Posted by Alex on .
A CD (radius 6.0 cm) is spinning freely with an angular velocity of 450 rpm when a bug drops onto the CD a distance 4.6 cm from the center. If the CD slows to 260 rpm, what is the ratio of the bug's mass to the mass of the CD? (Ignore the effect of the hole in the center of the CD.)

Physics 
drwls,
Angular momentum is conserved. The bug and CD end up with the same angular velocity.
Icd*w1 = Icd*w2 + Mbug*Rbug^2*w2
(1/2)*Mcd*6^2 * 450 = (1/2)Mcd*6^2*260 + (4.6)^2*260*Mbug
Solve for Mbug/Mcd.
You don't have to change rpm to angular velocity (radians/sec) in this case. The conversion factors cancel out.