How Does a Bug Affect the Spin of a CD?

[dorian_stokes]

Homework Statement

A CD (radius 6.0 cm) is spinning freely with an angular velocity of 420 rpm when a bug drops onto the CD a distance 4.4 cm from the center. If the CD slows to 280 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.)

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi dorian_stokes!

Use conservation of angular momentum …

what do you get?
​

 

[kerimek]

I would approach this problem by first identifying the relevant equations and variables. The equation that relates angular velocity (ω), radius (r), and tangential velocity (v) is v = ωr. In this case, the initial tangential velocity of the bug is v1 = (420 rpm)(6.0 cm) = 2520 cm/s, and the final tangential velocity is v2 = (280 rpm)(6.0 cm) = 1680 cm/s. We can use the conservation of angular momentum to find the ratio of the bug's mass (mbug) to the mass of the CD (mCD). The equation for conservation of angular momentum is L = Iω, where I is the moment of inertia. Since the CD is spinning freely, we can assume that the moment of inertia does not change. Therefore, we can set the initial angular momentum of the system (CD + bug) equal to the final angular momentum of the system. The initial angular momentum is L1 = mbugv1r + mCDv1r, and the final angular momentum is L2 = mbugv2r + mCDv2r. Setting these two equal and solving for the ratio of masses, we get mbug/mCD = (v1/v2)(1 - v2/v1) = (2520 cm/s)/(1680 cm/s) * (1 - (1680 cm/s)/(2520 cm/s)) = 3/2 * (1 - 2/3) = 1/2. Therefore, the ratio of the bug's mass to the mass of the CD is 1:2.