Reading and Review

The moment of inertia I:
The total kinetic energy of a rolling object is the sum of its linear and rotational kinetic energies:

Rolling Down
Two spheres start rolling down a ramp from the same height at the same time. One is made of solid gold , and the other of solid aluminum .
Which one reaches the bottom first?
a) solid aluminum b) solid gold c) same d) can’t tell without more information

Rolling Down
Two spheres start rolling down a ramp from the same height at the same time. One is made of solid gold , and the other of solid aluminum .
Which one reaches the bottom first?
a) solid aluminum b) solid gold c) same d) can’t tell without more information
Moment of inertia depends on initial PE: mgh mass and distance from axis final KE: squared. For a sphere:
I = 2/5 MR 2
But you don’t need to know that!
All you need to know is that it depends on MR 2
MR 2 cancels out!
Mass and radius don’t matter, only the distribution of mass (shape)!

Moment of Inertia
Two spheres start rolling down a ramp at the same time. One is made of solid aluminum , and the other is made from a hollow shell of gold .
Which one reaches the bottom first?
a) solid aluminum b) hollow gold c) same d) can’t tell without more information

Moment of Inertia
Two spheres start rolling down a ramp at the same time. One is made of solid aluminum , and the other is made from a hollow shell of gold .
Which one reaches the bottom first?
a) solid aluminum b) hollow gold c) same d) can’t tell without more information initial PE: mgh final KE:
Larger moment of inertia -> lower velocity for the same energy.
A solid sphere has more of its mass close to the center. A shell has all of its mass at a large radius.
A shell has a larger moment of inertia than a solid object of the same mass, radius and shape


Power output of the Crab pulsar
•Power output of the Crab pulsar, in radio and X-rays, is about 6 x 10 31 W
(which is about 150,000 times the power output of our sun) . Since the pulsar is out of nuclear fuel, where does all this energy come from ?
• The angular speed of the pulsar, and so the rotational kinetic energy, is going down over time. This kinetic energy is converted into the energy coming out of that star.
• calculate the rotational kinetic energy at the beginning and at the end of a second, by taking the moment of inertia to be 1.2x10
38 kg-m 2 and the initial angular speed to be 190 s -1 . Δω over one second is given by the angular acceleration.
KE i

1
I  2
2
KE f

1
2
 KE   I
I
      2 
1
I  2 
   
2
   I    2
1
2
I

2      
1
2
I    2

Power output of the Crab pulsar
•Power output of the Crab pulsar, in radio and X-rays, is about 6 x 10 31 W
(which is about 150,000 times the power output of our sun) . Since the pulsar is out of nuclear fuel, where does all this energy come from ?
• The angular speed of the pulsar, and so the rotational kinetic energy, is going down over time. This kinetic energy is converted into the energy coming out of that star.
• calculate the rotational kinetic energy at the beginning and at the end of a second, by taking the moment of inertia to be 1.2x10
38 kg-m 2 and the initial angular speed to be 190 s -1 . Δω over one second is given by the angular acceleration.

We know that the same force will be much more effective at rotating an object such as a nut or a door if our hand is not too close to the axis.
This is why we have longhandled wrenches, and why doorknobs are not next to hinges.

The torque increases as the force increases, and also as the distance increases.

F sinθ
F cosθ   
Right Hand Rule
You can think of this as either:
- the projection of force on to the tangential direction
OR
- the perpendicular distance from the axis of rotation to line of the force

If the torque causes a counterclockwise angular acceleration, it is positive; if it causes a clockwise angular acceleration, it is negative.

You are using a wrench to loosen a rusty nut. Which arrangement will be the most effective in tightening the nut?
Using a Wrench a c d e) all are equally effective b

You are using a wrench to loosen a rusty nut. Which arrangement will be the most effective in tightening the nut?
Using a Wrench a
Because the forces are all the same, the only difference is the lever arm. The arrangement with the largest lever arm ( case #2 ) will provide the largest torque .
c d e) all are equally effective b

The gardening tool shown is used to pull weeds. If a
1.23 N-m torque is required to pull a given weed, what force did the weed exert on the tool?
What force was used on the tool?

Consider a mass m rotating around an axis a distance r away.
Newton’s second law: a = r
Or equivalently,

Once again, we have analogies between linear and angular motion:

The L-shaped object shown below consists of three masses connected by light rods. What torque must be applied to this object to give it an angular acceleration of 1.2 rad/s 2 if it is rotated about
(a) the x axis,
(b) the y axis
(c) the z axis (through the origin and perpendicular to the page)
(a)
(b)
(c)

Only the tangential component of force causes a torque
F sinθ
F cosθ

 r F
 rF

 rF sin


Angular motion is analogous to linear motion

The L-shaped object shown below consists of three masses connected by light rods. What torque must be applied to this object to give it an angular acceleration of
1.2 rad/s 2 if it is rotated about an axis parallel to the y axis, and through the 2.5kg mass?

The L-shaped object shown below consists of three masses connected by light rods. What torque must be applied to this object to give it an angular acceleration of
1.2 rad/s 2 if it is rotated about an axis parallel to the y axis, and through the 2.5kg mass?

A 2.85-kg bucket is attached to a disk-shaped pulley of radius 0.121 m and mass 0.742 kg . If the bucket is allowed to fall,
(a) what is its linear acceleration?
(b) What is the angular acceleration of the pulley?
(c) How far does the bucket drop in 1.50 s?

A 2.85-kg bucket is attached to a disk-shaped pulley of radius 0.121 m and mass 0.742 kg . If the bucket is allowed to fall,
(a) what is its linear acceleration?
(b) What is the angular acceleration of the pulley?
(c) How far does the bucket drop in 1.50 s?
(a) Pulley spins as bucket falls
(b)
(c)

Static equilibrium describes an object at rest – neither rotating nor translating.
If the net torque is zero, it doesn’t matter which axis we consider rotation to be around; you choose the axis of rotation
This can greatly simplify a problem
X

axis of rotation center of mass m
1
...
m j m
1
...
m j
X x j
F j
= m j g x j axis of rotation
X x cm
F = Mg

If an extended object is to be balanced, it must be supported through its center of mass.

This fact can be used to find the center of mass of an object – suspend it from different axes and trace a vertical line. The center of mass is where the lines meet.

Balancing Rod
A 1-kg ball is hung at the end of a rod
1-m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod?
a) ¼ kg b) ½ kg c) 1 kg d) 2 kg e) 4 kg
1m
1kg

Balancing Rod
A 1-kg ball is hung at the end of a rod
1-m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod?
a) ¼ kg b) ½ kg c) 1 kg d) 2 kg e) 4 kg
The total torque about the pivot must be zero !!
The CM of the rod is at its center, 0.25 m to the right of the pivot . Because this must balance the ball, which is the same distance to the left of the pivot , the masses must be the same !!
1 kg same distance
X m
ROD
CM of rod
= 1 kg

When you arrive at Duke’s Dude Ranch, you are greeted by the large wooden sign shown below. The left end of the sign is held in place by a bolt, the right end is tied to a rope that makes an angle of 20.0
° with the horizontal. If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg, what is
(a) the tension in the rope, and
(b) the horizontal and vertical components of the force, exerted by the bolt?

When you arrive at Duke’s Dude Ranch, you are greeted by the large wooden sign shown below. The left end of the sign is held in place by a bolt, the right end is tied to a rope that makes an angle of 20.0
° with the horizontal. If the sign is uniform, 3.20 m long, and has a mass of 16.0 kg, what is
(a) the tension in the rope, and
(b) the horizontal and vertical components of the force exerted by the bolt?
Torque, vertical force, and horizontal force are all zero
But I don’t know two of the forces!
I can get rid of one of them, by choosing my axis of rotation where the force is applied.
Choose the bolt as the axis of rotation, then:
(b)

A force is applied to a dumbbell for a certain period of time, first as in
(a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed ?
Dumbbell I a) case (a) b) case (b) c) no difference d) it depends on the rotational inertia of the dumbbell

A force is applied to a dumbbell for a certain period of time, first as in
(a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed ?
Dumbbell I a) case (a) b) case (b) c) no difference d) it depends on the rotational inertia of the dumbbell
Because the same force acts for the same time interval in both cases, the change in momentum must be the same, thus the CM velocity must be the same.

F = ma implies Newton’s first law: without a force, there is no acceleration
Now we have
Linear momentum was the concept that tied together Newton’s Laws, is there something similar for rotational motion?

Consider a particle moving in a circle of radius r,
I = mr 2
L = Iω = mr 2 ω = rm(rω)
= rmv t
= rp t

For more general motion (not necessarily circular),
The tangential component of the momentum, times the distance

For an object of constant moment of inertia, consider the rate of change of angular momentum analogous to 2nd Law for
Linear Motion

If the net external torque on a system is zero, the angular momentum is conserved.
As the moment of inertia decreases, the angular speed increases, so the angular momentum does not change.

Angular momentum is also conserved in rotational collisions

Figure Skater
A figure skater spins with her arms extended. When she pulls in her arms, she reduces her rotational inertia a) the same b) larger because she’s rotating and spins faster so that her angular faster momentum is conserved. Compared to her initial rotational kinetic energy, her c) smaller because her rotational inertia is smaller rotational kinetic energy after she pulls in her arms must be:

Figure Skater
A figure skater spins with her arms extended. When she pulls in her arms, she reduces her rotational inertia a) the same b) larger because she’s rotating and spins faster so that her angular faster momentum is conserved. Compared to her initial rotational kinetic energy, her c) smaller because her rotational inertia is smaller rotational kinetic energy after she pulls in her arms must be:
KE rot
= I

2 = L 2
/
I (used L = I

).
Because L is conserved, smaller I means larger KE rot
. The “extra” energy comes from the work she does on her arms.

A torque acting through an angular displacement does work, just as a force acting through a distance does.
Consider a tangential force on a mass in circular motion:
= r F
Work is force times the distance on the arc:
W = s F s = r Δθ
W = (r Δθ) F = rF Δθ =
Δθ
The work-energy theorem applies as usual.

Power is the rate at which work is done, for rotational motion as well as for translational motion.
Again, note the analogy to the linear form (for constant force along motion):

The direction of the angular velocity vector is along the axis of rotation . A right-hand rule gives the sign.
Right-hand Rule: your fingers should follow the velocity vector around the circle

Similarly, the right-hand rule gives the direction of the torque vector , which also lies along the assumed axis or rotation
Right-hand Rule: your fingers should follow the force vector around the circle

The linear momentum of components related to the vector angular momentum of the system

Applied tangential force related to the torque vector

Applied torque over time related to change in the vector angular momentum.

Cassette Player
When a tape is played on a cassette deck, there is a tension in the tape that applies a torque to the supply reel.
Assuming the tension remains constant during playback, how does this applied torque vary as the supply reel becomes empty?
a) torque increases b) torque decreases c) torque remains constant

Cassette Player
When a tape is played on a cassette deck, there is a tension in the tape that applies a torque to the supply reel.
Assuming the tension remains constant during playback, how does this applied torque vary as the supply reel becomes empty?
As the supply reel empties, the lever arm decreases because the radius of the reel
(with tape on it) is decreasing. Thus, as the playback continues, the applied torque diminishes.