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Can anyone explain this angular momentum problem?
It was on a test, and I want to know if I got it right.
A man of mass m is initially standing on a solid disk, a distance R from the center of the disk (which has radius R). When he's standing there, the disk has angular velocity omega initial. He then walks in to a distance of R/3 from the center. What is the final angular velocity of the disk?
Moment of inertia for a solid disk is (1/2)MR^2.
1 Answer
- Anonymous1 decade agoFavorite Answer
initial angular momentum: (½MR² + mR)ω_i
final angular momentum: (½MR² + ⅓mR)ω_f
momentum is conserved, so set the expressions above equal and solving for ω_f gives
ω_f = (½MR² + mR)ω_i ÷ (½MR² + ⅓mR) = (3MR + 6m)ω_i ÷ (3MR + 2m)