Rotational Motion – II

Reading: Chapter 11

Rolling

2 points of view:

(1) Combined rotation and translation

blurred

clear

(a) Translation: the center of mass moves with velocity

vcm.
(b) Rotation: the wheel rotates about the center of mass.

If the wheel rolls without slipping, s = R, then

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(2) Pure Rotation

Rolling can also be considered as a pure rotation, with

angular speed , about an axis through the contact point.
e.g. velocity at the top: vtop = ()(2R) = 2(R) = 2vcm.

Kinetic Energy of Rolling

If we consider the motion as a pure rotation about the

contact point,

Using the parallel axis theorem,

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Hence , and

The kinetic energy consists of:

(a)the kinetic energy of the translational motion of the
center of mass,
(b)the kinetic energy of the rotation about the center of
mass.

Friction and Rolling

(a)When the cyclist applies a torque on the wheel

intending to make it rotate faster, the bottom of the
wheel tends to slide to the left at point P. A frictional
force at P, directed to the right, opposes the tendency
to slide.
(b)The frictional force acts on the wheel and produces
the acceleration of the bicycle.

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Rolling Down a Ramp

The gravitational force

tends to make the wheel
slide down the ramp.
There is a frictional force
opposing this sliding,
and is thus directed up
the ramp.

Using Newton’s second

law for translational
motion,

(1)

Using Newton’s second

law for rotational
motion,

(2)

Since a = R, we obtain from (2): .

Substituting into (1),

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Example

11-2 A uniform ball, of mass M = 6.00 kg and radius R,

rolls smoothly from rest down a ramp at angle  = 30.0o.
(a) The ball descends a vertical height h = 1.20 m to
reach the bottom of the ramp. What is its final speed?
(b) What are the magnitude and direction of the frictional
force on the ball as it rolls down the ramp?
(a) Method 1: Conservation of energy

Other terms: Uf
= Ki = 0, Ui = Mgh. Hence

(answer)
Method 2: Newton’s law
Translational motion: (1)
Rotational motion: (2)
where . Since a = R, (2): .
(1):
(answer)

(b)
= 8.4 N (answer)

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The Yo-Yo

Using Newton’s second

law for translational
motion,

(1)

Using Newton’s second

law for rotational
motion,
R
R0
(2)

Since a = R0, we
obtain from (2):

Substituting into (1),

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Angular Momentum
    
l  r  p  m(r  v ).

l  mrv sin.

Alternatively, l  rp  rmv or

l  r p  r mv .

Newton’s Second Law


 dl
  .
dt

The vector sum of all the torques

acting on a particle is equal to the
time rate of change of the
angular momentum of that
particle.

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Proof
  
l  m(r  v ).

Differentiating with respect to time,

  
dl  dv dr 
 m r    v  .
dt  dt dt 


dl    
 mr  a  v  v .
dt
  
v  v  0 because the angle between v and itself is zero.

dl    
 mr  a   r  ma.
dt
 
Using Newton’s law, F  ma. Hence

  
Since   r  F , we arrive at

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The Angular Momentum of a System of Particles

Total angular momentum for n particles:

   
L  l1  l2   li .

Newtons’ law for angular motion:

 
 dli d  dL
 i   l  .
dt dt i dt

 includes torques acting on all the n particles. Both
i
internal torques and external torques are considered.
Using Newton’s law of action and reaction, the internal
forces cancel in pairs. Hence

 dL
  .
ext
dt

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The Angular Momentum of a Rigid Body

For the ith particle, angular

momentum:

li  ri pi sin90 o  ri mi vi .

The component of angular

momentum parallel to the
rotation axis (the z component):

The total angular momentum for

the rotating body

This reduces to

L  I .

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Conservation of Angular Momentum

 dL
  .
ext
dt

If no external torque acts on the system,


dL
 0.
dt

L  constant.
 
Li  L f .

Law of conservation of angular momentum.

Examples

1. The spinning volunteer: When the student pulls in

the dumbbells, the rotational inertia I decreases. Since
Ii i  I f  f , the angular velocity increases.

2. The springboard diver: When the diver is in the tuck

position, the rotational inertia decreases, and the angular
velocity increases.

When the diver is in the layout position, the rotational

inertia increases, and the angular velocity decreases.

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3. Long jump: When an athlete takes off, her angular
momentum gives her a forward rotation around a
horizontal axis.

In the air, the jumper shifts the angular momentum to her

arms by rotating them in a windmill fashion. Then the
body carries little angular momentum, keeping her body
upright. She can then land with her legs extended
forward.

4. Tour jeté: The dancer/gymnast leaps with one leg

perpendicular to the body. In the air, the outstretched leg
is brought down and the other leg is brought up, with
both ending up at an angle  to the body. The rotational
inertia decreases and the angular speed increases.

On landing, a leg is again outstretched and the rotation

seems to vanish.

See Youtube “Chen Helps China Sweep”, “Irving

Saladino” and “tour jete”.
See demonstration “Bicycle wheel gyroscope” and
“swinging Atwood machine”.

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Examples

11-7 A student sits on a stool that can rotate freely about

a vertical axis. The student, initially at rest, is holding a
bicycle wheel whose rim is loaded with lead and whose
rotational inertia I about its central axis is 1.2 kg m 2. The
wheel is rotating at an angular speed wh of 3.9 rev/s; as
seen from overhead, the rotation is counterclockwise.
The axis of the wheel is vertical, and the angular
momentum Lwh of the wheel points vertically upward.
The student now inverts the wheel; as a result, the
student and stool rotate about the stool axis. The
rotational inertia Ib of the student + stool + wheel system
about the stool axis is 6.8 kg m2. With what angular
speed b and in what direction does the composite body
rotate after the inversion of the wheel?

Using the conservation of

angular momentum,

(ans)

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11-8 A cockroach with mass m rides on a disk of mass
6m and radius R. The disk rotates like a merry-go-round
around its central axis at angular speed i = 1.5 rad s1.
The cockroach is initially at radius r = 0.8R, but then it
crawls out to the rim of the disk. Treat the cockroach as a
particle. What then is the angular speed?

Using the conservation of angular momentum,

Rotational inertia:
The disk:
The cockroach: and

Therefore,
(ans)

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Precession of a Gyroscope

Torque due to the gravitational

force

Angular momentum

For a rapidly spinning gyroscope,

the magnitude of is not
affected by the precession,

Using Newton’s second law for

rotation,

where is the
precession rate

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Example Rolling of a Hexagonal Prism (1998 IPhO)

Consider a long, solid, rigid, regular hexagonal prism

like a common type of pencil. The mass of the prism is
M and it is uniformly distributed. The length of each side
of the cross-sectional hexagon is a. The moment of
inertia I of the hexagonal prism about its central axis is I
= 5Ma2/12.
a) The prism is initially at rest with its axis horizontal on
an inclined plane which makes a small angle  with the
horizontal. Assume that the surfaces of the prism are
slightly concave so that the prism only touches the plane
at its edges. The effect of this concavity on the moment
of inertia can be ignored. The prism is now displaced
from rest and starts an uneven rolling down the plane.
Assume that friction prevents any sliding and that the
prism does not lose contact with the plane. The angular
velocity just before a given edge hits the plane is i
while f is the angular velocity immediately after the
impact. Show that f = si and find the value of s.
b) The kinetic energy of the prism just before and after
impact is Ki and Kf. Show that Kf = rKi and find r.
c) For the next impact to occur, Ki must exceed a
minimum value Ki,min which may be written in the form
Ki,min = Mga. Find  in terms of  and r.
d) If the condition of part (c) is satisfied, the kinetic
energy Ki will approach a fixed value Ki,0 as the prism
rolls down the incline. Show that Ki,0 can be written as
Ki,0 = Mga and find .
e) Calculate the minimum slope angle 0 for which the
uneven rolling, once started, will continue indefinitely.

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 Pf

30o
Pi
 E

a) Angular momentum about edge E before the impact

Angular momentum about edge E after the impact

Using the conservation of angular momentum, Li = Lf

 
b)

 
c) After the impact, the center of mass of the prism raises
to its highest position by turning through an angle 90o 
( + 60o) = 30o  . Hence


d) At the next impact, the center of mass lowers by a
height of asin. Change in the kinetic energy

Kinetic energy immediately before the next impact

When the kinetic energy approaches Ki,0,

 
e) For the rolling to continue indefinitely,


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where A = r/(1  r) = 121/168 and

 u  35.36o.
  + u  41.94o.
 0  6.58o.

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