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• Rolling motion (axis of rotation is moving)

• Kinetic Energy of rolling motion

• Rolling motion on an incline

• Torque

• Angular momentum

• Angular momentum is conserved

Chapter 11: Rolling Motion, Torque and Angular Momentum

2/26

Rolling motion of a particle on a wheel

(Superposition of rolling and linear motion)

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 3/26

Smooth rolling:

There is no slipping

Linear speed of center of mass: COM

ds R dv R

dt dt

11-2 Rolling motion

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 4/26

The angular velocity of any point on the wheel is the same.

The linear speed of any point on the object changes as shown in the diagram!!

For one instant (bottom), point P has no linear speed.

For one instant (top), point P’ has a linear speed of 2·vCOM

11-2 Rolling motion

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 5/26

Superposition principle:

Rolling motion = Pure translation + Pure rotation

Kinetic energy

of rolling motion:

22

2

1

2

1 CMIMvK

11-3 Kinetic Energy of Rolling

6/26

Approximate each wheel on the car Thrust SSC as a disk of uniform thickness and mass M = 170 kg, and assume smooth rolling. When the car’s speed was 1233 km/h, what was the kinetic energy of each wheel?

1233 / 342.5 /comv km h m s /comv R21

2comI MR

.

234 (170 )(342.5 / )kg m s

71.50 10 J

2 21 12 2com comK I M v

2 21 12 2( )( )( / )comMR v R 21

2 comM v 234 comM v

Sample Problem 11-1

7/26

Friction and Rolling If the wheel rolls without sliding (smooth rolling) and is accelerating, then from ,

(smooth rolling)

where is the linear acceleration of the center of mass and α is the angular acceleration.

comv Rcom

comd v Rda

d t d t

coma R

coma

• The force to provide for macom is the static frictional force (assuming the wheel rolls without sliding).

• Therefore, for a wheel to roll without sliding, the maximum static frictional force, between the wheel and the ground must be greater than macom.

• and point to the right if the wheel if the wheel rotates faster, for example, at the start of a bicycle race.

• Do not assume that is equal to the maximum value of

s N

sf

coma

s Nsf

11-4 Forces of Rolling

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comsin (1)sMg f M a

Do not assume that is at its maximum value of . The value of self-adjusts so the body rolls without sliding. sf

s Nsf

x

yThe positive direction here is chosen to be down the plane.

(2)s comR f I

α is counterclockwise and positive.

coma R

where acom points down plane +ve

2com

s comaf IR

2sin

1 /comcom

gaI M R

Therefore from (2)

and substituting this in (1)

Note that a positive acom points down plane.

F Ma

I

Rolling Down a Ramp

04/18/23 9/26

A ring and and disk of equal mass and diameter are rolling down a frictionless incline.

Both start at the same position; which one will be faster at the end of the incline?

Demo

10/26

A uniform ball, of mass M = 6.00 kg and radius R, rolls smoothly from rest down a ramp at angle = 30.0°

(a)  The ball descends a vertical height h = 1.20 m to reach the bottom of the ramp. What is its speed at the bottom?

22com 5( )I ball MR

22 21 2 1

22 5com

comvMR M v M g h

R

210 107 7( ) ( )(9.8 / )(1.2 )comv g h m s m

4.1 /m s

A positive vcom points down plane.

ex thW K U E

0 0K U

2 21 12 20 0com comI M v Mg h

comv R

12

2 25

M R2

2comvR

212 comM v M g h

2710 comM v M g h

Sample Problem 11-2

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(b) What are the magnitude and direction of the friction force on the ball as it rolls down the ramp?

2sin

1 /comcom

gaI MR

2

25

(9.8 / ) sin30.01

m s

A positive acom points down plane.

225 (6.00 )(3.50 / )kg m s

A positive fs means that the direction we selected for fs (up) is correct! fsR is a clockwise torque (+ve)

2 225

sin1 /

gMR MR

23.50 /m s

225 2

comaMRR

25 comMa2

coms com

af IR

8.40 N

coms comR

af IR

I

12/26

The yo-yo can be considered as a rolling down a ramp:

• Instead of rolling down a ramp at angle with the horizontal, the yo-yo rolls down a string at angle = 90° with the horizontal.

• Instead of rolling on its outer surface at radius R, the yo-yo rolls on an axle of radius Ro.

• Instead of being slowed by frictional force fs, the yo-yo is slowed by the net force T on it from the string.

21 / ocom

com

gaI M R

So we would again get the same expression for the acceleration as for rolling with = 90°.

11-5 Yo-Yo

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 13/26

Fr

sinr F r is the component of F perpendicular F tF o r

( )

F is the moment arm of F the perpendiculardistance between O and the line of action o F

r rf

:magnitude

11-6 Torque and the vector product

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 14/26

In Fig. 11-11a, three forces, each of magnitude 2.0 N, act on a particle. The particle is in the xz plane at point A given by position vector , where r = 3.0 m and = 30°. Force is parallel to the x axis, force is parallel to the z axis, and force is parallel to the y axis

r

1F

2F

3F

What is the torque, with respect to the origin O, due to each force?

1 1 1sin (3.0 )(2.0 )(sin150 ) 3.0r F m N N m

2 2 2sin (3.0 )(2.0 )(sin120 ) 5.2r F m N N m

3 3 3sin (3.0 )(2.0 )(sin90 ) 6.0r F m N N m

To find the directions of the torques, we use the right hand rule and rotate into through the smaller of the two angles between their directions.

r

F

Sample Problem 11.3

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 15/26

Sample Problem 11.3

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 16/26

( )

l r p

m r v

• The SI unit of angular momentum l is .• Angular momentum is a “vector”, the direction is determined by the right

hand rule.• The magnitude of angular momentum is

• where φ is the angle between and when these two vectors are arranged tail to tail.

2 /kg m s J s

sinl r m v

r

p

r vr p ml

l pr r mv :magnitude

11-7 Angular momentum of a particle

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Figure 11-13 shows an overhead view of two particles moving at constant momentum along horizontal paths. Particle 1, with momentum magnitude p1 = 5.0 kg·m/s, has position vector and will pass 2.0 m from point O. Particle 2, with momentum magnitude p2 = 2.0 kg·m/s, has position vector and will pass 4.0 m from point O.

1r

2r

What is the net angular momentum about point O of the two-particle system? L

1 1 1l r p 210 /kg m s

The RHR indicates that l1 is positive. 21 10 /l kg m s

2 2 2l r p 28.0 /kg m s

The RHR indicates that l2 is negative. 22 8.0 /l kg m s

1 2L l l 22.0 /kg m s

(2.0 )(5.0 / )m kg m s

(4.0 )(2.0 / )m kg m s

210 /kg m s 2( 8.0 / )kg m s

RHR = right hand rule

Sample Problem 11-4

18/26

netd pFd t

netd ld t

• Note that the torque and angular momentum must be defined with respect to the same origin.

l

(single particle)

(single particle)

• Proof: ( x )l m r v

( x x ) ( x x )d l d v d rm r v m r a v vd t d t d t

Because , this leads tox 0v v

net net( x ) x x ( x )d l m r a r ma r F r Fd t

Therefore, netd ld t

11-8 Newton’s Second Law in Angular Form

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1 2 31

...n

n ii

L l l l l l

net,1 1

n ni

ii i

d ld Ld t d t

netd Ld t

( system of particles )

• The net external torque   acting on a system of particles is equal to the time rate of change of the system's total angular momentum .

net

L

• is the net torque on the ith particle. is the sum of all the torque(internal and external) on the system. However the internal torques sums to zero. Let represent the net external torque on the system.

net,i net,1

n

ii

net

(L = total angular momentum)

11-9 Angular momentum of a system of Particles

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We’ll consider an object that is rotating about the z-axis.

The angular momentum of the object is given by:

ILz

Note that in this case L and are along the z axis.

Also note the analog formula for linear momentum p = m·v

11-10 Angular momentum of a rigid object rotating about a fixed axis

Lecture notes by Dr. M. S. Kariapper KFUPM - PHYSICS

04/18/23 21/26

The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero.

constantL

If the system undergoes an internal “rearrangement”, then

constant fi LL

If the object is rotating about a fixed axis (say z-axis), then:

constant ffii II

11-11 Conservation of angular momentum

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Demo

A students stands still on a rotatable platform and holds a spinning wheel. The bicycle wheel is spinning in the clockwise direction when viewed from above.

He flips the wheel over.

What happens?

A students stands still on a rotating platform and holds two texts on outstretched arms. He brings the arms closer. What happens? Discuss

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TABLE 11-1     More Corresponding Variables and Relations for Translational and Rotational Motiona

Translational  

Force  

Rotational  

Conservation lawd  

Newton's second lawb  

Linear momentumb  

Linear momentumb  

Linear momentum  

F

p

Conservation lawd  

Newton's second lawb  

Angular momentumc  

Angular momentumb  

Angular momentum  

Torque  

)( ipP

comvMP

td

PdF

net

)x( Fr

)x( prl

constantaP

)( ilL

IL

td

Ld

net

constant aL

a  See also Table 10-3.b  For systems of particles, including rigid bodies.c  For a rigid body about a fixed axis, with L being the component along that axis.d  For a closed, isolated system.

24/26

Force F = 2i-3k acts on a particle with position vector r = 0.5j-2.0k relative to the origin. In unit vector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point (2.0, 0, -3.0)?

P22

A uniform solid ball rolls smoothly along a floor and up a ramp inclined at 15.0°. It is momentarily stops when it has rolled 1.50 m along the ramp. What was its initial speed?

P72

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P85

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 6.00 m long, weighs 10.0 N, and rotates at 240 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about the axis.

In fig. 11.-62, a constant horizontal force Fapp of magnitude 12 N is applied to a uniform solid cylinder by a fishing line wrapped around the cylinder. The mass of the cylinder is 10 kg, its radius is 0.10 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the com of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the com? (c) In unit vector notation, what is the frictional force acting on the cylinder?

P90

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P48

A cockroach of mass 0.17 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a veritcal axle) that has radius 15 cm, rotational inertia 5.0 x 10-3 kgm2, and frictionless bearings. The cockroach’s speed (relative to the ground) is 2.0 m/s, and the lazy Susan turns clockwise with angular velocity wo = 2.8 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops?