CT131 A Disk Is Spinning As Shown With Angular

ct13-1 a disk is spinning as shown with angular velocity . it begins to slow down. w hile it is slowing, what is the direction of its vect

CT13-1 A disk is spinning as shown with angular velocity . It begins
to slow down.
W hile it is slowing, what is the direction of its vector
angular acceleration ?
Pink:  Yellow: 
Green:  Blue: 
Purple: Some other direction.
Answer: Up .

CT13-2 A planet in elliptical orbit about the Sun is in the position
shown.
W
ith the origin located at the Sun, the vector torque on the planet..
Pink: is zero. Yellow: points along +z.
Green: is in the x-y plane. Purple: None of these.
A
nswer: torque is zero! The vectors are anti-parallel. So =180,
sin=0 in =rRsin. Central-forces always give zero torque.
How does the magnitudes of the angular momentum of the planet Lplanet
(with the origin at the Sun) at positions A and B compare?

Pink: LA=LB
Yellow: LA>LB
Green: LA Answer: The angular momentum is constant, so LA=LB . The torque is
zero (last question) so L must be constant , by conservation of
angular momentum. The speed v of the planet is not constant, and the
"moment arm" r (the line from the sun to the planet) is not constant.
But the combination remains constant.
CT13-3 Three identical wheels are all spinning with the same angular
velocity . The total angular momentum of the 3-wheel system
has magnitude L.
O
ne of the three wheels is flipped upside-down, while the magnitude of
its angular velocity remains constant.
The new angular momentum of the 3-wheel system has magnitude..
Pink: L (the same as before) Green: (2/3)L Yellow: (1/3)L
Purple: some other value.
Answer: (1/3)L
Before the flip. L=Ltotal=3L1 where L1 is the angular momentum of one
of the wheels. (Since all 3 ang.mom. vectors are in the same
direction, they just add like numbers).
When one of the wheels is flipped. The flipped L2 cancels L1 and the
net is just L3 = (1/3)Ltot.

CT13-4
C onsider a solid disk with an axis of rotation through the
center (perpendicular to the diagram). The disk has mass M and radius
R. A small mass m is placed on the rim of the disk. What is the moment
of inertia of this system?
Pink: (M+m)R2
Green: less than (M+m)R2
Yellow: greater than (M+m)R2
Answer: Less than (M+m)R2. Itot = Idisk + Imass = (1/2)MR2 + mR2.
Suppose that mass-disk system is rotating and the axle is
frictionless. Atom-Ant carries the mass m toward the center of the
rotating disk. As Atom-Ant moves inward, the magnitude of the angular
momentum L of the system..
Pink: increases Green: decreases Yellow: remains constant
Answer: If the system is isolated from outside torques, the total
ang.mom. L remains constant.
As Atom-Ant moves inward the kinetic energy of the system..
Pink: increases Green: decreases Yellow: remains constant
(Hint: does Atom-Ant do work? )
Answer: KE increases. Although L=constant, the ant does positive work
in pushing the mass toward the center. The work-energy theorem says KE=Wnet.
So the KE of the ant+disk increases.
Suppose the disk was on a phonograph player, so that it always turned
at 33 rpm. As Atom-Ant moves inward, the speed of the mass m Pink:
increases Green: decreases Yellow: remains constant
Answer: decreases. Speed v=(2r)/T. Constant rotation rate means
constant period T. As r gets smaller (closer to axis) , v decreases.
Now we can see, qualitatively, why (if the system is isolated) the
disk speeds up when the mass m moves toward the center. If the speed v
decreases as the mass moves toward the center , there must be a
(tangential, not radial) force on the mass causing it to slow down.
The disk is exerting a force tending to slow the mass m down. By
Newton's third law, there must be a equal and opposite force from the
mass m on the disk, tending to speed the disk up.
CT13-5.
A star is rotating with a period T. Over a period of a million years,
its radius decreases by a factor of 2. What is the new period of the
star? (Hint: )
Pink:T/2 Green: 2T
Blue: 4T Yellow: T/4
Purple: None of these.
Answer: The new, shorter period is T/4. Angular momentum L = I  =
(2/5)MR2 2/T = constant, therefore R2/T = constant. If R decreases by
a factor of 2, then R2 decreases by a factor of 4, so T must decrease
by a factor of 4.