A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg·

Question

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg·m2 and is rotating about a frictionless vertical axle. As a child of mass 25.0 kg stands at a distance of 1.00 m from the axle, the system (merry-go-round and child) rotates at the rate of 16.0 rev/min. The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

I can't figure out what I am doing wrong! I think it's just a simple mathematical error I keep making because I understand the problem conceptually. Please provide step-by-step explanation with plugged-in numbers + final result. Thanks!

Explanation / Answer

Moment of inertia I = m.i. of merry go round + m.i. of child

= 275 + m * r2

where m is mass of child and r is his distance from axis of rotation.

Initial m.i. I1   = 275 + 25.0 * 1.002

                               = 300 kg-m2

   and final m.i. I2   = 275   + 25.0 * 2.002

                             = 375 kg-m2

Initial angular speed 1   =   16.0   rev/min

                                      = 16.0 * 2 / 60

                                      = 1.676   rad/s

According to law of conservation of angular momentum

I1 * 1   = I2 * 2

Hence final angular speed 2   =  300 * 1.676 / 375

                                              = 1.34   rad/s

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