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.

Anonymous2011-05-03T12:59:11Z

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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)