Little confused on a physics problem.. any help?

Question

A uniform disk (like a record turntable) turns with a frequency of 4.8 rev/s around a frictionless spindle. A non-rotating rod of the same mass as the disk and length equal to the disk's diameter is placed on the freely spinning disk so both turn around the spindle.

a) First consider the idealized case where the rod sticks tightly to the disk immediately upon impact. How fast does the combination turn?

b) More realistically, there is a frictional force between the two surfaces such that the magnitude of the torque slowing down the disk is equal to the magnitude of the torque speeding up the rod (this is a consequence of Newton's 3rd Law.). Hence we have alpha*disk = -C alpha*rod, where alpha generically means angular acceleration. Find the constant C.

c) The disk slows and the rod speeds up until they are moving together. Find the frequency with which the combination turns in this case.

d) Suppose Idisk = .65 kg m2. What is the magnitude of the kinetic energy lost due to friction?

Answer 1

For the disk, I = 1/2 mr^2
For the rod, I = 1/12 md^2 = 1/3 mr^2

a) It's basically a conservation of angular momentum problem
ie I*ω must be the same before and after

Initiall I*ω = 1/2 mr^2 * 4.8
After I*ω = (1/2 + 1/3) mr^2 * rps
rps = 2.88 rev/sec

b) The friction between the two is constant, and results in a torque T, which acts equally and "oppositely" on each.
T = - I_d*α_d = + I_r*α_r
-α_d / α_r = I_r / I_d
= 1/3 / (1/2) = 2/3
C = 2/3

c) Let's suppose the friction acts for a time t until both objects attain an angular vel ω
ω = 0 + α_r*t
or C*ω = C*α_r*t = α_d*t
also
ω = 4.8*2π - α_d*t

Eliminating α_d*t,
5/3 * ω = 4.8*2π
ω = 2.88*2π rad/s
= 2.88 rev/sec

d) We can work out hte change in kinetic energy to find the frictional loss.
KE = 1/2 * I*ω^2
If I_d = 0.65
I_r = 2/3 * 0.65
Initial KE = 1/2 * 0.65 * (4.8*2*π)^2
= 295.6 J

Final KE = 1/2 * (0.65*5/3) * (2.88*2*π)^2
= 177.4 J

Change in KE = 118.2 J = loss due to friction