Chapter 11

Rolling, 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 of particles -Newton’s second law for rotational motion -Conservation of angular momentum -Applications of the conservation of angular momentum

(11-1)

t1 = 0 t2 = tConsider an object with circular cross section that rolls

along a surface without slipping. This motion, though

common, is complicated. We can simplify it

Rolling as Translation and Rotation Combined

s study by

treating it as a combination of translation of the center of

mass and rotation of the object about the center of mass.

com

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 of the wheel

move forward with a speed . The point

O

v

com

at which the wheel makes contact

with the road also moves with the same speed. During the time interval between

the two snapshots both and cover a distance , (eq. 1). During

P

t

dsO P s v t

dt

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

= (eq. 2). If we combine equation 1 with equation 2

we get the condition for rolling without slipping:

O

ds ds R R

dt dt

comv R(11-2)

com

com

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

with speed and a purely rotational motion about the center of mass

with angular velocity . The velocity of each p

v

v

R

com

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

v

nt to point. Its magnitude is equal to where is

the distance of the point from . Its direction is tangent to the circular orbit

(see fig. a). The net velocity is the vector sum of these two te

r r

O

com com

rms. For example,

the velocity of point is always zero. The velocity of the center of mass is

( 0). Finally, the velocity of the top point is equal to 2 .

P O

v r T v

comv R

(11-3)

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 point

between the w

P

Rolling as Pure Rotation

com

heel and the road. The angular

velocity of the rotation is .v

R

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

tangent to its circular orbit. For example, at point the velocity vector is

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

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

AA v

A B

v r r

com com

cular

point and the contact point . For example, at point 2 .

Thus 2 2 . For point , thus .

For point , 0 thus 0.T O

P

P T r R

v R v O r R v R v

P r v

(11-4)

Consider the rolling object shown in the figure.

It is easier to calculate the kinetic energy of the rolling

body by considering the motion as pure rolling

about the contac

The Kinetic Energy of Rolling

t point . The rolling object has mass

and radius .

P M

R2

2com

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

2rotational inertia of the rolling body about point . We can determine using

the parallel axis theorem.

P P

P

P

K K I I

P I

I I MR K

2 2com

2 2 2 2 2com com

1.

21 1 1

2 2 2The expression for the kinetic energy consists of two terms. The first term

corresponds to the rotation about the center of mass with angular velocity .

I MR

K I MR I MR

O

com

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

motion of every point with speed . v (11-5)

2 21 1

2 2com comK I Mv

.

com 0a

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

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

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

r

P

Friction and Rolling

comolling body it results in a nonzero 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 to the left.

a

P

,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

com

com c

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

acceleration of the center of mass and its angular acceleration ,

. We take time derivatives of both sides

a

v R a

comom .

dv dR R

dt dt

com a R(11-6)

acomcom

Consider a round uniform body of mass and radius

rolling down an inclined plane of angle . We will

calculate the acceleration of the center of mass

along the -axis using

M R

a

x

Rolling Down a Ramp

Newton's second law for the

translational and rotational motion.

com

com

com

Newton's second law for motion along the -axis: sin (eq. 1)

Newton's second law for rotation about the center of mass:

We substitute in the second equation and g

s

s

x f Mg Ma

Rf I

a

R

comcom

comcom 2

comcom com2

et

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

sin

s

s s

aRf I

Ra

f I fR

aI Mg Ma

R

(11-7)

comcom

2

sin

1

ga

I

MR

acom

22

1 2

1 22 21 2

1 2

2

sin sin

1 /

Cylinder

1 /

sin

1

/

Hoo

2

p

MRI I MR

g ga a

I MR I MR

ga

MR

22 2 2

1 2

1 2

sin

1 /sin sin

1 1/ 2 1 12 sin sin

(0.67) sin (0.5) sin3 2

ga

MR MR MRg g

a a

g ga g a g

comcom

2

sin| |

1

ga

I

MR

(11-8)

Example: A uniform cylinder rolls down a ramp inclined at an angle of θ to the horizontal. What is the linear acceleration of the cylinder at the bottom of the ramp? Remember that: The friction force is used to rotate the object.

Example: Consider a solid cylinder of radius R that rolls without slipping down an incline from some initial height h. The linear velocity of the cylinder at the bottom of the incline is vcm and the angular velocity is ω.

We can also solve for angular velocity using the equation 2

4

3cmv gh

R R

Example: A bowling ball has a mass of 4.0 kg, a moment of inertia of 1.6×10**(−2) kg ·m2 and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 4.0 m/s, what is its total energy?

Example: A bowling ball has a mass of 4.0 kg, a moment of inertia of 1.6×10**(−2) kg ·m2 and a radius of 0.10 m. If it rolls down the lane without slipping at a linear speed of 4.0 m/s, what is its total energy?

y

acom

0

com

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 Newton's

second law f

M R R

a y

The Yo - Yo

com

or the translational and rotational motion as we did

in the previous problem.

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

(eq. 1).

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

y

Mg T Ma

com0 com

0

s:

. Angular acceleration .

We substitute in the second equation and get

aR T I

R

comcom 2

0

comcom co com

com02

0

m2

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

1

.

aT I T

R

aM

ga

Ig I Ma

MR

R

(11-9)

B

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

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

definition of torque so t

Torque Revisited

hat it can describe the motion of a particle that moves

along 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 the

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

The magnitude

.

of the torque vect

r

F

z

F

r xy

or sin , where is the angle

between and . From triangle we have sin

, in agreement with the definition of Chapter 10.

rF

r F OAB r r

r F

r F

(11-10)

ˆ ˆ ˆi j k

( ) ( )o o o

x y z

r F x x y y z z

F F F

Torque Revisited

Example: (T72_Q17.) A force is applied to an object that is pivoted about a fixed axis aligned along the z-axis. If the force is applied at the point of coordinates (4.0, 5.0, 0.0) m, what is the applied torque (in N.m) about the z axis?

ˆ ˆ ˆi j k

(4 0) (5 0) 0 0

2 3 0

ˆ2 k N m

r F

ˆ ˆ2.0 i 3.0 j NF

Example: (T71_Q19.). At an instant, a particle of mass 2.0 kg has a position of and acceleration of . What is the net torque on the particle at this instant about the point having the position vector: ?

ˆ ˆ9.0 i 15.0 j mr 2ˆ3.0 i m/sa

ˆ9.0 i mor

Example: (T71_Q19.). At an instant, a particle of mass 2.0 kg has a position of and acceleration of . What is the net torque on the particle at this instant about the point having the position vector: ?

ˆ ˆ9.0 i 15.0 j mr 2ˆ3.0 i m/sa

ˆ9.0 i mor

ˆ ˆ ˆ ˆ ˆ ˆi j k i

( ) ( ) 2 (9 9) (15 0) 0 0

3 3 0

ˆ90 k N m

o o o

x y z

j k

r F m r a m x x y y z z

a a a

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 th

.

e f

r p

p mv

Angular Momentum

igure 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 p

xy

z

rmv

and . From triangle

we have: sin .

r p

OAB r r r mv

2

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

is shifted, in general we get a different value of .

SI unit for angular momentum: Sometimes the ekg m / s. quiv

Not

ale

e

J s

:

nt

O

is used.

r p m r v (11-11)r mv

Example: (T052 Q#18) A stone attached to a string is whirled at 3.0 rev/s around a horizontal circle of radius 0.75 m. The mass of the stone is 0.15 kg. The magnitude of the angular momentum of the stone relative to the center of the circle is:

22 20.15 0.75 3 2 1.6 kg.m /L mvr mr s

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

will derive the angular form of Newton's second law for a particle.

dpF

dt

d dm r v m r

dt dt

Newton's Second Law in Angular Form

n

net net

t ene t

0

Thus: C. ompare with: .

dv drv m r v m r a v v

dt dt

dv v m r a r ma

d

dt

r

d

Fd

dpF

t

t

net d

dt

(11-12)

O

m1

m3

m2

ℓ1

ℓ2

ℓ3

x y

z

1 2 3

We will now explore Newton's second law in

angular form for a system of particles that have

angular momentum , , ,..., .n

n

The Angular Momentum of a System of Particles

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 pa

n

n ii

ni

i

L L

ddL

dt dt

net,

net ,

net ,

rticle

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

ii

i

d

dt

dL

dt

net net1

. 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 an

n

ii

gular form takes the form:

(11-13)

net dL

dt

Angular Momentum of a Rigid Body Rotating About a Fixed Axis

We take the -axis to be the fixed rotation axis. We will determine

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

divided into elements of mass that have a position vector .

The a

i i

z

z

n m r

ngular momentum of the th element is .

Its magnitude is sin 90 = . The -compoment

of is sin sin .

The -component of the angular momen

i i i i

i i i i i i

iz i iz i i i i i i i

i r p

r p r m v z

r m v r m v

z

2

1 1 1 1

2

1

tum 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

(11-14) zL I

net

net

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

second law in angular form is .

If the net external torque 0 then we have: 0

dL

dt

dL

dt

Conservation of Angular Momentum

Net angular momentumNet angular momentum

at some later time at some initial tim

a constant. This result is known as the law of the

conservation of angular momentum. In words:

In

e

e

q

fi tt

L

uation form:

If the component of the external torque along a certain

axis is equal to zero, then the component of the angular momentum

of the system along this axis cannot cha

Note:

nge.

i fL L

(11-15)

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 , holds

two dumbbe

Example:

lls

in hi

s 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 the

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

No net external torque acts on the student-st

i 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 i

f f

t L I t L I

I IL L I I I I

I I

The rotation rate of the student in fig. b is fa te

.

s r.

f i

(11-16)

2

1.2 kg m

Samp

2 3.9 rad/s

6.8 rad/s

?

le Problem 11-7:

wh

wh

b

b

I

I

2

2 2 1.2 2 3.92 2 1.4 rad/s

6.8

i f wh wh b b wh

wh whb b wh wh b

b

L L L L L L L

II I

I

y-axis

(11-17)

Rotational Motion

Translational Motion

x

v

Analogies Between Translational and Rotational Motion

22

2

2

a

p

mvK

m

F

IK

ma

I

n netet

F

P Fv

dpF

dtp m

I

P

v

d

dt

L I(11-18)