Problems for 17.4 Mechanics of contact

17.4.1 A uniform disc of mass $m$ and radius $r$ rolls without slip at constant rate. What is the total kinetic energy of the disk?

17.4.2 A non-uniform disc of mass $m$ and radius $r$ rolls without slip at constant rate. The center of mass is located at a distance $\frac{r}{2}$ from the center of the disc. What is the total kinetic energy of the disc when the center of mass is directly above the center of the disc?

17.4.3 Falling hoop. A bicycle rim (no spokes, tube, tire, or hub) is idealized as a hoop with mass $m$ and radius $R$ . $G$ is at the center of the hoop. An inextensible string is wrapped around the hoop and attached to the ceiling. The hoop is released from rest at the position shown at $t=0$ .

(a)

Find $y_{G}$ at a later time $t$ in terms of any or all of $m$ , $R$ , $g$ , and $t$ .
(b)

Does $G$ move sideways as the hoop falls and unrolls?

Filename:pfigure-blue-89-1

Figure 17.99

17.4.4 A uniform disk with radius $R$ and mass $m$ has a string wrapped around it. The string is pulled with a force $F$ . The disk rolls without slipping.

(a)

What is the angular acceleration of the disk, ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{Disk}=-\ddot{\theta}\hat{% \boldsymbol{k}}$ ? Make any reasonable assumptions you need that are consistent with the figure information and the laws of mechanics. State your assumptions. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\alpha}}}_{Disk}=-\frac{4}{3}\,% \mathrm{rad}/\mathrm{s}^{2}\hat{\boldsymbol{k}}$ .
(b)

Find the acceleration of the point A in the figure. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{8}{3}\,\mathrm{m}/% \mathrm{s}^{2}\hat{\boldsymbol{\imath}}$ .

Filename:pfigure-blue-35-2

Figure 17.100

17.4.5 If a pebble is stuck to the edge of the wheel in problem 17.3, what is the maximum speed of the pebble during the motion? When is the force on the pebble from the wheel maximum? Draw a good FBD including the force due to gravity.

Filename:pfigure4-3Dpend

Figure 17.101

17.4.6 Spool Rolling without Slip and Pulled by a Cord. The light-weight spool is nearly empty but a lead ball with mass $m$ has been placed at its center. A force $F$ is applied in the horizontal direction to the cord wound around the wheel. Dimensions are as marked. Coordinate directions are as marked.

(a)

What is the acceleration of the center of the spool? Answer: $\boldsymbol{a}_{C}=(F/m)(1-R_{i}/R_{O})\hat{\boldsymbol{\imath}}$ .
(b)

What is the horizontal force of the ground on the spool?

Answer: $(-R_{i}/R_{O})F\hat{\boldsymbol{\imath}}$ .

Filename:pfigure-s94h6p3

Figure 17.102

17.4.7 A film spool is placed on a very slippery table. Assume that the film and reel (together) have mass distributed the same as a uniform disk of radius $R_{i}$ . What, in terms of $R_{i},\ R_{o},\ m,\ g,\ \hat{\boldsymbol{\imath}},\ \hat{\boldsymbol{\jmath}}$ , and $F$ are the accelerations of points C and B at the instant shown (the start of motion)?

Filename:pfigure-s94h6p4

Figure 17.103

17.4.8 Again, Spool Rolling without Slip and Pulled by a Cord. Reconsider the spool from problem 17.4. This time, a force $F$ is applied vertically to the cord wound around the wheel. In this case, what is the acceleration of the center of the spool? Is it possible to pull the cord at some angle between horizontal and vertical so that the angular acceleration of the spool or the acceleration of the center of mass is zero? If so, find the angle in terms of $R_{i}$ , $R_{0}$ , $m$ , and $F$ .

17.4.9 A napkin ring lies on a black velvet tablecloth. The thin ring (of mass $m$ , radius $r$ ) rolls without slip as a mischievous child pulls the tablecloth (mass $M$ ) out with acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ . The ring starts at the right end ( $x=d$ ). You can make a reasonable physical model of this situation with an empty soda can and a piece of paper on a flat table.

(a)

What is the ring’s acceleration as the tablecloth is being withdrawn?
(b)

How far has the tablecloth moved to the right from its starting point $x=0$ when the ring rolls off its left-hand end?
(c)

Clearly describe the subsequent motion of the ring. Which way does it end up rolling at what speed?
(d)

Would your answer to the previous question be different if the ring slipped on the cloth as the cloth was being pulled out?

Filename:pg84-3

Figure 17.104

17.4.10 A block of mass $M$ is supported by two rollers (uniform cylinders) each of mass $m$ and radius $r$ . They roll without slip on the block and the ground. A force $F$ is applied in the horizontal direction to the right, as shown in the figure. Given $F$ , $m$ , $r$ , and $M$ , find:

(a)

the acceleration of the block,
(b)

the acceleration of the center of mass of this block/roller system,
(c)

the reaction at the wheel bases,
(d)

the force of the right wheel on the block,
(e)

the acceleration of the wheel centers, and
(f)

the angular acceleration of the wheels.

Filename:pfigure-blue-36-1

Figure 17.105

17.4.11 Dropped spinning disk. 2-D . A uniform disk of radius $R$ and mass $m$ is gently dropped onto a surface and doesn’t bounce. When it is released it is spinning clockwise at the rate $\dot{\theta}_{0}$ . The disk skids for a while and then is eventually rolling.

(a)

What is the speed of the center of the disk when the disk is eventually rolling (answer in terms of $g$ , $\mu$ , $R$ , $\dot{\theta}_{0}$ , and $m$ )? Answer: speed $v=\frac{R\dot{\theta}}{3}$ .
(b)

In the transition from slipping to rolling, energy is lost to friction. How does the amount lost depend on the coefficient of friction (and other parameters)? How does this loss make or not make sense in the limit as $\mu\rightarrow 0$ and the dissipation rate $\rightarrow$ zero? Answer: The energy lost to friction is $E_{fric}=\frac{mR^{2}\dot{\theta}_{0}^{2}}{6}$ . The energy lost to friction is independent of $\mu$ for $\mu>0$ . Thus, the energy lost to friction is constant for given $m$ , $R$ , and $\dot{\theta}_{0}$ . As $\mu\rightarrow 0$ , the transition time to rolling $\rightarrow\infty$ . It is not true, however, that the energy lost to friction $\rightarrow 0$ as $\mu\rightarrow 0$ . Since the energy lost is constant for any $\mu>0$ , the disk will slip for longer and longer times so that the distance of slip goes to infinity. The dissipation rate $\rightarrow 0$ since the constant energy is divided by increasing transition time. The energy lost is zero only for $\mu=0$ .

Filename:pfigure-s95q8

Figure 17.106

17.4.12 Disk on a conveyor belt. A uniform metal cylinder with mass of $200\,\mathrm{kg}$ is carried on a conveyer belt which moves at $V_{0}=3\,\mathrm{m}/\mathrm{s}$ . The disk is not rotating when on the belt. The disk is delivered to a flat hard platform where it slides for a while and ends up rolling. How fast is it moving (i.e. what is the speed of the center of mass) when it eventually rolls? Answer: $V=2\,\mathrm{m}/\mathrm{s}$ .

Filename:pfigure4-rpi

Figure 17.107

17.4.13 A rigid hoop with radius $R$ and mass $m$ is rolling without slip so that its center has translational speed $v_{o}$ . It then hits a narrow bar with height $R/2$ . When the hoop hits the bar suddenly it sticks and doesn’t slide. It does hinge freely about the bar, however. The gravitational constant is $g$ . How big is $v_{o}$ if the hoop just barely rolls over the bar? Answer: $v_{o}=2\frac{\sqrt{2gR}}{3}$ .

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Figure 17.108

17.4.14 2-D rolling of an unbalanced wheel. A wheel, modeled as massless, has a point mass (mass $=m$ ) at its perimeter. The wheel is released from rest at the position shown. The wheel makes contact with coefficient of friction $\mu$ .

(a)

What is the acceleration of the point P just after the wheel is released if $\mu=0$ ?
(b)

What is the acceleration of the point P just after the wheel is released if $\mu=2$ ?
(c)

Assuming the wheel rolls without slip (no-slip requires, by the way, that the friction be high: $\mu=\infty$ ) what is the velocity of the point P just before it touches the ground?

Filename:pfigure-blue-88-1

Figure 17.109

17.4.15 Spool and mass. A reel of mass $M$ and moment of inertia $I^{cm}_{zz}=I$ rolls without slipping upwards on an incline with slope-angle $\alpha$ . It is pulled up by a string attached to mass $m$ as shown. Find the acceleration of point G in terms of some or all of $M,m,I,R,r,\alpha,g$ and any base vectors you clearly define.

Filename:pfigure4-2-rp3

Figure 17.110

17.4.16 Two objects are released on two identical ramps. One is a sliding block (no friction), the other a rolling hoop (no slip). Both have the same mass, $m$ , are in the same gravity field and have the same distance to travel. It takes the sliding mass 1 $\,{\rm s}$ to reach the bottom of the ramp. How long does it take the hoop? [Useful formula: “ $s=\frac{1}{2}at^{2}$ ”]

Filename:pfigure-s94q6p1

Figure 17.111

17.4.17 The hoop is rolled down an incline that is 30^∘ from horizontal. It does not slip. It does not fall over sideways. It is let go from rest at t = 0.

(a)

At t = 0⁺ what is the acceleration of the hoop center of mass?
(b)

At t = 0⁺ what is the acceleration of the point on the hoop that is on the incline?
(c)

At t = 0⁺ what is the acceleration of the point on the hoop that is furthest from the incline?
(d)

After the hoop has descended 2 vertical meters (and traveled an appropriate distance down the incline) what is the acceleration of the point on the hoop that is (at that instant) furthest from the incline?

Filename:pfigure-spinningbrick

Figure 17.112

17.4.18 A uniform cylinder of mass $m$ and radius $r$ rolls down an incline without slip, as shown below. Determine: (a) the angular acceleration $\alpha$ of the disk; (b) the minimum value of the coefficient of friction $\mu$ that will insure no slip.

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Figure 17.113

17.4.19 Race of rollers. A uniform disk with mass $M_{0}$ and radius $R_{0}$ is allowed to roll down the quite slip-resistant ( $\mu=1$ ) $30^{\circ}$ ramp shown. It is raced against four other objects ( $A$ , $B$ , $C$ and $D$ ), one at a time. Who wins the races, or are there ties? First try to construct any plausible reasoning. Good answers will be based, at least in part, on careful use of equations of mechanics. Answer: Accelerations of the center of mass, where $\hat{\boldsymbol{\imath}}$ is parallel to the slope and pointing down: (a) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}=g\sin\theta\hat{\boldsymbol% {\imath}}$ , (b) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}=\frac{2}{3}g\sin\theta\hat{% \boldsymbol{\imath}}$ , (c) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}=\frac{1}{2}g\sin\theta\hat{% \boldsymbol{\imath}}$ , (d) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}=\frac{2}{3}g\sin\theta\hat{% \boldsymbol{\imath}}$ . So, the block is fastest, all uniform disks are second, and the hollow pipe is third.

(a)

Block $A$ has the same mass and has center of mass a distance $R_{0}$ from the ramp. It rolls on massless wheels with frictionless bearings.
(b)

Uniform disk $B$ has the same mass ( $M_{B}=M_{0}$ ) but twice the radius ( $R_{B}=2R_{0}$ ).
(c)

Hollow pipe $C$ has the same mass ( $M_{C}=M_{0}$ ) and the same radius ( $R_{C}=R_{0}$ ).
(d)

Uniform disk $D$ has the same radius ( $R_{D}=R_{0}$ ) but twice the mass ( $M_{D}=2M_{0}$ ).

Can you find a round object which will roll as fast as the block slides? How about a massless cylinder with a point mass in its center? Can you find an object which will go slower than the slowest or faster than the fastest of these objects? What would they be and why? (This problem is harder.)

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Figure 17.114

17.4.20 A roller of mass $M$ and polar moment of inertia about the center of mass $I_{G}$ is connected to a spring of stiffness $K$ by a frictionless hinge as shown in the figure. Consider two kinds of friction between the roller and the surface it moves on:

1.

Perfect slipping (no friction), and
2.

Perfect rolling (infinite friction).

(a)

What is the period of oscillation in the first case?
(b)

What is the period of oscillation in the second case?

Filename:pg92-2

Figure 17.115

17.4.21 A uniform cylinder of mass $m$ and radius $R$ rolls back and forth without slipping through small amplitudes (i.e., the springs attached at point A on the rim act linearly and the vertical change in the height of point A is negligible). The springs, which act both in compression and tension, are unextended when A is directly over C.

(a)

Determine the differential equation of motion for the cylinder’s center.
(b)

Calculate the natural frequency of the system for small oscillations.

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Figure 17.116

17.4.22 Hanging disk, 2-D. A uniform thin disk of radius $R$ and mass $m$ hangs in a gravity field $g$ from a pair of massless springs each with constant $k$ . In the static equilibrium configuration the springs are vertical and attached to points A and B on the right and left edges of the disk. In the equilibrium configuration the springs carry the weight, the disk counter-clockwise rotation is $\phi=0$ , and the downwards vertical deflection is $y=0$ . Assume throughout that the center of the disk only moves up and down, and that $\phi$ is small so that the springs may be regarded as vertical when calculating their stretch ( $\sin\phi\approx\phi$ and $\cos\phi\approx 1$ ).

(a)

Find $\ddot{\phi}$ and $\ddot{y}$ in terms of some or all of $\phi,\dot{\phi},y,\dot{y},k,m,R$ , and $g$ .
(b)

Find the natural frequencies of vibration in terms of some or all of $k, m, R$ , and $g$ .

Filename:p-f96-p3-3

Figure 17.117

17.4.23 A disk rolls in a cylinder. For all of the problems below, the disk rolls without slip and rocks back and forth due to gravity.

(a)

Sketch. Draw a neat sketch of the disk in the cylinder. The sketch should show all variables, coordinates and dimension used in the problem.
(b)

FBD. Draw a free-body diagram of the disk.
(c)

Momentum balance. Write the equations of linear and angular momentum balance for the disk. Use the point on the cylinder which touches the disk for the angular momentum balance equation. Leave as unknown in these equations variables which you do not know.
(d)

Kinematics. The disk rolling in the cylinder is a one-degree-of-freedom system. That is, the values of only one coordinate and its derivatives are enough to determine the positions, velocities and accelerations of all points. The angle that the line from the center of the cylinder to the center of the disk makes from the vertical can be used as such a variable. Find all of the velocities and accelerations needed in the momentum balance equation in terms of this variable and its derivative. [Hint: you’ll need to think about the rolling contact in order to do this part.]
(e)

Equation of motion. Write the angular momentum balance equation as a single second order differential equation.
(f)

Simple pendulum? Does this equation reduce to the equation for a pendulum with a point mass and length equal to the radius of the cylinder, when the disk radius gets arbitrarily small? Why, or why not?

Filename:s97p3-3

Figure 17.118

A disk rolls without slip inside a bigger cylinder.

17.4.24 A uniform hoop of radius $R_{1}$ and mass $m$ rolls from rest down a semi-circular track of radius $R_{2}$ as shown. Assume that no slipping occurs. At what angle $\theta$ does the hoop leave the track and what is its angular velocity $\omega$ and the linear velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}$ of its center of mass at that instant?

If the hoop slides down the track without friction, so that it does not rotate, will it leave at a smaller or larger angle $\theta$ than if it rolls without slip (as above)? Give a qualitative argument to justify your answer.

HINT: Here is a geometric relationship between angle $\phi$ hoop turns through and angle $\theta$ subtended by its center when no slipping occurs: $\phi=[(R_{1}+R_{2})/R_{2}]\theta$ . (You may or may not need to use this hint.)

Filename:pfigure-s94h8p1

Figure 17.119

17.5 Rigid-object collision mechanics

This section extends the particle collisions discussion in Section 13.2 that starts on page 13.2.

2D collisions

For collisions between rigid bodies with more general motions before and after the collisions we depend on the three key ideas

I.

Collision forces are big,
II.

Collisions are quick, and
III.

The laws of mechanics apply during the collision.

There are two extra assumptions that are needed in simple analysis:

IV.

Collision forces are few. For a given rigid body there is one, or at most two non-negligible collision forces. This is the real import of idea (I) above. Because collision forces are big most other forces can be neglected.
V.

The collision force(s) act at a well defined point which does not move during the collision.

Based on these assumptions one then uses linear and angular momentum balance in their time-integrated form.

Filename:sfig4-7-DH1

Figure 17.120: Two bodies collide at point C. The only non-negligible collision impulse is

{\accentset{\rightharpoonup}{\boldsymbol{P}}}

acting on body 2 (and

-{\accentset{\rightharpoonup}{\boldsymbol{P}}}

on body 1) at point C. The material points on the contacting bodies are C₁ and C₂. The outward normal to body 1 at C₁ is

\hat{\boldsymbol{n}}

Example: Two bodies in space

Two bodies collide at point C. The impulse acting on body 2 is ${\accentset{\rightharpoonup}{\boldsymbol{P}}}=\int{\accentset{\rightharpoonup}% {\boldsymbol{F}}}_{\rm coll}\,dt$ . If the mass and inertia properties of both bodies is known, as are the velocities and rotation rates before the collision we have the following linear and angular momentum balance equations for the two bodies:

\begin{array}[]{rcl}-{\accentset{\rightharpoonup}{\boldsymbol{P}}}&=&m_{1}% \left({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{-}\right)\\[5.0pt] {\accentset{\rightharpoonup}{\boldsymbol{P}}}&=&m_{2}\left({\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm G2}^{+}-{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\rm G2}^{-}\right)\\[5.0pt] {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/G1}}\times(-{% \accentset{\rightharpoonup}{\boldsymbol{P}}})&=&{I^{\mathrm{cm}}_{zz}}^{1}% \left(\omega_{1}^{+}-\omega_{1}^{-}\right)\hat{\boldsymbol{k}}\\[5.0pt] {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/G2}}\times{% \accentset{\rightharpoonup}{\boldsymbol{P}}}&=&{I^{\mathrm{cm}}_{zz}}^{2}\left% (\omega_{2}^{+}-\omega_{2}^{-}\right)\hat{\boldsymbol{k}}.\end{array}

(17.73)

These make up 6 scalar equations (2 for each momentum equation, 1 for each angular momentum equation). There are 8 scalar unknowns: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}$ (2), ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G2}^{+}$ (2), $\omega_{1}^{+}$ (1), $\omega_{2}^{+}$ (1), and ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ (2). Thus the motion after the collision cannot be determined.

[Note that linear and angular momentum balance for the system would give equations which could be obtained by adding and subtracting combinations of the equations above. So adding system momentum balance equations does not add information (i.e., adds linearly dependent equations).]

So, as for 1-D collisions, momentum balance is not enough to determine the outcome of the collision. Eqns. 17.73 aren’t enough. A thousand different models and assumptions could be added to make the system solvable. But there are only two cases that are non-controversial and also relatively simple: 1) sticking collisions, and 2) frictionless collisions.

Sticking collisions

A ‘perfectly-plastic’ sticking collision is one where the relative velocities of the two contacting points are assumed to go suddenly to zero. That is

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C1}^{+}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{C2}^{+}

Writing ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm C1}^{+}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}+(\omega_{1}^{+}\hat{\boldsymbol% {k}})\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/G1}}$ and similarly for ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm C2}$ thus adds a vector equation (2 scalar equations) to the equation set 17.73. This gives 8 equations in 8 unknowns.

A little cleverness can reduce the problem to one of solving only 4 equations in 4 unknowns. Linear momentum balance for the system, angular momentum balance for the system and angular momentum balance for object 2 make up 4 scalar equations. None of these equations includes the impulse ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ . Because the system moves as if hinged at C₁ after the collision, the state of motion after the system is fully characterized by ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}$ , $\omega_{1}^{+}$ , and $\omega_{2}^{+}$ . Thus we have 4 equations in 4 unknowns.

Filename:sfig4-7-DH2

Figure 17.121: Sticking collision with an immovable object. The box sticks at A and then rotates about A. Angular momentum about point A is conserved in the collision.

Example: One body is hugely massive: collision with an immovable object

If body 2, say, is huge compared to body 1 then it can be taken to be immovable and collision problems can be solved by only considering body 1 (see fig. 17.121). In the case of a sticking collision the full state of the system after the collision is determined by $\omega_{1}^{+}$ . This can be found from the single scalar equation obtained from angular momentum balance about the collision point.

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm A}^{-}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm A}^{+}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}% \times m{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}_{\rm G}^{-}+I^{\mathrm% {cm}}_{zz}\omega^{-}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}% \times m{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}_{\rm G}^{+}+I^{\mathrm% {cm}}_{zz}\omega^{+}\hat{\boldsymbol{k}}$

Because the state of the system before the collision is assumed known (the left “-” side of the equation, and because the post-collision (+) state is a rotation about A, this equation is one scalar equation in the one unknown $\omega^{+}$ . Note that ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm A}^{+}$ could also be evaluated as ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm A}^{+}=\omega^{+}I^{A}_{zz}% \hat{\boldsymbol{k}}$ . So one way of expressing the post-collision state is as

\omega^{+}={\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A% }}\times m{{\accentset{\rightharpoonup}{\boldsymbol{v}}}}_{\rm G}^{-}+I^{% \mathrm{cm}}_{zz}\omega^{-}\hat{\boldsymbol{k}}\right)\cdot\hat{\boldsymbol{k}% }\over I^{A}_{zz}}\hbox to36.135pt{\hfil and\hfil}{{\accentset{\rightharpoonup% }{\boldsymbol{v}}}}_{\rm G}^{+}=\omega^{+}\hat{\boldsymbol{k}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}.

Note also that the same ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}$ is used on the right and left sides of the equation because only the velocity and not the position is assumed to jump during the collision.

The collision impulse ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ can then be found from linear momentum balance as

{\accentset{\rightharpoonup}{\boldsymbol{P}}}=m\left({{\accentset{% \rightharpoonup}{\boldsymbol{v}}}}_{\rm G}^{+}-{{\accentset{\rightharpoonup}{% \boldsymbol{v}}}}_{\rm G}^{-}\right).

Sticking collisions are used as models of projectiles hitting targets, of robot and animal limbs making contact with the ground, of monkeys and acrobats grabbing hand holds, and of some particularly dead and frictional collisions between solids (such as when a car trips on a curb).

Frictionless collisions

The second special case is that of a frictionless collision. Here we add two assumptions:

1.

There is no friction so ${\accentset{\rightharpoonup}{\boldsymbol{P}}}=P\hat{\boldsymbol{n}}$ . The number of unknowns is thus reduced from 8 to 7.
2.

There is a coefficient of (normal) restitution $e$ .

The normal restitution coefficient is taken as a property of the colliding bodies. It is a given number with $0<e<1$ with this defining equation:

({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm C2}^{+}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm C1}^{+})\cdot\hat{\boldsymbol{n}}=-e({% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm C2}^{-}-{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm C1}^{-})\cdot\hat{\boldsymbol{n}}.

This says that the normal part of the relative velocity of the contacting points reverses sign and its magnitude is attenuated by $e$ . This adds a scalar equation to the set Eqns. 17.73 thus giving 7 scalar equations (4 momentum, 2 angular momentum, 1 restitution) for 7 unknowns (4 velocity components, 2 angular velocities and the normal impulse).

The most popular application of the frictionless collision model is for billiard or pool balls, or carom pucks. These things have relatively small coefficients of friction.

We state without proof that a frictionless collision with $e=1$ conserves energy.

Filename:sfig4-7-DH3

Figure 17.122: Frictionless collision between two identical round objects. Ball one is initially moving to the right, ball 2 is initially stationary. The impulse of ball 1 on ball 2 is

{\accentset{\rightharpoonup}{\boldsymbol{P}}}

Example: Pool balls

Assume one ball approaches the other with initial velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}=v\hat{\boldsymbol{% \imath}}$ and has an elastic frictionless collision with the other ball at a collision angle of $\theta$ as shown in fig. 17.122. Defining $\hat{\boldsymbol{n}}\equiv\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}}$ we have that ${\accentset{\rightharpoonup}{\boldsymbol{P}}}=P\hat{\boldsymbol{n}}$ . To determine the outcome of the equation we have the angular momentum balance equations (about the center-of-mass) which trivially tell us that

\omega_{1}^{+}=\omega_{2}^{+}=0

because the balls start with no spin and the frictionless collision impulses ${\accentset{\rightharpoonup}{\boldsymbol{P}}}=P\hat{\boldsymbol{n}}$ and $-{\accentset{\rightharpoonup}{\boldsymbol{P}}}=-P\hat{\boldsymbol{n}}$ have no moment about the center-of-mass. Linear momentum balance for each of the balls

	$\displaystyle-P\hat{\boldsymbol{n}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}-mv% \hat{\boldsymbol{\imath}}$
	$\displaystyle P\hat{\boldsymbol{n}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G2}^{+}-{% \accentset{\rightharpoonup}{\boldsymbol{0}}}$

gives 4 scalar equations which are supplemented by the restitution equation (using $e=1$ )

	$\displaystyle\left(\Delta{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}% \right)\cdot\hat{\boldsymbol{n}}=-e\left(\Delta{\accentset{\rightharpoonup}{% \boldsymbol{v}}}^{-}\right)\cdot\hat{\boldsymbol{n}}$
	$\displaystyle\Rightarrow-v\cos\theta={\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{\rm G2}^{+}\cdot\hat{\boldsymbol{n}}-{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{\rm G1}^{+}\cdot\hat{\boldsymbol{n}}$

which together make 5 scalar equations in the 5 scalar unknowns ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}$ , ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G2}^{+}$ , and $P$ (each vector has 2 unknown components). These have the solution

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G1}^{+}$	$\displaystyle=$	$\displaystyle v\sin\theta(\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}}),$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm G2}^{+}$	$\displaystyle=$	$\displaystyle v\cos\theta(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}}),\quad\hbox{and}$
$\displaystyle P$	$\displaystyle=$	$\displaystyle mv\cos\theta.$

The solution can be checked by plugging back into the momentum and restitution equations. Also, as promised, this $e=1$ solution conserves kinetic energy. The solution has the interesting property that the outgoing trajectories of the two balls are orthogonal for all $\theta$ but $\theta=0$ in which case ball 1 comes to rest in the collision. [The solution can be found graphically by looking for two outgoing vectors which add to the original velocity of mass 1, where the sum of the squares of the outgoing speeds must add to the square of the incoming speed.]

Frictional collisions

For a collision with friction, but not so much that total sticking is accurate, the modeling is complex and subtle. As of this writing there are no standard acceptable ways of dealing with such situations. Commercial simulation packages should be used for such with skeptical caution. They are generally defective in that they can predict only a limited range of phenomena and/or they can create energy even with innocent input parameters.

Why is it hard to find a good collision law

Ideally one would like a rule to determine how bodies move after a collision from how they move before the collision. Such a rule would be called a collision law or a constitutive relation for collisions. That accurate collision laws are rare at best might be surmised from the basic problem that the phrase rigid body collisions is in some sense a contradiction in terms, an oxymoron. The force generated in the contact comes from material deformation, and deformation is just what we generally try to neglect when doing rigid body mechanics.

There is a temptation to say that one wants to continue to neglect deformation during the collision, but for an infinitesimal contact region. And some collision laws are formulated with this approach. Even then, there are no reliable models for the deformation in that small region, and such laws are doomed to inaccuracy in situations where the deformation is not so limited.

For complex shaped bodies touching at various points that are generally not known a priori, no collision law is reliably accurate.

SAMPLE 17.27 For two masses conservation of momentum is expressed by the vector equation $m_{1}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{1}+\,\mathrm{m}_{2}{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{2}=m_{1}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{1}^{+}+\,\mathrm{m}_{2}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{2}^{+}$ . Suppose ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{1}={\accentset{\rightharpoonup}% {\boldsymbol{0}}},\;{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{2}=-v_{0}% \hat{\boldsymbol{\jmath}},\;{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_% {1}=v^{+}_{1}\hat{\boldsymbol{\imath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{2}=v^{+}_{2_{t}}\hat{% \boldsymbol{e}}_{t}+v^{+}_{2_{n}}\hat{\boldsymbol{e}}_{n}$ , where $\hat{\boldsymbol{e}}_{t}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}$ and $\hat{\boldsymbol{e}}_{n}=-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}}$ .

1.

Obtain two independent scalar equations from the momentum equation corresponding to projections in the $\hat{\boldsymbol{e}}_{n}$ and $\hat{\boldsymbol{e}}_{t}$ directions.
2.

Assume that you are given another equation ${v^{+}}_{2_{t}}=-v_{0}\sin\theta$ . Set up a matrix equation to solve for $v^{+}_{1},v^{+}_{2_{t}}$ , and $v^{+}_{2_{n}}$ from the three equations.

Solution

1.

The given equation of conservation of linear momentum is

$\displaystyle m_{1}\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% 1}}_{\mbox{\bf 0}}+\,\mathrm{m}_{2}{\accentset{\rightharpoonup}{\boldsymbol{v}% }}_{2}$ $\displaystyle=$ $\displaystyle m_{1}{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{1}+\,% \mathrm{m}_{2}{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{2}$

$\displaystyle\hbox to36.135pt{or\hfil}-m_{2}v_{0}\hat{\boldsymbol{\jmath}}$ $\displaystyle=$ $\displaystyle m_{1}v^{+}_{1}\hat{\boldsymbol{\imath}}+m_{2}(v^{+}_{2_{t}}\hat{% \boldsymbol{e}}_{t}+v^{+}_{2_{n}}\hat{\boldsymbol{e}}_{n}).$ (17.74)

Dotting both sides of eqn. (17.74) with $\hat{\boldsymbol{e}}_{n}$ gives

$-m_{2}v_{0}(\overbrace{\hat{\boldsymbol{e}}_{n}\cdot\hat{\boldsymbol{\jmath}}}% ^{\cos\theta})=m_{1}v^{+}_{1}(\overbrace{\hat{\boldsymbol{e}}_{n}\cdot\hat{% \boldsymbol{\imath}}}^{-\sin\theta})+m_{2}v^{+}_{2_{t}}(\overbrace{\hat{% \boldsymbol{e}}_{n}\cdot\hat{\boldsymbol{e}}_{t}}^{0})+m_{2}v^{+}_{2_{n}}(% \overbrace{\hat{\boldsymbol{e}}_{n}\cdot\hat{\boldsymbol{e}}_{n}}^{1})$

$\hbox to36.135pt{or\hfil}-m_{2}v_{0}\cos\theta=-m_{1}v^{+}_{1}\sin\theta+m_{2}% v^{+}_{2_{n}}.$ (17.75)

Dotting both sides of eqn. (17.74) with $\hat{\boldsymbol{e}}_{t}$ gives

$-m_{2}v_{0}(\overbrace{\hat{\boldsymbol{e}}_{t}\cdot\hat{\boldsymbol{\jmath}}}% ^{\sin\theta})=m_{1}v^{+}_{1}(\overbrace{\hat{\boldsymbol{e}}_{t}\cdot\hat{% \boldsymbol{\imath}}}^{\cos\theta})+m_{2}v^{+}_{2_{t}}(\overbrace{\hat{% \boldsymbol{e}}_{t}\cdot\hat{\boldsymbol{e}}_{t}}^{1})+m_{2}v^{+}_{2_{n}}(% \overbrace{\hat{\boldsymbol{e}}_{t}\cdot\hat{\boldsymbol{e}}_{n}}^{0})$

$\hbox to36.135pt{or\hfil}-m_{2}v_{0}\sin\theta=m_{1}v^{+}_{1}\cos\theta+m_{2}v% ^{+}_{2_{t}}.$ (17.76)

Answer: $-m_{2}v_{0}\cos\theta=-m_{1}v^{+}_{1}\sin\theta+m_{2}v^{+}_{2_{n}},\;\;-m_{2}v% _{0}\sin\theta=m_{1}v^{+}_{1}\cos\theta+m_{2}v^{+}_{2_{t}}$
2.

Now, we rearrange eqn. (17.75) and 17.76 along with the third given equation, ${v^{+}}_{2_{t}}=-v_{0}\sin\theta$ , so that all unknowns are on the left hand side and the known quantities are on the right hand side of the equal sign. These equations, in matrix form, are as follows.

$\left[\begin{array}[]{ccc}-m_{1}\sin\theta&0&m_{2}\\ -m_{1}\cos\theta&m_{2}&0\\ 0&1&0\end{array}\right]\left\{\begin{array}[]{c}v^{+}_{1}\\ v^{+}_{2_{t}}\\ v^{+}_{2_{n}}\end{array}\right\}=\left\{\begin{array}[]{c}-m_{2}v_{0}\cos% \theta\\ -m_{2}v_{0}\sin\theta\\ -v_{0}\sin\theta\end{array}\right\}.$

This equation can be easily solved on a computer for the unknowns.

SAMPLE 17.28 Cueing a billiard ball.

Filename:sfig4-7-1

Figure 17.123:

A billiard ball is cued by striking it horizontally at a distance $d=10\,{\rm mm}$ above the center of the ball. The ball has mass $m=0.2\,\mathrm{kg}$ and radius $r=30\,{\rm mm}$ . Immediately after the strike, the center-of-mass of the ball moves with linear speed $v=1\,\mathrm{m}/\mathrm{s}$ . Find the angular speed of the ball immediately after the strike. Ignore friction between the ball and the table during the strike.

Solution

Filename:sfig4-7-1a

Figure 17.124: FBD of the ball during the strike. The nonimpulsive forces

m g, N,

and

f

can be ignored in comparison to the strike force

F

Let the force imparted during the strike be $F$ . Since the ball is cued by giving a blow with the cue, $F$ is an impulsive force. Impulsive forces, such as $F$ , are in general so large that all non-impulsive forces are negligible in comparison during the time such forces act. Therefore, we can ignore all other forces $(mg,N,f)$ acting on the ball from its free-body diagram during the strike.

Now, from the linear momentum balance of the ball we get

F\hat{\boldsymbol{\imath}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}% \quad\mbox{or}\quad(F\hat{\boldsymbol{\imath}})\textrm{d}t=\textrm{d}{% \accentset{\rightharpoonup}{\boldsymbol{L}}}\hbox to36.135pt{\hfil$\Rightarrow% $\hfil}\int(F\hat{\boldsymbol{\imath}})\textrm{d}t={\accentset{\rightharpoonup% }{\boldsymbol{L}}}_{2}-{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{1}

where $L_{2}-L_{1}=\Delta{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ is the net change in the linear momentum of the ball during the strike. Since the ball is at rest before the strike, ${\accentset{\rightharpoonup}{\boldsymbol{L}}}_{1}=m\underbrace{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{1}}_{0}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$ . Immediately after the strike, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=v\hat{\boldsymbol{\imath}}=1\,% \mathrm{m}/\mathrm{s}$ .

	$\displaystyle\mbox{Thus}\quad{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{2}% =m{\accentset{\rightharpoonup}{\boldsymbol{v}}}=0.2\,\mathrm{kg}\mathord{\cdot% }1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}=0.2\,{\mbox{N}}\mathord{% \cdot}\,\mathrm{s}\hat{\boldsymbol{\imath}}.$
	$\displaystyle\mbox{Hence}\quad\int(F\hat{\boldsymbol{\imath}})\textrm{d}t=0.2% \,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}\hat{\boldsymbol{\imath}}\quad\mbox{or}% \quad\int F\textrm{d}t=0.2\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}.$		(17.77)

To find the angular speed we apply the angular momentum balance. Let $\omega$ be the angular speed immediately after the strike and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=\omega\hat{\boldsymbol{k}}$ . Now,

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}\int\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_% {\mathrm{cm}}\,\textrm{d}t=\int\textrm{d}{\accentset{\rightharpoonup}{% \boldsymbol{H}}}_{\mathrm{cm}}=({\accentset{\rightharpoonup}{\boldsymbol{H}}}_% {\mathrm{cm}})_{2}-({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}% })_{1}.

$\displaystyle\mbox{Because}\quad{\accentset{\rightharpoonup}{\boldsymbol{H}}}_% {\mathrm{cm}}$	$\displaystyle=$	$\displaystyle I_{cm}^{zz}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}% \quad\mbox{ and just before the strike, }{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}},$
$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}})_{1}$	$\displaystyle\equiv$	$\displaystyle\mbox{ angular momentum just before the strike }={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}})_{2}$	$\displaystyle\equiv$	$\displaystyle\mbox{ angular momentum just after the strike }=I_{cm}^{zz}{% \omega\hat{\boldsymbol{k}}},$
$\displaystyle\int\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm% }}\,\textrm{d}t$	$\displaystyle=$	$\displaystyle I_{cm}^{zz}\omega\hat{\boldsymbol{k}}=\frac{2}{5}mr^{2}\omega% \hat{\boldsymbol{k}}\mbox{ (since for a sphere, $I_{cm}^{zz}=\frac{2}{5}mr^{2}% $).}$

	But		$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}=-% Fd\hat{\boldsymbol{k}},$
	therefore		$\displaystyle-\int(Fd)\textrm{d}t\hat{\boldsymbol{k}}=\frac{2}{5}mr^{2}\omega% \hat{\boldsymbol{k}}$
	or		$\displaystyle-\underbrace{d}_{{\rm constant}}\int F\textrm{d}t=\frac{2}{5}mr^{% 2}\omega\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\omega=-\frac{5d}{2mr^{2}}% \int F\textrm{d}t.$

Substituting the given values and $\int F\textrm{d}t=0.2\,{\mbox{N}}\mathord{\cdot}\,\mathrm{s}$ from equation 17.77 we get

\omega=-\frac{5(0.01\,\mathrm{m})}{2\mathord{\cdot}0.2\,\mathrm{kg}\mathord{% \cdot}(0.03\,\mathrm{m})^{2}}\mathord{\cdot}0.2\,{\mbox{N}}\mathord{\cdot}\,% \mathrm{s}=-27.78\,\mathrm{rad}/\mathrm{s}.

The negative value makes sense because the ball will spin clockwise after the strike, but we assumed that $\omega$ was anticlockwise.

Answer: $\omega=-27.78\,\mathrm{rad}/\mathrm{s}.$

SAMPLE 17.29

Filename:sfig4-7-1b

Figure 17.125:

Falling stick. A uniform bar of length $\ell$ and mass $m$ falls on the ground at an angle $\theta$ as shown in the figure. Just before impact at point C, the entire bar has the same velocity $v$ directed vertically downwards. Assume that the collision at C is plastic, i.e., end C of the bar gets stuck to the ground upon impact.

1.

Find the angular velocity of the bar just after impact.
2.

Assuming $\theta$ to be small, find the velocity of end B of the bar just after impact.

Solution We are given that the impact at point C is plastic. That is, end C of the bar has zero velocity after impact. Thus end C gets stuck to the ground. Then we expect the rod to rotate about point C as the rest of the bar moves (perhaps faster) to touch the ground. The free-body diagram of the bar is shown in fig. 17.126 during the impact at point C. Note that we can ignore the force of gravity in comparison to the large impulsive force $F_{c}$ due to impact at C.

Filename:sfig4-7-2

Figure 17.126: The free-body diagram of the bar during collision. The impulsive force at the point of impact C is so large that the force of gravity can be completely ignored in comparison.

Now, if we carry out angular momentum balance about point C, there will be no net moment acting on the bar, and therefore, angular momentum about the impact point C is conserved. Distinguishing the kinematic quantities before and after impact with superscripts ‘-’ and ‘+’, respectively, we get from the conservation of angular momentum about point C,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{C}$
	$\displaystyle I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}^{-}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}% \times m{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{-}_{G}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}^{+}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}% \times m{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{G}.$

Now, we know that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{-}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ since every point on the bar has the same vertical velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=-v\hat{\boldsymbol{\jmath}}$ , and that just after impact, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{G}={\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}^{+}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{G/C}}$ where we can take ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}=\omega(-\hat{% \boldsymbol{k}})$ . Thus,

^†^†margin: Since C is a fixed point for the motion of the bar after impact, we could calculate

{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{C}

as follows.

{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{C}=I^{C}_{zz}{\accentset{% \rightharpoonup}{\boldsymbol{\omega}}}=\underbrace{\frac{1}{3}m\ell^{2}}_{I^{C% }_{zz}}\omega(-\hat{\boldsymbol{k}}).

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{C}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}% \times m{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{-}_{G}=(\ell/2)\hat{% \boldsymbol{\lambda}}\times mv(-\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-\frac{mv\ell}{2}\cos\theta\hat{\boldsymbol{k}}\qquad(\mbox{since% }\hat{\boldsymbol{\lambda}}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta% \hat{\boldsymbol{\jmath}})$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{C}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}^{+}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}}% \times m({\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/C}})$
	$\displaystyle=$	$\displaystyle-I^{\mathrm{cm}}_{zz}\omega\hat{\boldsymbol{k}}+(\ell/2)\hat{% \boldsymbol{\lambda}}\times m\underbrace{(-\omega\hat{\boldsymbol{k}}\times% \ell/2\hat{\boldsymbol{\lambda}})}_{\omega\ell/2(-\hat{\boldsymbol{n}})}$
	$\displaystyle=$	$\displaystyle-\frac{1}{12}m\ell^{2}\omega\hat{\boldsymbol{k}}-\frac{1}{4}m\ell% ^{2}\omega\hat{\boldsymbol{k}}=-\frac{1}{3}m\ell^{2}\omega\hat{\boldsymbol{k}}.$

Now, equating ${\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{C}$ and ${\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{C}$ we get

\omega=\frac{3v}{2\ell}\cos\theta,\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\frac{3v}{2\ell}\cos\theta% \hat{\boldsymbol{k}}.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}=-\frac{3v}{2\ell}\cos\theta% \hat{\boldsymbol{k}}$

2.

The velocity of the end B is now easily found using ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{C}}+{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/C}}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/C}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/C}}={% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/C}}$ . Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B/C}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/C}}=-\omega\hat{% \boldsymbol{k}}\times\ell\hat{\boldsymbol{\lambda}}$

$\displaystyle=$ $\displaystyle-\omega\ell\hat{\boldsymbol{n}}=-\frac{3v}{2}\cos\theta(-\sin% \theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{\boldsymbol{\jmath}})$

but, for small $\theta$ , $\cos\theta\approx 1$ , and $\sin\theta\approx 0$ . Therefore, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/C}}=-% \frac{3v}{2}\hat{\boldsymbol{\jmath}}$ .

Thus, end B of the bar speeds up by one and a half times its original speed due to the plastic impact at C.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{B/C}}=-(3/2% )v\hat{\boldsymbol{\jmath}}$

SAMPLE 17.30

Filename:sfig4-7-2a

Figure 17.127:

Tipping box. A box of mass $m$ and dimensions $2a$ and $2b$ moves along a horizontal surface with uniform speed $v$ . Suddenly, it bumps into an obstacle at A. Assume that the impact is plastic and point A is at the lowest level of the box. What is the maximum $v$ the box can have so that it does not tip over after the impact.

Solution Whether the box can tip or not depends on whether it gets sufficient initial angular speed just after collision to overcome the restoring moment due to gravity about the point of rotation A. So, first we need to find the angular velocity of the box immediately following the collision. The free-body diagram of the box during collision is shown in fig. 17.128. There is an impulse ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ acting at the point of impact. If we carry out the angular momentum balance about point A, we see that the impulse at A produces no moment impulse about A, and therefore, the angular momentum about point A has to be conserved. That is, ${{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}}^{+}={{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}}^{-}$ .

Filename:sfig4-7-2b

Figure 17.128: The free-body diagram of the box during collision.

Now,

{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}}^{-}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}\times m{{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{G}}}^{-}=(-b\hat{% \boldsymbol{\imath}}+a\hat{\boldsymbol{\jmath}})\times mv\hat{\boldsymbol{% \imath}}=-mav\hat{\boldsymbol{k}}

Let the box have angular velocity ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}=\omega\hat{\boldsymbol{% k}}$ just after impact. Then,

$\displaystyle{{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}}^{+}$	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}{\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}^{+}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G/A}}% \times m{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{G}% }}^{+}=I^{\mathrm{cm}}_{zz}\omega\hat{\boldsymbol{k}}+r\hat{\boldsymbol{% \lambda}}\times m(\omega\hat{\boldsymbol{k}}\times r\hat{\boldsymbol{\lambda}})$
	$\displaystyle=$	$\displaystyle I^{\mathrm{cm}}_{zz}\omega\hat{\boldsymbol{k}}+mr^{2}\omega\hat{% \boldsymbol{k}}=\frac{1}{12}(4a^{2}+4b^{2})m\omega\hat{\boldsymbol{k}}+m(a^{2}% +b^{2})\omega\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle\frac{4}{3}(a^{2}+b^{2})m\omega\hat{\boldsymbol{k}}.$

Now equating the two momenta, we get

\omega^{+}=-\frac{3a}{4(a^{2}+b^{2})}\,v\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}=-\frac{3a}{4(a^{2% }+b^{2})}\,v\hat{\boldsymbol{k}}.

(17.78)

Thus we know the angular velocity immediately after impact. Now let us find out if it is enough to get over the hill, so to speak. Once the impact is over (in a few milliseconds), the usual forces show up on the free-body diagram (see fig. 17.129).

Filename:sfig4-7-2c

Figure 17.129: The free-body diagram of the box just after the collision is over.

In this post-collision part of the motion energy is conserved. Calling just after collision ‘+’ and the top position ‘t’ (where G is directly above A), using that $I^{A}_{zz}=\frac{4}{3}m(a^{2}+b^{2})$ and solving for the critical condition, that the block has just enough energy to get to the top position, we have,

	$\displaystyle{E_{\mathrm{P}}}^{+}+{E_{\mathrm{K}}}^{+}$	$\displaystyle={E_{\mathrm{K}}}^{t}+{E_{\mathrm{P}}}^{t}$
	$\displaystyle{E_{\mathrm{P}}}^{t}-{E_{\mathrm{P}}}^{+}$	$\displaystyle=-({E_{\mathrm{K}}}^{t}-{E_{\mathrm{K}}}^{+})$
	$\displaystyle mg(\sqrt{(b^{2}+a^{2})}-b)$	$\displaystyle=-\left(0-\frac{1}{2}(\omega^{+})^{2}I^{A}_{zz}\right)$
	$\displaystyle(\omega^{+})^{2}$	$\displaystyle=\frac{3}{2}g\frac{\left(1-\frac{b}{\sqrt{(b^{2}+a^{2})}}\right)}% {\sqrt{(b^{2}+a^{2})}}.\par$

This tells us how big $\omega^{+}$ has to be. Substituting this $\omega^{+}$ into eqn. (17.78) we find how big $v$ has to be. We find that the box does not tip over so long as $v$ is small enough, as given by

Answer: $v\leq\frac{2}{a}\sqrt{2g(b^{2}+a^{2})^{3/2}\left(1-\frac{b}{\sqrt{(b^{2}+a^{2}% )}}\right)/3}.$

SAMPLE 17.31

Filename:pfigure-s94q7p1

Figure 17.130:

Ball hits the bat. A uniform bar of mass $m_{2}=1\,\mathrm{kg}$ and length $2\ell=1\,\mathrm{m}$ hangs vertically from a hinge at A. A ball of mass $m_{1}=0.25\,\mathrm{kg}$ comes and hits the bar horizontally at point D with speed $v=5\,\mathrm{m}/\mathrm{s}$ . The point of impact D is located at $d=0.75\,\mathrm{m}$ from the hinge point A. Assume that the collision between the ball and the bar is plastic.

1.

Find the velocity of point D on the bar immediately after impact.
2.

Find the impulse on the bar at D due to the impact.
3.

Find and plot the impulsive reaction at the hinge point A as a function of $d$ , the distance of the point of impact from the hinge point. What is the value of $d$ which makes the impulse at A zero?

Solution The free-body diagram of the ball and the bar as a single system is shown in fig. 17.131 during impact. There is only one external impulsive force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}}$ acting at the hinge point A. We take the ball and the bar together here so that the impulsive force acting between the ball and the bar becomes internal to the system and we are left with only one external force at A. Then, the angular momentum balance about point A yields $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ since there is no net moment about A. Thus the angular momentum about A is conserved during the impact.

Let us distinguish the kinematic quantities just before impact and immediately after impact with superscripts ‘-’ and ‘+’, respectively. Then, from the conservation of angular momentum about point A, we get ${\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{A}={\accentset{% \rightharpoonup}{\boldsymbol{H}}}^{+}_{A}$ . Now,

Filename:pfigure-s94h7p2

Figure 17.131: The free-body diagram of the ball and the bar together during collision. The impulsive force at the point of impact is internal to the system and hence, does not show on the free-body diagram.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{A}$	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{A})_{\rm ball% }+({\accentset{\rightharpoonup}{\boldsymbol{H}}}^{-}_{A})_{\rm bar}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/A}}% \times m_{1}{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{-}+I^{A}_{zz}{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{-}$
	$\displaystyle=$	$\displaystyle d\hat{\boldsymbol{\jmath}}\times m_{1}v(-\hat{\boldsymbol{\imath% }})+{\accentset{\rightharpoonup}{\boldsymbol{0}}}=m_{1}dv\hat{\boldsymbol{k}}.$
Similarly,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/A}}% \times m_{1}{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}+I^{A}_{zz}{% \accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}$
but, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}={\accentset{\rightharpoonup}% {\boldsymbol{\omega}}}^{+}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{D/A}}=-\omega d\hat{\boldsymbol{\imath}}$ , where ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}=\omega\hat{\boldsymbol{% k}}$ (let). Hence,
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}^{+}_{A}$	$\displaystyle=$	$\displaystyle d\hat{\boldsymbol{\jmath}}\times m_{1}(-\omega d\hat{\boldsymbol% {\imath}})+\frac{1}{3}m_{2}(2\ell)^{2}\omega\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(m_{1}d^{2}+\frac{4}{3}m_{2}\ell^{2})\omega\hat{\boldsymbol{k}}.$

Equating the two momenta, we get

$\displaystyle\omega$	$\displaystyle=$	$\displaystyle\frac{m_{1}dv}{m_{1}d^{2}+(4/3)m_{2}\ell^{2}}$
	$\displaystyle=$	$\displaystyle\frac{v}{d\left(1+\frac{4}{3}\frac{m_{2}}{m_{1}}\left(\frac{\ell}% {d}\right)^{2}\right)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{D}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}^{+}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/A}}=\omega d(-\hat{% \boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle-\frac{v}{1+\frac{4}{3}\frac{m_{2}}{m_{1}}\left(\frac{\ell}{d}% \right)^{2}}\hat{\boldsymbol{\imath}}.$

Now, substituting the given numerical values, $v=5\,\mathrm{m}/\mathrm{s}$ , $m_{1}=0.25\,\mathrm{kg}$ , $m_{2}=1\,\mathrm{kg}$ , $\ell=0.5\,\mathrm{m}$ , and $d=0.75\,\mathrm{m}$ , we get ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{D}}=-2.08\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu\mathrm{D}}=-2.08\,% \mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}}$

2.

To find the impulse at D due to the impact, we can consider either the ball or the bar separately, and find the impulse by evaluating the change in the linear momentum of the body. Let us consider the ball since it has only one impulse acting on it. The free-body diagram of the ball during impact is shown in fig. 17.132. From the linear impulse-momentum relationship we get,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{D}=\int{\accentset% {\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{D}}\,\textrm{d}t$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{L}}}^{+}-{\accentset{% \rightharpoonup}{\boldsymbol{L}}}^{-}=m_{1}({\accentset{\rightharpoonup}{% \boldsymbol{v}}}^{+}-{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{-})$

$\displaystyle=$ $\displaystyle m_{1}\left(-\frac{v}{1+\frac{4}{3}\frac{m_{2}}{m_{1}}\left(\frac% {\ell}{d}\right)^{2}}\hat{\boldsymbol{\imath}}+v\hat{\boldsymbol{\imath}}\right)$

$\displaystyle=$ $\displaystyle m_{1}v\left(1-\frac{1}{1+\frac{4}{3}\frac{m_{2}}{m_{1}}\left(% \frac{\ell}{d}\right)^{2}}\right)\hat{\boldsymbol{\imath}}.$

Filename:ch4-7-c

Figure 17.132: Separate free-body diagrams of the ball and the bar during collision.

Substituting the given numerical values, we get ${\accentset{\rightharpoonup}{\boldsymbol{P}}}_{D}=0.73\,{\mathrm{kg}}\mathord{% \cdot}{\mathrm{m}}/{\mathrm{s}}\hat{\boldsymbol{\imath}}$ . The impulse on the bar is equal and opposite. Therefore, the impulse on the bar is $-{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{D}=-0.73\,{\mathrm{kg}}% \mathord{\cdot}{\mathrm{m}}/{\mathrm{s}}\hat{\boldsymbol{\imath}}$ .

Answer: $\mbox{Impulse at D}=-0.73\,{\mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{\mathrm{s% }}\hat{\boldsymbol{\imath}}$

Now that we know the impulse at D, we can easily find the impulse at A by applying impulse-momentum relationship to the bar. Since, the bar is stationary just before impact, its initial momentum is zero. Thus, for the bar,

\int({\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}}-{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{D}})\textrm{d}t={% \accentset{\rightharpoonup}{\boldsymbol{L}}}^{+}-{\accentset{\rightharpoonup}{% \boldsymbol{L}}}^{-}={\accentset{\rightharpoonup}{\boldsymbol{L}}}^{+}=m_{2}{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}^{+}.

Denoting the impulse at A with ${\accentset{\rightharpoonup}{\boldsymbol{P}}}_{A}$ , the mass ratio $m_{2}/m_{1}$ by $m_{r}$ , and the length ratio $\ell/d$ by $q$ , and noting that ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}^{+}=\omega\hat{% \boldsymbol{k}}\times\ell\hat{\boldsymbol{\jmath}}=-\omega\ell\hat{\boldsymbol% {\imath}}$ , we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{A}$	$\displaystyle\equiv$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}% }\,\textrm{d}t=\int{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{% D}}\,\textrm{d}t+m_{2}(-\omega\ell\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle m_{1}v\left(1-\frac{1}{1+\frac{4}{3}m_{r}q^{2}}\right)\hat{% \boldsymbol{\imath}}-m_{2}\ell\frac{v}{d\left(1+\frac{4}{3}m_{r}q^{2}\right)}% \hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle m_{1}v\left(\frac{\frac{4}{3}m_{r}q^{2}}{1+\frac{4}{3}m_{r}q^{2}% }\right)\hat{\boldsymbol{\imath}}-m_{2}v\left(\frac{q}{1+\frac{4}{3}m_{r}q^{2}% }\right)\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\frac{(4/3)m_{2}q^{2}-m_{2}q}{1+\frac{4}{3}m_{r}q^{2}}\,v\hat{% \boldsymbol{\imath}}=\frac{q(4q-3)}{3\left(1+\frac{4}{3}m_{r}q^{2}\right)}\,m_% {2}v\hat{\boldsymbol{\imath}}.$

Now, we are ready to graph the impulse at A as a function of $q\equiv\ell/d$ . However, note that a better quantity to graph will be $P_{A}/(m_{1}v)$ , that is, the nondimensional impulse at A, normalized with respect to the initial linear momentum $m_{1}v$ of the ball. The plot is shown in fig. 17.133. It is clear from the plot, as well as from the expression for ${\accentset{\rightharpoonup}{\boldsymbol{P}}}_{A}$ , that the impulse at A is zero when $q=3/4$ or $d=4\ell/3=2/3(2\ell)$ , that is, when the ball strikes at two thirds the length of the bar. Note that this location of the impact point is independent of the mass ratio $m_{r}$ .

Filename:pfigure-blue-87d-1

Figure 17.133: Plot of normalized impulse at A as a function of

q=\ell/d

Answer: $d=2/3(2\ell)\mbox{ for }{\accentset{\rightharpoonup}{\boldsymbol{P}}}_{A}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}$

Comment: This particular point of impact D (when $d=2/3(2\ell)$ ) which induces no impulse at the support point A is called the center of percussion. If you imagine the bar to be a bat or a racquet and point A to be the location of your grip, then hitting a ball at D gives you an impulse-free shot. In sports, point D is called a sweet spot.

SAMPLE 17.32

Filename:Mikef91p3

Figure 17.134:

Flying dish collides with solar panel. A uniform rectangular plate of dimensions $2a=2\,\mathrm{m}$ and $2b=1\,\mathrm{m}$ and mass $m_{P}=2\,\mathrm{kg}$ drifts in space at a uniform speed $v_{p}=10\,\mathrm{m}/\mathrm{s}$ (relative to some local Newtonian reference frame) in the direction shown. A circular disk of radius $R=0.25\,\mathrm{m}$ and mass $m_{D}=1\,\mathrm{kg}$ is heading towards the plate at a linear speed $v_{0}=v_{D}=1\,\mathrm{m}/\mathrm{s}$ directed normal to the facing edge of the plate. In addition, the disk is spinning at $\omega_{D}=5\,\mathrm{rad}/\mathrm{s}$ in the clockwise direction. The plate and the disk collide at point A of the plate, located at $d=0.8\,\mathrm{m}$ from the center of the long edge. Assume that the collision is frictionless and purely elastic. Find the linear and angular velocities of the plate and the disk immediately after the collision.

Solution We are given the linear and angular velocities of the two bodies before their collision. We have to find their linear and angular velocities after the collision. Let the linear velocities of the plate and the disk after the collision be ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{+}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}^{+}$ , respectively. We assume that the collision takes place in the plane of the two bodies (that is, all motions before and after collision are planar). Then the unknown angular velocities of the plate and the disk after the collision can be assumed as ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{P}^{+}=\omega_{P}^{+}\hat{% \boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}^{+}=\omega_{D}^{+}\hat{% \boldsymbol{k}}$ . Thus, in total, we have 6 scalar unknowns here — 4 for linear velocities of the disk and the plate (each velocity has two components) and 2 for the two angular velocities.

Since we are trying to find velocities after a collision, given the velocities before the collision, we can use linear and angular momentum-impulse relations. Let ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ be the impulse acting during the collision. Note that the collision is frictionless and hence ${\accentset{\rightharpoonup}{\boldsymbol{P}}}$ must act normal to the two colliding surfaces. Therefore, the direction of the impulse is known, only its magnitude is not known. Thus, in total, we have 7 unknowns now. To solve for them, we need 7 independent equations.

We have 6 independent equations from the linear and angular impulse-momentum balance for the two bodies (3 each). We need one more equation. That equation is the relationship between the normal components of the relative velocities of approach and departure with the coefficient of restitution $e$ (=1 for elastic collision). Thus we have 7 equations for 7 unknowns. First we set them up, then we solve them with a calculator or computer.

Filename:pfigure-s94h7p3

Figure 17.135: Separate free-body diagrams of the disk and the plate during collision. The impulse

{\accentset{\rightharpoonup}{\boldsymbol{P}}}

acts normal to the colliding surfaces and hence we can assume that on the plate,

{\accentset{\rightharpoonup}{\boldsymbol{P}}}=P\hat{\boldsymbol{\imath}}

. Therefore, the impulse on the disk is

-P\hat{\boldsymbol{\imath}}

Filename:p-f96-p2-3

Figure 17.136: The free-body diagram of the disk and the plate together during collision. The impulsive force at the point of impact is internal to the system and hence, does not show on the free-body diagram .

The free-body diagrams of the disk and the plate together and the two separately are shown in fig. 17.136 and 17.135, respectively. Using an $x y$ coordinate system oriented as shown in fig. 17.136, we can write

\begin{array}[]{lrcl}\mbox{LMB for disk:}&m_{D}({\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{D}^{+}-{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}^{-}% )&=&-P\hat{\boldsymbol{\imath}}\\ \mbox{LMB for plate:}&m_{P}({\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^% {+}-{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{-})&=&P\hat{\boldsymbol% {\imath}}\\ \mbox{AMB for disk:}&I^{\rm cm}_{D}({\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{D}^{+}-{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}^{-})&% =&{\accentset{\rightharpoonup}{\boldsymbol{0}}}\\ \mbox{AMB for plate:}&I^{\rm cm}_{P}({\accentset{\rightharpoonup}{\boldsymbol{% \omega}}}_{P}^{+}-{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{P}^{-})&% =&{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/P}}\times P\hat{% \boldsymbol{\imath}}\\ \mbox{kinematics:}&{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{D}}^{+}-{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{P}}^{+}&=&e({\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{A_{P}}^{-}-{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{A_{D}}^{-})\end{array}

where, in the last equation ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{D}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{P}}$ refer to the velocities of the material points located at A on the disk and on the plate, respectively. Other linear velocities in the equations above refer to the velocities at the center-of-mass of the corresponding bodies. We are given that ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}^{-}=v_{D}\hat{\boldsymbol{% \imath}},\,{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{-}=-v_{P}\hat{% \boldsymbol{\imath}},\,{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}^% {-}=-\omega_{D}\hat{\boldsymbol{k}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{P}^{-}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ . Let us assume that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}^{+}=\omega_{D}^{+}\hat{% \boldsymbol{k}},\,{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{P}^{+}=% \omega_{P}^{+}\hat{\boldsymbol{k}},\,{\accentset{\rightharpoonup}{\boldsymbol{% v}}}_{D}^{+}=v_{D_{x}}^{+}\hat{\boldsymbol{\imath}}+v_{D_{y}}^{+}\hat{% \boldsymbol{\jmath}}$ , and similarly, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{+}=v_{P_{x}}^{+}\hat{% \boldsymbol{\imath}}+v_{P_{y}}^{+}\hat{\boldsymbol{\jmath}}$ . Then,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{D}}^{-}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}^{-}+{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{D}^{-}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/O}}=v_{D}\hat{\boldsymbol{% \imath}}-\omega_{D}R\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{D}}^{+}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}^{+}+{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{D}^{+}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/O}}=v_{D_{x}}^{+}\hat{% \boldsymbol{\imath}}+(v_{D_{y}}^{+}+\omega_{D}^{+}R)\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{P}}^{-}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{-}=-v_{P}\hat{% \boldsymbol{\imath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A_{P}}^{+}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{+}+{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{P}^{+}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/P}}=(v_{P_{x}}^{+}-\omega_{P}^{% +}d)\hat{\boldsymbol{\imath}}+(v_{P_{y}}^{+}-\omega_{P}^{+}b)\hat{\boldsymbol{% \jmath}}.$

Substituting these quantities in the kinematics equation above and dotting with $\hat{\boldsymbol{\imath}}$ , the unit normal at A, we get

v_{D_{x}}^{+}-v_{P_{x}}^{+}+\omega_{P}^{+}d=\underbrace{e}_{1}(-v_{P}-v_{D})=-% v_{P}-v_{D}.

(17.79)

Now, let us extract the scalar equations from the impulse-momentum equations for the disk and the plate by dotting with appropriate unit vectors. Dotting LMB for the disk with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , respectively, we get

$\displaystyle m_{D}(v_{D_{x}}^{+}-v_{D})$	$\displaystyle=$	$\displaystyle-P$	(17.80)
$\displaystyle m_{D}v_{D_{y}}^{+}$	$\displaystyle=$	$\displaystyle 0.$	(17.81)
Dotting LMB for the plate with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , respectively, we get
$\displaystyle m_{P}(v_{P_{x}}^{+}+v_{P})$	$\displaystyle=$	$\displaystyle P$	(17.82)
$\displaystyle m_{P}v_{P_{y}}^{+}$	$\displaystyle=$	$\displaystyle 0.$	(17.83)
Dotting AMB for the disk and the plate with $\hat{\boldsymbol{k}}$ , we get
$\displaystyle I^{\rm cm}_{D}(\omega_{D}^{+}+\omega_{D})$	$\displaystyle=$	$\displaystyle 0$	(17.84)
$\displaystyle I^{\rm cm}_{P}\omega_{P}^{+}$	$\displaystyle=$	$\displaystyle-Pd.$	(17.85)

We have all the equations we need. Let us rearrange these equations in a matrix form, taking the known quantities to the right and putting all unknowns to the left side. We then, write eqns. (17.80)–(17.85), and then eqn. (17.79) as

\left[\begin{array}[]{ccccccr}m_{D}&0&0&0&0&0&1\\ 0&m_{D}&0&0&0&0&0\\ 0&0&m_{P}&0&0&0&-1\\ 0&0&0&m_{P}&0&0&0\\ 0&0&0&0&I^{\rm cm}_{D}&0&0\\ 0&0&0&0&0&I^{\rm cm}_{P}&d\\ 1&0&-1&0&0&d&0\end{array}\right]\left\{\begin{array}[]{c}v_{D_{x}}^{+}\\ v_{D_{y}}^{+}\\ v_{P_{x}}^{+}\\ v_{P_{y}}^{+}\\ \omega_{D}^{+}\\ \omega_{P}^{+}\\ P\end{array}\right\}=\left\{\begin{array}[]{c}m_{D}v_{D}\\ 0\\ -m_{P}v_{P}\\ 0\\ -I^{\rm cm}_{D}\omega_{D}\\ 0\\ -v_{P}-v_{D}\end{array}\right\}.

Substituting the given numerical values for the masses and the pre-collision velocities, and the moments of inertia, $I^{\rm cm}_{D}=(1/2)m_{D}R^{2}$ and $I^{\rm cm}_{P}=(1/12)m_{P}(4a^{2}+4b^{2})$ , and then solving the matrix equation on a computer, we get,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{D}^{+}$	$\displaystyle=-8.70\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}},\quad{% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}^{+}=-5.15\,\mathrm{m}/\mathrm% {s}\hat{\boldsymbol{\imath}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{D}^{+}$	$\displaystyle=-5\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}},\quad{\accentset% {\rightharpoonup}{\boldsymbol{\omega}}}_{P}^{+}=-9.31\,\mathrm{rad}/\mathrm{s}% \hat{\boldsymbol{k}}$
	$\displaystyle P$	$\displaystyle=9.70\,{\mathrm{kg}}\mathord{\cdot}{\mathrm{m}}/{\mathrm{s}}.$

You can easily check that the results obtained satisfy the conservation of linear momentum for the plate and the disk taken together as one system.

Comments: In this particular problem, the equations are simple enough to be solved by hand. For example, eqns. (17.81), (17.83), and (17.84) are trivial to solve and immediately give, $v_{D_{y}}^{+}=0,v_{P_{y}}^{+}=0$ , and $\omega_{D}^{+}=\omega^{D}=-5\,\mathrm{rad}/\mathrm{s}$ . The rest of the equations can be solved by usual eliminations and substitutions, etc. The key is that you have 7 linearly independent equations for the 7 unknowns.

Problems for 17.5 Collisions

17.5.1 The two blocks shown in the figure are identical except that one rests on two springs while the other one sits on two massless wheels. Draw free-body diagrams of each mass as each is struck by a hammer. Here we are interested in the free-body diagrams only during collision. Therefore, ignore all forces that are much smaller than the impulsive forces. State in words why the forces you choose to show should not be ignored during the collision.

Filename:s97p3-2

Figure 17.137

17.5.2 These problems concern two colliding masses. In (a) the smaller mass hits the hanging mass from above at an angle $45^{\circ}$ with the vertical. In (b) the second case the smaller mass hits the hanging mass from below at the same angle. Assuming perfectly elastic impact between the balls, find the velocity of the hanging mass just after the collision. [Note, these problems are not well posed and can only be solved if you make additional modeling assumptions.]

Filename:pfigure-dynbalancers

Figure 17.138

17.5.3 A narrow pole is in the middle of a pond with a $10\,\mathrm{m}$ rope tied to it. A frictionless ice skater of mass $50\,\mathrm{kg}$ and speed $3\,\mathrm{m}/\mathrm{s}$ grabs the rope. The rope slowly wraps around the pole. What is the speed of the skater when the rope is 5 $\,\mathrm{m}$ long? (A tricky question.)

Filename:pfigure-s94h7p5

Figure 17.139

17.5.4 The masses $m$ and 3 $m$ are joined by a light-weight bar of length 4 $\ell$ . If point A in the center of the bar strikes fixed point B vertically with velocity $V_{0}$ , and is not permitted to rebound, find $\dot{\theta}$ of the system immediately after impact.

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Figure 17.140

Neglect gravity!

17.5.5 Two equal masses each of mass $m$ are joined by a massless rigid rod of length $\ell$ . The assembly strikes the edge of a table as shown in the figure, when the center of mass is moving downward with a linear velocity $v$ and the system is rotating with angular velocity $\dot{\phi}$ in the counter-clockwise sense. The impact is ’elastic’. Find the immediate subsequent motion of the system, assuming that no energy is lost during the impact and that there is no gravity. Show that there is an interchange of translational and rotational kinetic energy.

Filename:pfigure-blue-137-2

Figure 17.141

17.5.6 In the absence of gravity, a thin rod of mass $m$ and length $\ell$ is initially tumbling with constant angular speed $\omega_{o}$ , in the counter-clockwise direction, while its mass center has constant speed $v_{o}$ , directed as shown below. The end A then makes a perfectly plastic collision with a rigid peg O (via a hook). The velocity of the mass center happens to be perpendicular to the rod just before impact.

(a)

What is the angular speed $\omega_{f}$ immediately after impact?
(b)

What is the angular speed 10 seconds after impact? Why?
(c)

What is the loss in energy in the plastic collision?

Filename:pfigure-blue-78-2

Figure 17.142

17.5.7 A gymnast of mass $m$ and extended height $L$ is performing on the uneven parallel bars. About the $x$ , $y$ , $z$ axes which pass through her center of mass, her radii of gyration are $k_{x}$ , $k_{y}$ , and $k_{z}$ , respectively. Just before she grasps the top bar, her fully extended body is horizontal and rotating with angular rate $\omega$ ; her center of mass is then stationary. Neglect any friction between the bar and her hands and assume that she remains rigid throughout the entire stunt.

(a)

What is the gymnast’s rotation rate just after she grasps the bar? State clearly any approximations/assumptions that you make.
(b)

Calculate the linear speed with which her hips (CM) strike the lower bar. State all assumptions/approximations.
(c)

Describe (in words, no equations please) her motion immediately after her hips strike the lower bar if she releases her hands just prior to this impact.

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Figure 17.143

17.5.8 An acrobat modeled as a rigid body with uniform rigid mass $m$ of length $l$ . She falls without rotation in the position shown from height $h$ where she was stationary. She then grabs a bar with a firm but slippery grip. What is $h$ so that after the subsequent motion the acrobat ends up in a stationary handstand? [ Hint: What quantities are preserved in what parts of the motion?] Answer: $h=2L/3$

Filename:pfigure-s94h10p4

Figure 17.144

17.5.9 A crude see-saw is built with two supports separated by distance $d$ about which a rigid plank (mass $m$ , length $L$ ) can pivot smoothly. The plank is placed symmetrically, so that its center of mass is midway between the supports when the plank is at rest.

(a)

While the left end is resting on the left support, the right end of the plank is lifted to an angle $\theta$ and released. At what angular velocity $\omega_{1}$ will the plank strike the right hand support?
(b)

Following the impact, the left end of the plank can pivot purely about the right end if $d/L$ is properly chosen and the right end does not bounce. Find $\omega_{2}$ under these circumstances.

Filename:pfigure-blue-81-2

Figure 17.145

17.5.10 Baseball bat. In order to convey the ideas without making the calculation too complicated, some of the simplifying assumptions here are highly approximate. Assume that a bat is a uniform rigid stick with length $L$ and mass $m_{s}$ . The motion of the bat is a pivoting about one end held firmly in place with hands that rotate but do not move. The swinging of the bat occurs by the application of a constant torque $M_{s}$ at the hands over an angle of $\theta=\pi/2$ until the point of impact with the ball. The ball has mass $m_{b}$ and arrives perpendicular to the bat at an absolute speed $v_{b}$ at a point a distance $\ell$ from the hands. The collision between the bat and the ball is completely elastic.

(a)
To maximize the speed $v_{hit}$ of the hit ball How heavy should a baseball bat be? Where should the ball hit the bat? Here are some hints for one way to approach the problem.
- •
  
  Find the angular velocity of the bat just before collision by drawing a FBD of the bat etc.
- •
  
  Find the total energy of the ball and bat system just before the collision.
- •
  
  Draw a FBD of the ball and of the bat during the collision (with this model there is an impulse at the hands on the bat). Call the magnitude of the impulse of the ball on the bat (and vice versa) $\int Fdt$ .
- •
  
  Use various momentum equations to find the angular velocity of the bat and velocity of the ball just after the collision in terms of $\int Fdt$ and other quantities above. Use these to find the energy of the system just after collision.
- •
  
  Solve for the value of $\int Fdt$ that conserves energy. As a check you should see if this also predicts that the relative separation speed of the ball and bat (at the impact point) is the same as the relative approach speed (it should be).
- •
  
  You now can calculate $v_{hit}$ in terms of $m_{b},m_{s},M_{s},L,\ell$ , and $v_{b}$ .
- •
  
  Find the maximum of the above expression by varying $m_{s}$ and $\ell$ . Pick numbers for the fixed quantities if you like.
(b)

Can you explain in words what is wrong with a bat that is too light or too heavy?
(c)

Which aspects of the model above do you think lead to the biggest errors in predicting what a real ball player should pick for a bat and place on the bat to hit the ball?
(d)

Describe as clearly as possible a different model of a baseball swing that you think would give a more accurate prediction. (You need not do the calculation).

	$\displaystyle\left[{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{C}}\right]_{xy}$	$\displaystyle=$	$\displaystyle\left[\begin{array}[]{c}0.3\,\mathrm{m}/\mathrm{s}\\ 0.2\,\mathrm{m}/\mathrm{s}\end{array}\right]+\left[\begin{array}[]{rr}0&-2\,% \mathrm{rad}/\mathrm{s}\\ 2\,\mathrm{rad}/\mathrm{s}&0\end{array}\right]\left[\begin{array}[]{rr}\frac{1% }{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2}\end{array}\right]\left[\begin{array}[]{c}0.2\,% \mathrm{m}\\ 0\end{array}\right]$
		$\displaystyle=$	$\displaystyle\left[\begin{array}[]{c}0.3\,\mathrm{m}/\mathrm{s}\\ 0.2\,\mathrm{m}/\mathrm{s}\end{array}\right]+\left[\begin{array}[]{c}-0.35\,% \mathrm{m}/\mathrm{s}\\ 0.2\,\mathrm{m}/\mathrm{s}\end{array}\right]=\left[\begin{array}[]{c}-0.05\,% \mathrm{m}/\mathrm{s}\\ 0.4\,\mathrm{m}/\mathrm{s}\end{array}\right].$

	$\displaystyle x(t)$	$\displaystyle=$	$\displaystyle x_{0}+\dot{x}_{0}t+\frac{1}{2}\frac{F}{m}t^{2}$
	$\displaystyle y(t)$	$\displaystyle=$	$\displaystyle y_{0}+\dot{y}_{0}t.$

	$\displaystyle\omega$	$\displaystyle=$	$\displaystyle\frac{\dot{x}}{R}=\frac{0.5\,\mathrm{m}/\mathrm{s}}{0.25\,\mathrm% {m}}=2\,\mathrm{rad}/\mathrm{s},$
	$\displaystyle\dot{\omega}$	$\displaystyle=$	$\displaystyle\frac{\ddot{x}}{R}=\frac{0.1\,\mathrm{m}/\mathrm{s}^{2}}{0.25\,% \mathrm{m}}=0.4\,\mathrm{rad}/\mathrm{s}^{2}.$

$\displaystyle\theta(t)$	$\displaystyle=$	$\displaystyle\omega t+\frac{1}{2}\dot{\omega}t^{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta(t=2\,\mathrm{s})$	$\displaystyle=$	$\displaystyle(2\,\mathrm{rad}/\mathrm{s})\cdot(2\,\mathrm{s})+\frac{1}{2}(0.4% \,\mathrm{rad}/\mathrm{s})\cdot(4\,\mathrm{s}^{2})$
	$\displaystyle=$	$\displaystyle 4.8\,{\rm rad}=\frac{4.8}{2\pi}\,\mbox{rev}=0.76\,\mbox{rev}.$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}$	$\displaystyle=$	$\displaystyle-3r\left[\sin\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{% \imath}}+\cos\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{\jmath}}\right],$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C}+r\left[-sin% \left(\omega t\right)\hat{\boldsymbol{\imath}}+\cos\left(\omega t\right)\hat{% \boldsymbol{\jmath}}\right].$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{C}$	$\displaystyle=$	$\displaystyle-\omega r\left[\cos\left(\frac{\omega t}{3}\right)\hat{% \boldsymbol{\imath}}-\sin\left(\frac{\omega t}{3}\right)\hat{\boldsymbol{% \jmath}}\right],$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}$	$\displaystyle=$	$\displaystyle-\omega r\left[\left\{\cos\left(\frac{\omega t}{3}\right)+\cos% \left(\omega t\right)\right\}\hat{\boldsymbol{\imath}}-\left\{\sin\left(\frac{% \omega t}{3}\right)-\sin\left(\omega t\right)\right\}\hat{\boldsymbol{\jmath}}% \right].$

$\displaystyle P$	$\displaystyle=$	$\displaystyle\frac{{E_{\mathrm{K}}}_{1}-{E_{\mathrm{K}}}_{2}}{\Delta t}$
	$\displaystyle=$	$\displaystyle\frac{1700\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}-0}{5\,\mathrm{s% }}=340\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}/\,\mathrm{s}$
	$\displaystyle=$	$\displaystyle 340\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}/\,\mathrm{s}\cdot% \frac{1\mbox{ hp}}{550\,\mathrm{ft}\mathord{\cdot}\,{\rm lbf}/\,\mathrm{s}}$
	$\displaystyle=$	$\displaystyle 0.62\mbox{ hp}$

$\displaystyle\int_{0}^{\omega}\omega\,d\omega$	$\displaystyle=$	$\displaystyle-\int_{\theta_{0}}^{\theta}\frac{3g}{2\ell}\cos\theta\,d\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\frac{1}{2}\omega^{2}$	$\displaystyle=$	$\displaystyle-\frac{3g}{2\ell}(\sin\theta-\sin\theta_{0})$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\omega$	$\displaystyle=$	$\displaystyle\pm\sqrt{\frac{3g}{\ell}(\sin\theta_{0}-\sin\theta)}.$

	$\displaystyle m_{1}\underbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% 1}}_{\mbox{\bf 0}}+\,\mathrm{m}_{2}{\accentset{\rightharpoonup}{\boldsymbol{v}% }}_{2}$	$\displaystyle=$	$\displaystyle m_{1}{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{1}+\,% \mathrm{m}_{2}{\accentset{\rightharpoonup}{\boldsymbol{v}}}^{+}_{2}$
	$\displaystyle\hbox to36.135pt{or\hfil}-m_{2}v_{0}\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle m_{1}v^{+}_{1}\hat{\boldsymbol{\imath}}+m_{2}(v^{+}_{2_{t}}\hat{% \boldsymbol{e}}_{t}+v^{+}_{2_{n}}\hat{\boldsymbol{e}}_{n}).$		(17.74)

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{A}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mkern 0.5mu% \mathrm{B}}+{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}$
	$\displaystyle=$	$\displaystyle(0.3\hat{\boldsymbol{\imath}}+0.6\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}+2\,\mathrm{rad}/\mathrm{s}\hat{\boldsymbol{k}}\times(-0.% 2\hat{\boldsymbol{\imath}})\,\mathrm{m}$
	$\displaystyle=$	$\displaystyle(0.3\hat{\boldsymbol{\imath}}+0.2\hat{\boldsymbol{\jmath}})\,% \mathrm{m}/\mathrm{s}.$