PH 221-1D Spring 2013

ROLLING, TORQUE, and ANGULAR MOMENTUM

Lectures 27-29

Chapter 11 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)

1

Chapter 11Rolling, Torque, and Angular Momentum

In this chapter we will cover the following topics:

-Rolling of circular objects and its relationship with friction -Redefinition of torque as a vector to describe rotational problems that are more complicated than the rotation of a rigid body about a fixed axis -Angular Momentum of single particles and systems or particles -Newton’s second law for rotational motion -Conservation of angular Momentum -Applications of the conservation of angular momentum

2

t1 = 0 t2 = tConsider an object with circular cross section that rollsalong a surface without slipping. This motion, though c

Rolling as Translation and R

ommon, is complicated. We

otation Combin

can simplif

ed

y its study bytreating it as a combination of translation of the center ofmass and rotation of the object about the center of mass

Consider the two snapshots of a rolling bicycle wheel shown in the figure. An observer stationary with the ground will see the center of mass O of the wheel move forward with a speed . The pointcomv P at which the wheel makes contactwith the road also moves with the same speed. During the time interval between

the two snapshots both O and P cover a distance . (eqs.1) During

t

com

tdss v tdt

he bicycle rider sees the wheel rotate by an angle about O so that

= (eqs.2) If we cambine equation 1 with equation 2

we get the condition for rolling without slipping.

ds ds R R Rdt dt

comv R3

We have seen that rolling is a combination of purely translational motion with speed and a purely rotaional motion about the center of mass

with angular velocity . The velocity of each p

com

com

vvR

oint is the vector sum

of the velocities of the two motions. For the translational motion the velocity vector is the same for every point ( ,see fig.b ). The rotational velocity varies from poi

comv

nt to point. Its magnitude is equal to where isthe distance of the point from O. Its direction is tangent to the circular orbit(see fig.a). The net velocity is the vector sum of these two ter

r r

ms. For examplethe velocity of point P is always zero. The velocity of the center of mass O is

( 0). Finally the velocity of the top point T is wqual to 2 .com comv r v

comv R

4

Problem 2. An automobile traveling at 80.0 km/h has tires of 75.0 cm diameter.(a) What is the angular speed of the tires about their axes? (b) If the car is brought to a stop uniformly in 30.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels?(c) How far does the car move during the braking?

The initial speed of the car is

80 km/h (1000 m/km)(1 h/3600 s) 22.2 m/sv . The tire radius is R = 0.750/2 = 0.375 m. (a) The initial speed of the car is the initial speed of the center of mass of the tire, so

com00

22.2 m/s 59.3 rad/s.0.375 m

vR

(b) With = (30.0)(2) = 188 rad and = 0, Eq. 10-14 leads to

2

2 2 20

(59.3 rad/s)2 9.31 rad/s .2 188 rad

(c) R = 70.7 m for the distance traveled. 5

A

B

vA

vB

vT

vO

Another way of looking at rolling is shown in the figure We consider rolling as a pure rotation about an axis of rotation that passes through the contact

Rollin

point

g as Pure Rotatio

P between th

n

e wheel and the road. The angular

velocity of the rotation is comvR

In order to define the velocity vector for each point we must know its magnitudeas well as its direction. The direction for each point on the wheel points along the tangent to its circular orbit. For example at point A the velocity vector is perpendicular to the dotted line that connects pont A with point P. The speedof each point is given by: . Here is the distance between a parti

Av

v r r

cularpoint and the contact point P. For example at point T 2 . Thus 2 2 . For point O thus For point P 0 thus 0

T com O com

P

r Rv R v r R v R v

r v

6

Consider the rolling object shown in the figureIt is easier to calculate the kinetic energy of the rollingbody by considering the motion as p

The Kinetic E

ure rotationa

ne

bo

rgy o

ut th

f Rollin

e co

g

ntact point P. The rolling object has mass and radius .

MR

2

2

1The kinetic energy is then given by the equation: . Here is the2

rotational inertia of the rolling body about point P. We can determine using

the parallel axis theorem.

P P

P

P com

K K I I

I

I I MR

2 2

2 2 2 2 2

12

1 1 12 2 2

The expression for the kinetic energy consists of two terms. The first term corresponds to the rotation about the center of mass O with angular velocity .

com

com com

K I MR

K I MR I MR

The second term is associated with the kinetic energy due to the translational motion of evey point with speed comv

2 21 12 2com comK I Mv

7

P rob lem 9 . A so lid cylinder of rad iu s 10 cm and m ass 12 kg sta rts from rest and ro lls w ithou tslip p ing a d istance L = 6 .0 m dow n a roof tha t is inc lin ed a t the angle = 30 .( ) W ha t is the angu la r speed oa

f the cylin der abou t its cen te r a s it leaves the roof?(b ) T he roof 's edge is a t he igh t H = 5 .0 m . H ow fa r horizon ta lly from the roof 's edge does the cylinde r h it the leve l ground?

(a) We find its angular speed as it leaves the roof using conservation of energy. Its initialkinetic energy is Ki = 0 and its initial potential energy is Ui = Mgh where

6.0sin 30 3.0 mh (we are using the edge of the roof as our reference level forcomputing U). Its final kinetic energy (as it leaves the roof) is

K Mv If 12

2 12

2 . Here we use v to denote the speed of its center of mass and is its angular speed — at the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding wecan set v = R = v where R = 0.10 m. Using I MR 1

22 (Table 10-2(c)), conservation of

energy leads to

2 2 2 2 2 2 2 21 1 1 1 3 .2 2 2 4 4

Mgh Mv I MR MR MR

The mass M cancels from the equation, and we obtain

1 4

31

01043

9 8 3 0 63R

gh.

. . .m

m s m rad s2c hb g

8

(b) Now this becomes a projectile motion of the type examined in Chapter 4. We put theorigin at the position of the center of mass when the ball leaves the track (the “initial”position for this part of the problem) and take +x leftward and +y downward. The result of part (a) implies v0 = R = 6.3 m/s, and we see from the figure that (with these positivedirection choices) its components are

0 0

0 0

cos30 5.4 m ssin 30 3.1 m s.

x

y

v vv v

The projectile motion equations become

x v t y v t gtx y 0 021

2and .

We first find the time when y = H = 5.0 m from the second equation (using the quadratic formula, choosing the positive root):

20 0 2

0.74s.y yv v gHt

g

Then we substitute this into the x equation and obtain x 54 0 74 4 0. . .m s s m.b gb g

9

0coma

When an object rolls with constant speed (see top figure) it has no tendency to slide at the contact point P and thus no frictional force acts there. If a net

F

f

riction and

orce acts o

Roll

n t

ing

he rolling body it results in a non-zero acceleration for the center of mass (see lower figure). If the rolling object accelerates to the right it has the tendency to slide

at point P to the left.

coma

,max

Thus a static frictional force opposes the tendency to slide. The motion is smooth rolling as long as

s

s s

f

f f

The rolling condition results in a connection between the magnitude of theacceleration of the center of mass and its angular acceleration

We take time derivatives of both sides

com

com com

a

v R a

comdv dR Rdt dt

coma R 10

Problem 7. A constant horizontal force of magnitude 10N is applied to a wheel of mass

10 kg and radius 0.30m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its cent

appF

2er of mass has magnitude 0.60 m/s .( ) In unit-vector notation, what is the frictional force on the wheel?(b) W hat is the rotational inertia of the wheel about the rotation axis through its center of a

mass?

(a) Newton’s second law in the x direction leads to

2app 10N 10kg 0.60 m s 4.0 N.s sF f ma f

In unit vector notation, we have ˆ( 4.0 N)isf

which points leftward.

(b) With R = 0.30 m, we find the magnitude of the angular acceleration to be

|| = |acom| / R = 2.0 rad/s2, The only force not directed towards (or away from) the center of mass is

f s , and the

torque it produces is clockwise:

20.30 m 4.0 N 2.0 rad sI I which yields the wheel’s rotational inertia about its center of mass: I 0 60. .kg m2

11

acom

Consider a round uniform body of mass and radius rolling down an inclined plane of angle . We willcalculate the acceleration of the center of ma

Rolling

ss along the x-axis usi

Down R

n

a am

g

p

com

M R

a

Newton's second law for the translational and rotational motion

Newton's second law for motion along the -axis: sin (eqs.1)Newton's second law for rotation about the center of mass:

We substitute in the second equation and g

s com

s com

com

x f Mg MaRf I

aR

2

2

et:

(eqs.2) We substitute from equation 2 into equation 1

sin

coms com

coms com s

comcom com

aRf IR

af I fR

aI Mg MaR

2

sin

1com

com

ga IMR

12

acom

22

1 2

1 22 21 2

1 2

2

sin sin 1 /

Cylinder

1 /sin

1

/

Hoo

2

pMRI I MR

g ga aI MR I MR

gaMR

22 2 2

1 2

1 2

sin 1 /

sin sin 1 1/ 2 1 12 sin sin(0.67) sin (0.5) sin

3 2

gaMR MR MR

g ga a

g ga g a g

2

sin| |1

comcom

ga IMR

13

y

acom

Consider a yo-yo of mass , radius , and axle radius rolling down a string . We will calculate the acceleration

of the center of its mass along the -axis using Ne

The

wton's second law

Yo-Yo

o

com

M R R

a y for the translational and rotational motion as we did

in the previous problemNewton's second law for motion along the -axis:

(eqs.1)Newton's second law for rotation about the center of mas

com

yMg T Ma

s:

. Angular acceleration

We substitute in the second equation and get:

como com

o

aR T IR

2

2

2

(eqs.2) We substitute from equation 2 into equation 1

1

comcom

o

ccom

com

omcom c

o

omo

aT I TR

aMg I M aaR

gI

MR

14

B

In chapter 10 we defined the torque of a rigid body rotating about a fixed axiswith each particle in the body moving on a circular path. We now expand the definitio

Torque Rev

n of torqu

isite

t

d

e so

hat it can describe the motion of a particle that movesalong any path relative to a fixed point. If is the position vector of a particle

on which a force is acting, the torque is defined as:

r

F

In the example shown in the figure both and lie in the -plane. Using theright hand rule we can see that the direction of is along the -axis.The magnitude of the torque

vect

r F

r F xyz

or sin , where is the angle

between and . From triangle OAB we have: sin , in agreement with the definition of chapter 10.

rF

r F r rr F

r F

15

1 2

Problem 21. In unit-vector notation, what is the net torque about the origin on a flea ˆ ˆlocated at coordinates (0, -4.0m, 5.0 m) when forces (3.0 ) and ( 2.0 )

act on the flea?F N k F N j

If we write r x y z i + j + k, then we find

r F is equal to

yF zF zF xF xF yFz y x z y x d i b g d i i + j + k. With (using SI units) x = 0, y = – 4.0, z = 5.0, Fx = 0, Fy = –2.0 and Fz = 3.0 (these latter terms being the individual forces that contribute to the net force), the expression aboveyields

ˆ( 2.0N m)i.r F

16

ˆˆ ˆˆˆ ˆ

ˆˆ ˆ( ) ( ) ( )

y z x yx zx y z

y z x yx zx y z

y z y z z x z x x y x y

i j ka a a aa a

a b b a a a a i j kb b b bb b

b b b

a b b a i a b b a j a b b a k

B

The counterpart of linear momentum in rotational motion is a new vector known as angular momentum.

The new vector is defined as follows: In the example shown in t

Angular Momentum

he fir p

p mv

gure both and

lie in the -plane. Using the right hand rule we

can see that the direction of is along the -axis.The magnitude of angular momentum sin , where is the angle between

r pxy

zrmv

and . From triangle

OAB we have: i v ms nr p

r rr

2

Angular momentum depends on the choice of the origin O. If the origin

is shifted in general we get a different value of SI unit for angular momentum: Sometimes the equivakg.m / lent

Note:

s J.s is

used

r p m r v mvr

17

ˆP ro b le m 2 9 . A t o n e in s ta n t, fo rc e 4 .0 N a c ts o n a 0 .2 5 k g o b je c t th a t h a s p o s itio nˆ ˆˆ ˆv e c to r ( 2 .0 2 .0 ) m a n d v e lo c ity v e c to r ( 5 .0 5 .0 ) m /s . A b o u t th e o r ig in

a n d in u n it v e c to r n o ta tio n ,

F j

r i k v i k

w h a t a re(a ) th e o b je c t's a n g u la r m o m e n tu m a n d(b ) th e to rq u e a c tin g o n th e o b je c t.

(a) We use mr v , where r is the position vector of the object, v is its velocity

vector, and m is its mass. Only the x and z components of the position and velocityvectors are nonzero, so Eq. 3-30 leads to r v xv zvz z b g j. Therefore, ˆ ˆj 0.25 kg 2.0 m 5.0 m s 2.0 m 5.0 m s j 0.z xm xv zv

(b) If we write r x y z i j k, then (using Eq. 3-30) we find

r F is equal to

yF zF zF xF xF yFz y x z y x d i b g d i .i j k

With x = 2.0, z = –2.0, Fy = 4.0 and all other components zero (and SI units understood)the expression above yields

r F 8 0 8 0. . i k N m.e j

ˆˆ ˆˆˆ ˆ

ˆˆ ˆ( ) ( ) ( )

y z x yx zx y z

y z x yx zx y z

y z y z z x z x x y x y

i j ka a a aa a

a b b a a a a i j kb b b bb b

b b b

a b b a i a b b a j a b b a k

18

Newton's second law for linear motion has the form: . Below we

will derive the angular form of Newton's sec

Newton's Second

ond law for a pa

Law in Angul

rticle

r Form

.

a

netdpFdt

d dm r v m rdt dt

0

Thus: Compar e with:

net ne

n n

t

e ett

dv drv m r v m r a v vdt dt

dv v m r

dpFd

a r ma r Ft

dd

dt t

netddt

19

O

m1

m3m2

ℓ1

ℓ2

ℓ3

x y

z

1 2 3

We will now explore Newton's second law in angular form for a system of n particles that

The Angular Momentum of a

have

angular momentum

System of P

,

articl

, ,...,

es

n

1 2 31

1

The angular momentum of the system is ...

The time derivative of the angular momentum is =

The time derivative for the angular momentum of the i-th part

n

n ii

ni

i

L L

ddLdt dt

,

,

,

icle

Where is the net torque on the particle. This torque has contributions from external as well as internal forces between the particles of the system. Thus

inet i

net i

net i

ddt

dLdt

1

Here is the net torque due to all the external forces.

By virtue of Newton's third law the vector sum of all internal torques is zero.Thus Newton's second law for a system in ang

n

net neti

ular form takes the form: net extdLdt

20

Angular Momentum of a Rigid Body Rotating about a Fixed Axis

We take the z-axis to be the fixed rotation axis. We will determine the z-component of the net angular momentum. The body isdivided n elements of mass that have a position vector

The angulai im r

r momentum of the i-the element is: Its magnitude sin90 = The z-compoment

of is: sin sin The z-component of the angular momentu

i i i i

i i i i i i

iz i iz i i i i i i i

r pr p r m v

r m v r m v

2

1 1 1 1

2

1

m is the sum:

The sum is the rotational inertia of the rigid body

Thus:

z

n n n n

z iz i i i i i i i ii i i i

n

i ii

z

L

L r m v r m r m r

m r I

L I

zL I21

1

2

P r o b le m 3 4 . A p a r t i c le i s a c te d o n b y tw o to r q u e s a b o u t th e o r ig in : h a s a m a g n i tu d eo f 2 .0 N m a n d i s d i r e c te d in th e p o s i t iv e d i r e c t io n o f th e x a x i s , a n d h a s a m a g n i tu d e o f 4 .0 N m a n d i s d i r e c

t e d in th e n e g a t iv e d i r e c t io n o f th e y a x i s . I n u n i t - v e c to r n o ta t io n ,

f in d , w h e r e i s th e a n g u la r m o m e n tu m o f th e p a r r t i c l e a b o u th e o r ig in .d l ld t

The rate of change of the angular momentum is

1 2ˆ ˆ(2.0 N m)i (4.0 N m) j.d

dt

Consequently, the vector d dt has a magnitude 22(2.0 N m) 4.0 N m 4.5 N m

and is at an angle (in the xy plane, or a plane parallel to it) measured from the positive xaxis, where

1 4.0 N mtan 632.0 N m

,

the negative sign indicating that the angle is measured clockwise as viewed “from above”(by a person on thez axis).

22

2

2

Problem 42. A disk with a rotational inertia of 7.00 kgm rotates like a merry-go-round while undergoing a torque given by (5.00 2.00 ) . At time t=1.00 s, its angularmomentum is 5.00 kgm / . What

t Nms

is its angular momentum at t=3.00s?

Torque is the time derivative of the angular momentum. Thus, the change in the angularmomentum is equal to the time integral of the torque. With (5.00 2.00 ) N mt , the angular momentum as a function of time is (in units 2kg m /s ) 2

0( ) (5.00 2.00 ) 5.00 1.00L t dt t dt L t t Since 25.00 kg m /sL when 1.00 st , the integration constant is 0 1L . Thus, the complete expression of the angular momentum is

2( ) 1 5.00 1.00L t t t . At 3.00 st , we have 2 2( 3.00) 1 5.00(3.00) 1.00(3.00) 23.0 kg m /s.L t

23

For any system of particles (including a rigid body) Newton's

second law in angular form is:

If

Con

th

servati

e net e

on of Angular

xternal torque

moment

0 then we hav : 0

m

e

u

net

net

dLdt

dLdt

a constant This result is known as the law of the

conservation of angular momentum. In wordsNet angular momentumNet angular momentumat some later time at some initial time

:

I

fi tt

L

n equation form:

If the component of the external torque along a certain axis is equal to zero, then the componet of the angular momentumof the system along this axis cannot cha

Note:

nge

i fL L

24

i

The figure shows a student seated on a stool that can rotate freely about a vertical axis. The student who has been set into rotation at an initial angular speed , holdstwo dumbb

Example:

ells

in h

is outstretched hands. His

angular momentum vector lies along the rotation axis, pointing upward.

L

The student then pulls in his hands as shown in fig.b. This action reduces therotational inertia from an initial value to a smaller final value .

No net external torque acts on the student-sti fI I

ool system. Thus the angular momentum of the system remains unchanged. Angular momentum at : Angular momentum at :

Since 1

i i i i f f f f

i ii f i i f f f i f

ffi

f

t L I t L I

I IL L I I I II I

The rotation rate of the student in fig.b is faster

i

25

2

2

1.2 kg.m2

Sample Probl

3.9 rad/

em 11-7:

s

6.8 kg.m?

wh

wh

b

b

I

I

2

2 2 1.2 2 3.92 2 1.4 rad/s6.8

i f wh wh b b wh

wh whb b wh wh b

b

L L L L L L L

II II

y-axis

26

Problem 60. A horizontal platform in the shape of a circular disk rotates on a frictionlessbearing about a vertical axle through the center of the disk. The platform has a mass of 150 kg, a radius of 22.0 m, and a rotational inertia of 300 kgm about the axis of rotation.A 60 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.5 rad/s when the student starts at the rim, what is the angular speed when she is 0.5 m from the center?

The initial rotational inertia of the system is Ii = Idisk + Istudent, where Idisk = 300 kg m2

(which, incidentally, does agree with Table 10-2(c)) and Istudent = mR2 where m = 60 kg and R = 2.0 m. The rotational inertia when the student reaches r = 0.5 m is If = Idisk + mr2. Angular momentum conservation leads to

I I I mRI mri i f f f i

disk

disk

2

2

which yields, for i = 1.5 rad/s, a final angular velocity of f = 2.6 rad/s.

27

Rotational Motion

Analogies betwee

Translati

n translational and r

onal Moti

otational M

on

otio

n

xv

22

2

2

ap

mvK

mF

IK

maI

net net

FP Fv

dpFdt

p m

dt

v

I

P

d

L I28