Lectures 18-19 Chapter 11 Fall 2010

7/30/2019 Lectures 18-19 Chapter 11 Fall 2010

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PH 221-3A Fall 2010

ROLLING, TORQUE, and

Lectures 18-19

Chapter 11

Hallida /Resnick/Walker Fundamentals of Ph sics 8th edition

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Chapter 11

Rolling, Torque, and Angular Momentum

In this chapter we will cover the following topics:

-Rollin of circular ob ects and its relationshi 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

-Angular Momentum of single particles and systems or particles

-Newtons second law for rotational motion

-Conservat on o angu ar Momentum-Applications of the conservation of angular momentum

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t1 = 0 t2 = t

Consider an object with circular cross section that rolls

Rolling as Translation and Rotation Combined

along a surface without slipping. This motion, though

common, is complicated. We can simplify its study by

mass and rotation of the object about the center of mass

Consider the two sna shots of a rollin bic cle wheel shown in the fi ure.

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) Duringcom

t

dss v t

dt

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

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

s R R R

we get the condition for rolling without slipping.

t t

comv R

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comv R

We have seen that rolling is a combination of purely translational motion

with s eed and a urel rotaional motion about the center of massv

with angular velocity . The velocity of each pcomv

R

oint is the vector sum

o e ve oc es o e wo mo ons. or e rans a ona mo on e

velocity vector is the same for every point ( ,see fig.b ). The rotational

velocit varies from oi

comv

nt to oint. Its ma nitude is e ual to where isr r

the 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 terms. For example

the 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

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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?

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

com0

0

22.2 m/s59.3 rad/s.

0.375 m

v

R

. , . -

22 2 2

0

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

2 188 rad

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


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AvA

vT

Another way of looking at rolling is shown in the figure

Rolling as Pure Rotation

B

vB

vO e cons er ro ng as a pure ro a on a ou an ax s

of rotation that passes through the contact point P

between the wheel and the road. The an ular

velocity of the rotation is comvR

In order to define the velocity vector for each point we must know its magnitude

as well as its direction. The direction for each point on the wheel points along the.

perpendicular to the dotted line that connects pont A with point P. The speed

of each point is given by: . Here is the distance between a parti

A

v r r cular

point and the contact point P. For example at point T 2 .

Thus 2 2 . For point O thusT com O com

r R

v R v r R v R v

For point P 0 thus 0Pr v

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Consider the rolling object shown in the figure

The Kinetic Energy of Rolling

It is easier to calculate the kinetic energy of the rolling

body by considering the motion as pure rotation

.

and radius .R21The kinetic energy is then given by the equation: . Here is theP PK K I I

2

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

2 21 . P com

2 2 2 2 22

1 1 1

com

com comK I MR I MR 2 21 1

com comK I Mv

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 .

The second term is associated with the kinetic energy due to the translational

motion of evey point with speed comv 7


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Problem 9 . A sol id cyl inder of radius 10 cm and m ass 12 kg s tarts from res t and rol ls wi thout

s l ipping a d is tance L=6.0 m dow n a roof that i s inc lined a t the angle =30 .

( ) W h a t is th e a ng u la r s pe e d oa

f the cylinder ab out i ts center as i t leave s the roof?

' = '. .

does the cylinder hit the level ground ?

(a) We find its angular speed as it leaves the roof using conservation of energy. Its initial

= =

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 I f 12

2 12

2 .

Here we use v to denote the speed of its center of mass and is its angular speed atthe moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we

can set v =R = v whereR = 0.10 m. Using I MR 12

2(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 massMcancels from the equation, and we obtain

1 4 1 4

9 8 3 0 63gh . . .m s m rad s2c hb g .

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(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 (theinitial

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 positive

direction choices) its components are

0 0

0 0

cos30 5.4 m s

sin 30 3.1 m s.

x

y

v v

v v

The projectile motion equations become

x v t y v t gtx y 0 021and .

We first find the time wheny =H= 5.0 m from the second equation (using the quadratic

formula, choosing the positive root):

2

0 02

0.74s.y y

v v gH t

g

Then we substitute this into thex equation and obtainx 54 0 74 4 0. . .m s s m.b gb g

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When an object rolls with constant speed (see top figure)

Friction and Rolling

it has no tendency to slide at the contact point P and thus

no frictional force acts there. If a net force acts on the

com -

for the center of mass (see lower figure). If the rolling

object accelerates to the right it has the tendency to slide

com

at point P to the left. Thus a static frictional force

opposes the tendency to slide. The motion is smooth

sf

,maxrolling as long as s sf f

The rolling condition results in a connection beteen the magnitude of the

acceleration of the center of mass and its angular acceleration

We take time derivatives of both sides

com

com com

a

v R a

comdv d

R R

coma R10


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

10 kg a nd radius 0.30m. T he wh eel rolls smoothly on the horizontal surface, and

the acceleration of its cent

ap pF

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 wh eel abou t the rotation axis through its center of

a

mass?

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

In unit vector notation, we have ( 4.0 N)isf which points leftward.

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

|| = |acom| /R = 2.0 rad/s ,

The only force not directed towards (or away from) the center of mass is

fs , and the

torque it produces is clockwise:

20.30 m 4.0 N 2.0 rad sI I

which ields the wheels rotational inertia about its center of mass: I 0 60. .k m2

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Consider a round uniform body of mass and radius

Rolling Down Ra amp

R

acom

rolling down an inclined plane of angle . We will

calculate the acceleration of the center of masscoma

'a ong e x-ax s us ng ew on s secon aw or e

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:

s comx f Mg Ma

Rf I

We substitute in the second equation and gcoma

R et: coms com

aRf I

R

2 (eqs.2) We substitute from equation 2 into equation 1coms com s

com

f I fR

a

sing

a

2

com comR

21

comcom

R

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sin

| |comg

a

acom 21com

R

Cylinder Hoop

22

1 2

2

MRI I MR

1 22 2

1 2

1 / 1 /

sin

a aI MR I MR

g

sing

21 / 2MR

2 2 2

1 2

1 /

sin sin

1 1/ 2 1 1

MR MR MRg g

a a

1 2

2 sin sin

(0.67) sin (0.5) sin3 2

g g

a g a g

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Consider a yo-yo of mass , radius , and axle radius

The Yo-Yo

oM R R

rolling down a string . We will calculate the acceleration

of the center of its mass along the -axis using Newton'scoma y

acom

in the previous problem

Newton's second law for motion along the -axis:y

y

(eqs.1)

Newton's second law for rotation about the center of mas

comMg T Ma

s:

. Angular acceleration

We substitute in the second equation and get:

com

o com

o

R T IR

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

o

aT I TR

a

2

2

1

ccom

com

omcom c

o

om

o

g aaR I

MR

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In chapter 10 we defined the torque of a rigid body rotating about a fixed axis

Torque Revisited

with each particle in the body moving on a circular path. We now expand the

definition of torqu te so hat it can describe the motion of a particle that moves

.

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

In the example shown in the figure both and lie in the -plane. Using the

r F

r F xy

right hand rule we can see that the direction of is along the -axis.

The magnitude of the torque vect

z

or sin , where is the anglerF between and . From triangle OAB we have: sin

, in agreement with the definition of chapter 10.

r F r r

r F

r

B

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1 2

Problem 21. In unit-vector notation, what is the net torque about the origin on a flea

located at coordinates (0, -4.0m, 5.0 m) when forces (3.0 ) and ( 2.0 )F N k F N j

act on t e ea

y z x yx z

x y z

y z x yx z

x y z

i j ka a a aa a

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

b b b

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

r F is equal to

y z y z z x z x x y x ya a a a a a

yF zF zF xF xF yFz y x z y x 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,

yields

( 2.0N m)i.r F

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Angular Momentum

e coun erpar o near momen um n ro a ona

motion is a new vector known as angular momentum.

The new vector is defined as follows: r

p mv

In the example shown in the figure both andlie in the -plane. Using the right hand rule we

r pxy

can see that the direction of is along the -axis.

The magnitude of angular momentum sin ,

z

rmv

.

OAB we have: i vms nr rr

An ular momentum de ends on the choice of the ori in O. If the ori inNote:

2

is shifted in general we get a different value of

SI unit for angular momentum: Sometimes the equivakg.m / lents J.s is

used

r p m r v

mvr

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P r o b l e m 2 9 . A t o n e i n st a n t, f o rc e 4 . 0 N a c ts o n a 0 . 2 5 k g o b j e c t t h a t h a s p o s it io n

v 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

and in un i t vec to r no ta t ion ,

F j

r i k v i k

w h a t a re

'

i k

( b ) t h e t o r q u e a c t i n g o n t h e o b j e c t .

( ) ( ) ( )

y z x yx z

x y z

y z x yx z

x y z

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

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

b b b bb bb b b

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

(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 velocity

vectors are nonzero, so Eq. 3-30 leads to r v xv zvz z b g j. Therefore,

j 0.25 kg 2.0 m 5.0 m s 2.0 m 5.0 m s j 0.z xm xv zv

- , .

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

Withx = 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.

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Newton's Second Law in Angul r Forma

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

will derive the an ular form of Newton's second law for a article.

net

dpF

dt

d dm r v m r dt dt

dv dr v m r v m r a v vdt dt

0 net net d

v v m r a r ma r F td

Thus: Compare with:n ne ett Fddt t

netd

dt

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m11 3

zWe will now explore Newton's second law in

The Angular Momentum of a System of Particles

O

m3m2

2

x y

1 2 3

angular momentum , , ,..., n

1 2 3

1

The angular momentum of the system is ...

=

n i

i

ni

L L

ddL

1

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

idt dt

,iclei

net i

d

dt

,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

net i

, net idt

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.

net net

i

Thus Newton's second law for a system in angular form takes the form:net ext

dt

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Angular Momentum of a Rigid Body Rotating about a Fixed Axis

- .the z-component of the net angular momentum. The body is

divided n elements of mass that have a position vectori im r

The angula

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

i i i i

i i i i i i

r p

r p r m v

of is: sin sin

The z-component of the angular momentu

iz i iz i i i i i i ir m v r m v

m is the sum:zn n n n

L

21 1 1 1

2

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

n

L r m v r m r m r

1

Thus:

i ii

zL I

zL I21


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1

2

P r o b l e m 3 4 . A p a r t ic l e i s a c t e d o n b y t w o to r q u e s a b o u t th e o ri g in : h a s a m a g n i t u d e

o f 2 . 0 N m a n d is d i r e c te d in t h e p o s i t iv e d i r e c ti o n o f t h 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 ir e c

t e d i n t h e n e g a t i v e d i r e c t i o n o f t h e y a x i s . In u n i t -v e c t o r n o t a t io n ,

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

The rate of change of the angular momentum is

d

1 2 . . .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 thexy plane, or a plane parallel to it) measured from the positivexaxis, where

1 4.0 N mtan 63

,

.

the negative sign indicating that the angle is measured clockwise as viewed from above

(by a person on the z axis).

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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 angulart Nm

. . .

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

22

0( ) (5.00 2.00 ) 5.00 1.00L t dt t dt L t t

Since 25.00 k m /sL when 1.00 st the inte ration constant is 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

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

Conservation of Angular moment mu

second law in angular form is: netdL

dt

If the net external torque 0 then we hav : 0enet

dt

a constant This result is known as the law of theL

conservation of angular momentum. In words

Net angular momentumNet angular momentum

:

I

fi

n equation form: L L

If the component of the external torque along a certainNote:

ax s s equa to zero, t en t e componet o t e angu ar momentum

of the system along this axis cannot change 24


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The figure shows a student seated

on a stool that can rotate freely about a

Example:

i

vert ca ax s. e stu ent w o as een set nto

rotation at an initial angular speed , holds

two dumbbells in h

is outstretched hands. His

angular momentum vector lies along therotation axis, pointing upward.

L

The student then pulls in his hands as shown in fig.b. This action reduces the

rotational inertia from an initial value to a smaller final value .I I

No net external torque acts on the student-stool 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 fi

t L I t L I I I

L L I I I I

iff

The rotation rate of the student in fig.b is faster25


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2

Sample Problem 11-7:

2

. .2 3.9 rad/s

6.8 kg.m

wh

wh

bI

?b

y-axis

2i f wh wh b b whL L L L L L L

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

wh whb b wh wh b

b

II II

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Problem 60. A horizontal platform in the shape of a circular disk rotates on a frictionless

bearing 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 towrd 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, whereIdisk = 300 kg mwhich incidentall does a ree with Table 10-2 c andI = mR

2where m = 60 k

andR = 2.0 m.

The rotational inertia when the student reaches r= 0.5 m is If = Idisk + mr2. Angular

momentum conservation leads to

I II mR

I mri i f f f i

disk

disk

2

2

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

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Rotational Motion

Analogies betwee

Translati

n translational and r

onal Moti

otational M

on

otion

x

v

22

a

p

2

2

K

m

K

I

F ma

F

P Fv

I

P

net net

dpF

dt

d

td

p mv L I

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Lectures 18-19 Chapter 11 Fall 2010