IIL, II

INTRODUCTIONSo far we've studied only translational motion: objects sliding, falling, or rising, but in none of ourexamples have we considered spinning objects. We will now look at rotation, which will completeour study of motion. All motion is some combination of translation and rotation, which are illus­trated in the figures below. Consider any two points in the object under study (on the left) andimagine connecting them by a straight line. If this line always remains parallel to itself while theobject moves, then the object is translating only. However, if this line does not always remain parallelto itself while the object moves, then the object is rotating.

puretranslation

purerotation translation

)0-

lSJ and B>(ati~

91

translationand

rotation

puretranslation

[C] ROTATIONAL KINEMATICSMark several dots along a radius on a disk, and call this radius the reference line. If the disk rotatesabout its center, we can use the movement of these dots to talk about angular displacement, angularvelocity, and angular acceleration.

I

+~'/: 'LlSI ,

tx~~ ...---------lI>

r

referenceline

If the disk rotates as a rigid body, then all three dots shown have the same angular displacement,~ e. In fact, this is the definition of a rigid body: All points along a radial line always have the sameangular displacement.

Just as the time rate-of-change of displacement gives velocity, the time rate-of-change of angulardisplacement gives angular velocity, denoted by OJ (omega). The definition of the average angularvelocity is:

_ ~e

OJ=-M

Notice that if we let the time interval Lit approach 0, then the equation above leads to the definitionof the instantaneous angular velocity: .

deOJ=-

dt

92 II CRACKING THE AP PHYSICS EXAM

And, finally, just as the time rate-of-change of velocity gives acceleration, the time rate-of-changeof angular velocity gives angular acceleration, or a (alpha). The definition of the average angularacceleration is:

_ 11ma=-

MIf we let the time interval !1 t approach 0, then the equation above leads to the definition of theinstantaneous angular velocity:

dma=-

dtOn the rotating disk illustrated on the previous page, we said that all points undergo the same

angular displacement in any given time interval; this means that all points on the disk have the sameangular velocity, (f), but not all points have the same linear velocity, v. This follows from the definitionof radian measure. Expressed in radians, the angular displacement, 11 e, is related to the arc length,!1s, by the equation

118 =!1sr

Rearranging this equation and dividing by M, we find that

!1s 118!1s=rI18 => -=r- => v=r(O

I1t M

Or, using the equations v =dsldt and m= de Idt,

v=rm

Therefore, the greater the value of r, the greater the value of v. Points on the rotating body fartherfrom the rotation axis move more quickly than those closer to the rotation axis.

From the equation v =rm, we can derive the relationship that connects angular acceleration andlinear acceleration. Differentiating both sides with respect to t (holding r constant), gives us

dv dm-=r- => a=radt dt

(It's important to realize that the acceleration a in this equation is not centripetal acceleration; it'stangential acceleration, which arises from a change in speed caused by an angular acceleration. Bycontrast, centripetal acceleration does not produce a change in speed.) Often, tangential accelerationis written as at to distinguish it from centripetal acceleration (a).

Solution. One complete revolution is equal to an angular displacement of 21t radians, so the body'saverage angular velocity is

_ 118 21t radm =- = 1t rad I s

I1t 2 s

ROTATIONAL MOTION III 93

Solution. By definition,

a= !Y.w = (5- 2) rad I s = 6 rad I S2

!Y.t 0.5 s

Solution. The linear speed, v, is related to the angular speed, 0), by the equation v =r 0). Therefore,

v =rio =(0.20 m)(6 rad/s) =1.2 m/s

Notice that although we typically write the abbreviation rad when writing angular measurements,the radian is actually a dimensionless unit, since, by definition, 8= sir. So !Y. 8= 6 means the samething as !Y. 8 =6 rad.

Solution. The linear acceleration a is related to the angular acceleration a by the equation a =ra.Since a =6 rad/s2 (as calculated in Example 6.2), we find that

a=ra =(0.50 m)(6 rad/s2) =3 m/s2

Solution. For an object revolving with linear speed v at a distance r from the center of rotation, thecentripetal acceleration is given by the equation ac =v2I r. Using the fundamental equation v =r 0),we find that

v2 (rO))2 2ac =7=-r-=0) r

94 II CRACKING THE AP PHYSICS EXAM

[C] THE BIG FIVE FOR ROTATIONAL MOTIONThe simplest type of rotational motion to analyze is motion in which the angular acceleration isconstant (possibly equal to zero). Another restriction that will make our analysis easier (and whichdoesn't diminish the power and applicability of our results too much) is to consider rotational motionaround afixed axis of rotation. In this case, there are only two possible directions for motion. Onedirection, counterclockwise, is called positive (+), and the opposite direction, clockwise, is callednegative (-).

Let's review the quantities we've seen so far. The fundamental quantities for rotational motion areangular displacement (~e), angular velocity (0t and angular acceleration (a). Because we'redealing with angular acceleration, we know about changes in angular velocity, from initial velocity(0; or 0

0) to final velocity (0 f or simply 0 -with no subscript). And, finally, the motion takes place

during some elapsed time interval, ~ t.Therefore, we have five kinematics quantities: ~ e, 0 0, OJ, a,and M.

These five quantities are interrelated by a group of five equations which we call The Big Five. Theywork in cases in which the angular acceleration is uniform. These equations are identical to The BigFive we studied in chapter 2 but, in these cases, the translational variables (s, v, or a) are replaced bythe corresponding rotational variables (e, 0, or a, respectively).

Big Five #1:

Big Five #2:

Big Five #3:

Big Five #4:

Big Five #5:

~e=OJM

~OJ=aM

~e =0oM+ta(M)2

MJ=OJM-ta(~t)2

OJ2 =0~ +2a~e

Variable that's missing

a

~e

~t

In BigFive #1,because angular acceleration is constant, the average angular velocity is simply the

average of the initial angular velocity and the final angular velocity: m=t(0o+0). Also, if we

decide that t; =0, then M = tf - t; =t - 0 =t; so we can just write "i" instead of" M" in the first fourequations. This simplification in notation makes the equations a little easier to memorize.

Each of the Big Five equations is missing exactly one of the five kinematics quantities and, likewith the other BigFive you learned, the way you decide which equation to use is to determine whichof the kinematics quantities is missing from the problem, and use the equation that's also missing thatquantity. For example, if the problem never mentions the final angular velocity- 0 is neither givennor asked for-then the equation that will work is the one that's missing 0; that's Big Five #3.

Notice that Big Five #1 and #2 are simply the definitions of mand a written in forms that don'tinvolve fractions.

ROTATIONAL MOTION 95

Solution. We're given m0' t, and m, and asked forLl fJ. So a is missing, and we use Big Five #1:

LlfJ =mM =t(mo+m)M =t(1 rad/ s+5 rad / s)(3 s)=9 rad

Solution. We're given m0' Ll fJ, and to, and asked for a. Since t is missing, we use Big Five#5:

m2 =m~ +2aLlfJ => m2 =2aLlfJ (since mo=0)

a = m2 =(10 rad / S)2 =2.5 rad / S2

2LlfJ 2(20 rad)

Tosummarize, here's a comparison of the fundamental quantities of translational and rotationalmotion and of the BigFive (assuming constantacceleration and a fixed axisof rotation):

Translational Rotational Connection

displacement: Lls LlfJ Lls =rLl fJ

velocity: v m v=rm

acceleration: a a a =ra

Big Five #1: Lls = vM Ll()=mM

Big Five #2: Llv= aLlt Llm= aM

BigFive #3: Lls = voM + ta(M)2 Ll() = moLlt + ta(Llt)2

Big Five #4: Lls = vM - ta(Lltf Ll() = mM - ta(Llt)2

BigFive#5: v2= v~ +2aLls m2= m~ + 2aA.()

ROTATIONAL DYNAMICSThedynamics of translational motioninvolves describing the acceleration of an object in termsof itsmass (inertia) and the forces that act on it; Fnet =mao Byanalogy, the dynamics of rotational motioninvolves describing the angular (rotational) acceleration of an object in terms of its rotational inertiaand the torques that act on it.

96 III CRACKING THE AP PHYSICS EXAM

TORQUEIntuitively, torque describes the effectiveness of a force in producing rotational acceleration. Con­sider a uniform rod that pivots around one of its ends, which is fixed.For simplicity, let's assume thatthe rod is at rest. What effect, if any, would each of the four forces in the figure below have on thepotential rotation of the rod?

Our intuition tells us that FI, Fz, and F3

would not cause the rod to rotate, but F4

would. What'sdifferent about F/ It has torque. ,

The torque of a force can be defined as follows. Let r be the distance from the pivot (axis ofrotation) to the point of application of the force F, and let 8 be the angle between vectors rand F.

pivot+-.---__---'--....

Then the torque of F, denoted byr (tau), is defined as:

t =rF sin 8

In the figure above, the angle between the vectors rand F is 8. Imagine sliding r over so that itsinitial point is the same as that of F. The angle between two vectors isthe angle between them when they startat the same point. However, the supplementary angle 8' can be used in place of 8 in the definition oftorque. This is because torque depends on sin 8, and the sine of an angle and the sine of itssupplement are always equal. Therefore, when figuring out torque, use whichever of these angles ismore convenient.

We will now see if this mathematical definition of torque supports our intuition about forces FI,

Fz, F3, and F

4•

ROTATIONAL MOTION 97

Theanglebetweenrand F1 is 0, and 8 = 0 implies sin 8 = 0,so by the definitionof torque,t: = 0as well. The angle betweenrand F2 is 180°, and 8= 180° givesus sin 8= 0, so r =O. For F

3, r =0

(because F3 actsat the pivot, so the distance from the pivot to the point of application of F3

is zero);since r =0, the torque is 0 as well. However, for F

4, neither r nor sin 8 is zero, so F

4has a nonzero

torque. Of the four forces shown in that figure, only F4

has torque and would produce rotationalacceleration.

There'sanotherway to determine the valueofthe torque. Ofcourse,it givesthe sameresultas themethod given above, but this method is ofteneasier to use. Look at the same object and force:

pivot

Instead of determining the distancefrom the pivot point to the point of application of the force,we willnow determinethe (perpendicular) distance fromthe pivot point to what's calledthe line ofactionof the force. Thisdistance is the lever arm (ormomentarm)of the force Frelative to the pivot,and is denoted by 1.

pivot

The torque of F is defined as the product

r =IF

(Justas the lever arm is oftencalled the moment arm, the torque is called the moment of the force.)Thatthesetwo definitions oftorque, t =rFsin 8 and t =IFareequivalentfollows immediately fromthe fact that 1=r sin 8:

98 • CRACKING THE AP PHYSICS EXAM

Since 1is the component of r that's perpendicular to F, it is also denoted by r1. ("r perp"). So thedefinitionof torque can be written as 1: =r1.F.

These two equivalent definitions of torque make it clear that only the component of F that'sperpendicular to r produces torque.ThecomponentofF that's parallel to r does not produce torque.Notice that 1: = rF sin 8= rF1.' where F1. ("F perp") is the component of F that's perpendicular to r:

~.ro Uj FF..L = F sin () I

pivot e------:..--_~-I-~

So the definitionof torque can alsobe written as 1: =rF1.'[C] Remark. Because only the component of F (perpendicular. to r) produces torque, it is not

surprising that the torque can be written as the crossproduct of rand F:

1:=rxF

Solution. Since the tensionforce, FT

, is tangent to the pulley, it is perpendicular to the radius vectorr at the point of contact:

ROTATIONAL MOTION II 99

Therefore, the torque produced by this tension force is simply

T= rFT = (0.05 m)(40 N) = 2 Nrrt

Solution. Each ofthe two forces producesa torque,but thesetorques opposeeachother.ThetorqueofF1 is counterclockwise, and the torqueofF,is clockwise. Thiscanbe visualizedeitherby imaginingthe effect of each force, assuming that the other was absent, or by using the vector definition oftorque, T =r x F. In the caseof F1, we have, by the right-hand rule,

T1 =f 1 X F1 points out of the plane of the page: 0, while

Tz=rz x Fzpoints into the plane of the page: 0

100 II CRACKING THE AP PHYSICS EXAM

If the torque vector points out of the plane of the page, this indicates a tendency to producecounterclockwise rotation and, if it points into the plane of the page, this indicates a tendency toproduce clockwise rotation.

The net torque is the sum of all the torques. Counting a counterclockwise torque as positive anda clockwise torque as negative, we have

1'1 =+rl1 =+(0.12 m)(lOO N) =+12 N·m

and

t Z =-riz =-(0.08 m)(80 N) =-6.4 Nrn

so

1'net = L t =1'1 + 1'z =(+12 Non) + (-6.4Nrn) = +5.6 Nrn

[C] ROTATIONAL INERTIAOur goal is to develop a rotational analog of Newton's Second Law, F net = ma. We're almost there;torque is the rotational analog of force and, therefore, t net is the rotational analog of Fnet" Therotational analog of translational acceleration, a, is rotational (or angular) acceleration, a. We willnow look at the rotational analog of inertial mass, m.

Consider a small point mass m at a distance r from the axis of rotation, being acted upon by atangential force F.

FIIIIIIIIIIIIIIIII

-----------------+---------------- mr

From Newton's Second Law, we have F = ma. Substituting a = ra, this equation becomesF= mr a. Now multiply both sides of this last equation by r to yield

rF =mrza

or, since rF =1',

l' =mrza

In the equation F=ma, the quantity m is multiplied by the acceleration produced by the force F,while in the equation t =mt"a, the quantity mrzis multiplied by the rotational acceleration producedby the torque 1'.

ROTATIONAL MOTION II 101

So for a point mass at a distance r from the axis of rotation, its rotational inertia (also calledmoment of inertia) is defined as mr2

• If we now take into account all the point masses that comprisethe object under study, we can get the total rotational inertia, I, of the body by adding them:

1= I"mir/

For a continuous solid body, this sum becomes the integral

1=fr2dm

This formula can be used to calculate expressions for the rotational inertia of cylinders (disks),spheres, slender rods, and hoops. Notice that the rotational inertia depends not only on m, but alsoon r; both the mass and how it'sdistributed about the axis ofrotation determine 1. By summing over allpoint masses and all external torques, the equation 'T= mr2 a becomes

~ 'T. = (~m.r?)a or 'T = I a£." I £." I I net

We've reached our goal:

Translational motion

force, F

acceleration, a

mass, m

F =manet

Rotational motion

torque, 'T

rotational acceleration, a

rotational inertia, I

'T =Ianet

102 • CRACKING THE AP PHYSICS EXAM

Solution.

(a) In the first case, both the left bead and the right bead are at a distance of L/2 fromthe axis of rotation, while the center bead is at distance zero from the axis of rota­tion. Therefore,

(b) In the second case, the left bead is at distance zero from the rotation axis, the centerbead is at distance L/2, and the right bead is at distance 1. Therefore,

l' =I,mi1i2=m(O)2 +m(tt +m(L)2 =~mL2

Notice that, although both assemblies have the same mass (namely, 3m), their rotational inertiasare different, because of the different distribution of mass relative to the axis of rotation.

If the rotational inertia of a body is known relative to an axis that passes through the body's centerof mass, then the rotational inertia,!', relative to any other rotation axis (parallel to the first one) canbe calculated as follows. Let I

ernbe the rotational inertia of a body relative to a rotation axis that passes

through the body's center of mass, let the mass of the body be M, and let x be the distance from theaxis through the center of mass to the rotation axis. Then

l' =I + Mx2em

This is called the parallel-axis theorem. Let's use this result to calculate the rotational inertia ofthe three-bead assembly in part (b)from the value obtained in part (a), which is I

ern• SinceM =3mand

x =L/2, we have

I' =I +Mx2=1.mL2+(3m)(1.)2 =i meem 2 2 4

which agrees with the value calculated above.

ROTA TI0NAL MOTI 0N II 103

Solution. A solid is homogeneous if its density is constant throughout. Let p be the density of thecylinder. In order to compute I using the formula I =fr2 dmI we choose our mass element to be aninfinitesimally thin cylindrical ring of radius r:

,,~

...,,,,IIIII:HIIII,I,II,I

of

104 1\1 CRACKING THE AP PHYSICS EXAM

Themass of this ring is equal to its volume, 2nr dr x H, times the density; that is,

dm = p x 2rtrH dr

Therefore,

1=Jr2dm= J;=~r2(p. 2wHdr)

=2rtpHJ~r3 dr

=2rtPH[ir4]:

=trtpHR4

To eliminate p, use the fact that the total mass of the cylinder is

M=pV=p·rtR2H

Putting this into the expression derived for I, we find that

1= trtpHR4 =t(p ·rtR2H)R2= t MR2

Theheight of the cylinder (the dimension parallel to the axis of rotation) is irrelevant. Therefore,the formula I = tMR2 gives the rotational inertia of any homogeneous solid cylinder revolvingaround its centralaxis. This includesa disk (which is just a reallyshort cylinder).

Solution. In Example 6.9, we figured out that 'X'net =+5.6 Nrn, Therotational inertia of the cylinder is

I = tMR2 = t (50 kg)(O.l2 m)2 =0.36 kg-m'

ROTATIONAl MOTION III 105

Therefore, from the equation T net =I a, we find that

_ Tnet _ 5.6 N· m -16 d / 2a--- 2 - ra sI 0.36 kg·m

This angular acceleration will be counterclockwise, because Tnet is counterclockwise.

Solution. In chapter 3, we treated pulleys as if they were massless, and no force was required tomake them rotate. Now, however, we know how to take the mass of a pulley into account, byincluding its rotational inertia in our analysis. The pulley is a disk, so its rotational inertia is given bythe formula I =-tMR2.

First we draw a free-body diagram for the falling block:

and apply Newton's Second Law:

rng-FT =rna (1)

Now the tension FT

in the cord produces a torque, T =RFT, on the pulley:

~

106 • CRACKING THE AP PHYSICS EXAM

Since this is the only torque on the pulley, the equation 'r net = [ a becomes RFr = [ a. But [= tMR2and a =a/R. (This last equation says that the cord doesn't slip as it slides over the pulley; the linearacceleration of a point on the rim, a =Ra, is equal to the acceleration of the connected block.)Therefore,

which tells us that

't = [a => RFT = IMRz . !!:.... = IMRaZ R Z

Fr = tMa (2)

Substituting Equation (2) into Equation (1), we find that

mg--z1 Ma=ma => (-Zl M+m)a=mg => a- m g

-lM+mZ

Solution. First we draw a free-body diagram for the cylinder:

We know that the cylinder rolls without slipping, so the force of friction is not kinetic friction.Since the speed of the point on the cylinder in contact with the ramp is zero with respect to the ramp,static friction supplies the torque that allows the cylinder to roll smoothly.

ROTATIONAL MOTION II 107

The linear acceleration of the cylinder, a, is given by the equation Fnet =Ma:

Fnet = Ma => Mgsine-Ff = Ma

The rotational acceleration of the cylinder, a, is given by the equation 1:net =I a: Since 1: =F"I =-tMR2, and a =aiR, we have

.1Ma2

Substituting this result into the earlier equation gives

Mgsine--tMa = Ma => fMa = Mgsine => a = ~gsine

[C] KINETIC ENERGY Of ROTATIONA rotating object has rotational kinetic energy, just as a translating object has translational kineticenergy. The formula for kinetic energy is, of course, K = -tmv2, but this can't be directly used tocalculate the kinetic energy of rotation because each point mass that makes up the body can have adifferent v. Fot this reason, we need a definition of Krotational that involves OJ instead of v.

K - "K - ,,1 ( 2 ) 2 _ 1 (" 2 ) . 2 _ 1 I 2rotational - £.J i - £.J"2 mi1j CO -"2' £.Jmi1j CO -"2 CO

Notice that this expression for rotational kinetic energy follows the general pattern displayed byour previous results: I is the rotational analog of mand OJ is the rotational analog of v.Therefore, therotational analog of -t mv2 should be F OJ2.

108 CRACKING THE AP PHYSICS EXAM

Solution. We will attack this problem using Conservation of Mechanical Energy. As the cylinderrolls down the ramp, its initial gravitational potential energy is converted into kinetic energy, whichis a combination of translational kinetic energy (since the cylinder's center of mass is translatingdown the ramp) and rotational kinetic energy:

Ki+Ui=Kf+Uf

O+Mgh=(tMv~m +tlai)+O

Since I = tMR2 and (j) = vem/ R, this equation becomes

Therefore,

Vcm =~tgh

We can verify this result using the result of the previous example. There we found that the

acceleration of the cylinder's center of mass as it rolled down the ramp was a = i gsinS. Applying

Big Five #5 gives us:

2aLh= 2a-­

sinS2 2 • S h= '-gsm .-- =

3 sinS = ~tgh

[C] WORK AND POWERConsider a small point mass m at distance r from the axis of rotation, acted upon by a tangentialforce F.

F,,,,,,,,: v,,,,,,,__ ~ + __ M _

: r,,,,,,,,,,,,,,,

ROTATIONAL MOTION III 109

As it rotates through an angular displacement of ,1e, the force does work on the point mass:W =F,15, where ,15 =r,1e. Therefore,

If the force is not purely tangential to the object's path, then only the tangential component of theforce does work; the radial component does not (since it's perpendicular to the object's displace­ment). Therefore, for a general constant force F, the equation above would read W =Ft,1 e= r,1 e,where F

Idenotes the tangential component of F.

Ifwe want to allow for a varying F-and a varying r-then the work done is equal to the definiteintegral:

Again, notice the analogy between this equation and the one that defines work by F:

JX2

W= FdxXl

The work-energy theorem (W= ,1K) also holds in the rotational case, where Wis the work doneby net torque and ,1Kis the resulting change in the rotational kinetic energy.

The rate at which work is done, or the power (P), is defined by the equation

p=dWdt

Over an infinitesimal angular displacement de, the torque r does an amount of work dW givenby dW= rde. This implies that

dW de- = 't- => P = 'toodt dt

Once again, notice the parallel to P = Fv, the translation version of this equation.

110 CRACKING THE AP PHYSICS EXAM

Solution.

(a) Apply Conservation of Mechanical Energy. The initial gravitational potential energyof the block is transformed into the purely rotational kinetic energy of the pulleyand translational kinetic energy of the falling block:

. Ki+Ui=Kf+UfO+mgh =(tmv2 +tlco2)+O

-1.mv 2 +1..1. MR 2 • (J!.)2- 2 2 2 R

=tmv2+iMv2

=(tm+iM)v2

mghV=

1. m+1.M2 4

Substituting in the given numerical values, we get

v = (5)(9.8)(2) =4.7 m / s}(5)+t(8)

ROTATIONAL MOTION 111

(b) The rotational kinetic energy of the pulley as the block strikes the floor is

K=FQ)2=ltMR2{f~f=lMv2=1(8)(4.7)2=44 J

(c) The rate at which work is done on the pulley is the power produced by the torque.One way to compute this is to first use the work-energy theorem to determine thework done by the torque and then divide this by the time during which the blockfell. So,

W=D.K".=Kf-Ki =Kf =44J

The time during which this work was done is the time required for the block todrop to the ground. Using Big Five #1 and the result of part (a), we find that

& =vt => t = & = h h = 12 m - 0.85 s

v t(vo+v) tv 2(4.7 m/ s)

Therefore,

p = W = 44J =52Wt 0.85 s

[C] ANGULAR MOMENTUMSo far we've developed rotational analogs for displacement, velocity, acceleration, force, mass, andkinetic energy. We will finish by developing a rotational analog for linear momentum; it's calledangular momentum.

Consider a small point mass mat distance r from the axis of rotation, moving with velocity v andacted upon by a tangential force F.

F,,,,,,I

: vI,II,I,

-----------------+----------------: r,,I,,,I,,I,I,I,

112 II CRACKING THE AP PHYSICS EXAM

..- -- .."",;.'"

,,r,,

Then, by Newton's Second Law,

F =t:.p =t:.(mv)t:.t t:.t

If we multiply both sides of this equation by r and notice that rF = "C, we get

t:.(rmv)"C =

,M

Therefore, to form the analog of the law F = I1pll1t (force equals the rate-of-change of linearmomentum), we say that torque equals the rate-of-change of angular momentum, and the angularmomentum (denoted by L) of the point mass m is defined by the equation

L=rmv

If we now take into account all the point masses which comprise the object under study, we can getthe angular momentum of the body by adding up all the individual contributions. This gives

L = [co

Notice that this expression for angular momentum follows the general pattern we saw previously:I is the rotational analog of m,and co is the rotational analog of v. Therefore, the rotational analog ofmv should be I co .

If the point mass m does not move in a circular path, we can still define its angular momentumrelative to any reference point.

reference • V-,"J(Vpoint r,'

,,,,,

<. ,/ path of object

If r is the vector from the reference point to the mass, then the angular momentum is

L = rmo;

where v.1 is the component of the velocity that's perpendicular to r. The fact that it's the perpendicularcomponent of v relative to r that's important for figuring out angular momentum motivates thegeneral vector definition with the cross product. The equation L =(r)(mvJ =(r)(p.1) becomes

L=rxp

ROTATIONAL MOTION 113

Solution. Apply the definition:

If you want to specify the direction of the angular momentum vector, L, use the right-hand rule. Let, the fingers of your right hand curl in the direction of rotation of the body. Your thumb gives thedirection of L, pointing along the rotation axis:

L

,,,,,,,,,,r,,

~

,,,,,,,,,,,,

~,,,,,,,,,y ,, L

[C] CONSERVATION OF ANGULAR MOMENTUMNewton's Second Law says that

_dpFnet -dF

so if Fnet =0, then p is constant. This is Conservation of Linear Momentum.The rotational analog of this is: '

dL"Cnet=dF

so if "Cnet =0, then L is constant. This is Conservation of Angular Momentum. Basically, this says thatif the torques on a body balance. so that the net torque is zero, then the body's angular momentumcan't change.

An often cited example of this phenomenon is the spinning of a figure skater. As she pulls herarms inward, she moves more of her mass closer to the rotation axis and decreases her rotationalinertia. Since the external torque on her is negligible, her angular momentum must be conserved.Since L =I ill , a decrease in I causes an increase in ill, and she spins faster.

114 l1li CRACKING THE AP PHYSICS EXAM

Solution. The childwalking toward the center of the merry-go-round does not provide an externaltorqueto thechild+disk system, soangularmomentum is conserved. Let'sdenotethechildas a pointmass, and consider the following two views of the merry-go-round (looking down from above):

,,,,,,·,,,,: R

----------------~---------------­:-- - - --- ---- - - - --~

: ri

··········

child

,,,,··,··,····----------------~---~------------·... ---~1 rf

···,·,,,··In the first picture, the totalrotationalinertia,Ii' is equal to the sum ofthe rotational inertiaof the

merry-go-round (MGR) and the child:

I j =IMGR + Ichild =-tMR2+mrj2 =-tMR2+mR2 =(-tM +m)R2

In the second picture, the total rotational inertia has decreased to

If = IMGR + I~hild = -tMR2 +mrl

So,by Conservation of AngularMomentum, we have

Lj =Lf

I/IJj =Ifcof

(tM + m)R2coj =(tMR2 + mrl)cof

(t M +m)R2

cof = .1MR2 + mr,z CO j

2 f

and, substituting the givennumerical values givesus

(t M +m)R2

COf =tMR2 +mrl COj

(t· 100+ 30)(2.5)2 1 d=t· 100· (2.5)2 +30.(0.5)2 ( ra / s)

= 1.6 rad/ s

Notice that co increasedas I decreased, just as Conservation of AngularMomentum predicts.

ROTATIONAL MOTION II I1S

EQUILIBRIUMAn object is said to be in translational equilibrium if the sum of the forces actingon it is zero; thatis, if Fnet =O. Similarly, an object is said tobe in rotational equilibrium if thesum ofthe torquesactingon it is zero; that is, if 'rnet =O. Theterm equilibrium by itselfmeansboth translational and rotationalequilibrium. Abody in equilibrium maybe in motion; Fnet =0 doesnot meanthat the velocity is zero;it onlymeans that the velocity is constant. Similarly, t net =0 does not mean that the angular velocityis zero;it onlymeans that it's constant. If an object is at rest, then it is said to be in static equilibrium.

Solution. Let Fc denote the (contact) force exertedby the wall on the bar. In order to simplify ourwork,we canwrite Fc in terms ofits horizontalcomponent, FCx' and its vertical component, FCy' Also,if FT is the tensionin the wire, then FTx =FT cos55° and FTy =FT sin 55° are its components. Thisgivesus the following force diagram:

,

F~x = FT cos 55°,,

," ,'-'

mg

116 III CRACKING THE AP PHYSICS EXAM

The first condition for equilibrium requires that the sum of the horizontal forces is zero and thesum of the vertical forces is 0:

I,F =0:x

I,F =0:y

Fex -r; cos 55° =0 (1)

Fey +Fr sin 55° - mg- Mg =0 (2)

We notice immediately that we have more unknowns (Fex/ Fey/ Fr) than equations/ so this systemcannot be solved as is. The second condition for equilibrium requires that the sum of the torquesabout any point is equal to zero. Choosing the contact point between the bar and the wall as our pivot/only three of the forces in the diagram above produce torque: Fry produces a counterclockwisetorque/ and both mg and Mg produce clockwise torques/ which must balance. From the definition"C =IF/ and taking counterclockwise torque as positive and clockwise torque as negative/ we have

I,"C =0: (L/2)Fry- (L/2)(mg) - LMg = 0 (3)

This equation contains only one unknown and can be solved immediately:

~FTY =~mg+LMg

FTy =mg+2Mg=(m+2M)g

Since Fry =F; sin 55°/ we can find that

Fr sin 55°=(m+2M)g· ~ F _ (m+2M)gr - sin 55°

= (8+2·12)(9.8)sin 55°

=380N

Substituting this result into Equation (1) gives us Fex:

(m+2M)gFex =Fr cos55°=. cos55°= (8+2.12)(9.8)cot 55°= 220 N

sro55°

And finally/ from Equation (2)/we get

Fey =mg+Mg-Fr sin 55°

(m+2M)g .=mg+Mg- sro55°sin 55°

=-Mg= -(12)(9.8)

=-118N

The fact that Fe turned out to be negative simply means that in our original force diagram/ thevector Fe points in the direction opposite to how we drew it. That is/ Fe points downward.

y • • yTherefore/ the magmtude of the total force exerted by the wall on the bar IS

ROTATIONAl MOTION iii 117

CHAPTER 6 REVIEW QUESTIONS

SEalON I: MULTIPLE CHOICE

1. [C] A compact disc has a radius of 6 cm.If the disc rotates about its central axis atan angular speed of 5 revIs, what is thelinear speed of a point on the rim of thedisc?

(A) 0.3 mls(B) 1.9 mls(C) 7.4 mls(D) 52 mls(E) 83m/s

2. [C] A compact disc has a radius of 6 cm.If the disc rotates about its central axis ata constant angular speed of 5 revI s, whatis the total distance traveled by a point onthe rim of the disc in 40 min?

(A) 180 m(B) 360 m(C) 540 m(D) 720 m(E) 4.5 km

3. [C] An object of mass 0.5 kg, moving in acircular path of radius 0.25 m, experiencesa centripetal acceleration of constantmagnitude 9 m /s". What is the object'sangular speed?

(A) 2.3 radls(B) 4.5 rad/s(C) 6rad/s(D) 12 rad/s(E) Cannot be determined from the

information given

118 II CRACKING THE AP PHYSICS EXAM

4. [C] An object, originally at rest, beginsspinning under uniform angular accelera­tion. In 10 8, it completes an angulardisplacement of 60 rad. What is thenumerical value of the angular accelera­tion?

(A) 0.3rad/s2

(B) 0.6 rad/s2

(C) 1.2 rad z's"(D) 2.4 rad I S2

(E) 3.6 rad I S2

5.

In an effort to tighten a bolt, a force F isapplied as shown in the figure above. Ifthe distance from the end of the wrench tothe center of the bolt is 20 cm andF = 20 R what is the magnitude of thetorque produced by F?

(A) 0 Non(B) 1 Non(C) 2 N·m(D) 4 N·m(E) 10N·m

8. [C] What is the rotational inertia of thefollowing body about the indicatedrotation axis? (The masses of the connect­ing rods are negligible.)

rotationaxis

6.

suspensionpoint ~

m

m

,,,G, m

In the figure above, what is the torqueabout the pendulum's suspension pointproduced by the weight of the bob, giventhat the length of the pendulum, L, is80 ern and m =0.50 kg?

(A) 0.49 Nrn(B) 0.98 N·m(C) 1.7N·m(0) 2.0 N·m(E) 3.4 N·m

~L3

,,,,,,,,,,,,,,y

m m""if------------- nn __ n)o,

L L

(A) ~*gh

(B) ~tgh

(C) ~fgh

(0) ~tgLsine

(E) )iogLsine

9. [C] The moment of inertia of a soliduniform sphere of mass M and radius R isgiven by the equation I = i MR2. Such asphere is released from rest at the top ofan inclined plane of height h, length L,and incline angle (). If the sphere rollswithout slipping, find its speed at thebottom of the incline.

7.

A uniform meter stick of mass 1 kg ishanging from a thread attached at thestick's midpoint. One block of massm = 3 kg hangs from the left end of thestick, and another block, of unknownmass M, hangs below the 80 em mark onthe meter stick. If the stick remains at restin the horizontal position shown above,whatisM?

(A) 4kg(B) 5 kg(C), 6 kg(0) 8 kg(E) 9 kg

(A)

(B)

(C)

(0)

(E)

4mU

32 mU3

64 mU9

128 mU9

256 mU9

ROTATIONAl MOTION 119

10. [C] An object spins with angular velocitym. If the object's moment of inertiaincreases by a factor of 2 without theapplication of an external torque, whatwill be the object's new angular velocity?

(A) m/4

(B) m/2

(C) to /.J2(D) .J2 m(E) zra

SEUION II: FREE RESPONSE

1. [C] In the figure below, the pulley is a solid disk of mass M and radius R, with rotationalinertia MR2/2. Two blocks, one of mass m1 and one of mass m2, hang from either side of thepulley by a light cord. Initially the system is at rest, with Block 1 on the floor and Block 2 heldat height h above the floor. Block 2 is then released and allowed to fall.

Block2

Block1

IIIIIII

;hIIIIII

(a) What is the speed of Block 2 just before it strikes the ground?

(b) What is the angular speed of the pulley at this moment?

(c) What's the angular displacement of the pulley?

(d) How long does it take for Block 2 to fall to the floor?

120 III CRACKING THE AP PHYSICS EXAM

2. The diagram below shows a solid uniform cylinder of radius R and mass M rolling (withoutslipping) down an inclined plane of incline angle e. A thread wraps around the cylinder as itrolls down the plane and pulls upward on a block of mass m. Ignore the rotational inertia ofthe pulley.

(a) Show that "rolling without slipping" means that the speed of the cylinder's center ofmass, vern' is equal to Rill, where ill is its angular speed.

(b) Show that, relative to P (the point of contact of the cylinder with the ramp), the speed ofthe top of the cylinder is 2vem•

(c) What is the relationship between the magnitude of the acceleration of the block and thelinear acceleration of the cylinder?

(d) What is the acceleration of the cylinder?

(e) What is the acceleration of the block?

ROTATIONAL MOTION 121

3. Two slender uniform bars, each of mass M and length 2L, meet at right angles at their mid­points to form a rigid assembly that's able to rotate freely about an axis through the intersec­tion point, perpendicular to the page. Attached to each end of each rod is a solid ball of clayof mass m. A bullet of mass mb is shot with velocity v as shown in the figure (which is a viewfrom above of the assembly) and becomes embedded in the targeted clay ball.

,

bullet /

~~~:_--------m

k-- M

L··,···,,,'t m

~--------L ---------~~--------L ---------~

(a) Show that the moment of inertia of each slender rod about the given rotation axis isMUj3.

(b) Determine the angular velocity of the assembly after the bullet has become lodged inthe targeted clay ball.

(c) What is the resulting linear speed of each clay ball?

(d) Determine the ratio of the final kinetic energy of the assembly to the kinetic energy ofthe bullet before impact.

122 III CRACKING THE AP PHYSICS EXAM