Physics Help! angular speed?

Question

A playground merry-go-round of radius 2.00 m has a moment of inertia I = 335 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 11.0 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?

Answer 1

Conservation of angular momentum:
(335+25.0*1.00^2)*11.0 = (335+25.0*2.00^2)*W
where w is the final angular speed of the system when the child reaches the edge (in rev/min).
Hence W = 9.10 (rev/min)

Answer 2

a million) b) factor B Angular velocity = radius* attitude / time As attitude and time for the two factors is comparable so we in basic terms take radius distinction. Angular velocity ~ radius As factor B is farther far flung from axis of rotation than factor A so factor B has better angular velocity. 2) b) factor B evaluate the rotation of blades as a circle. factor A is forming a smaller circle and factor B is forming an better circle. the two take comparable time to end 10 revolutions yet factor B is shifting with the aid of better distance.. than factor A.. velocity = distance / time V = d/t As factor B is shifting with the aid of better distance and velocity has a right away relation with the gap so, its tangential or linear velocity would be better. 3) factor B with the aid of fact the angular velocity of factor B is better than factor A and so the radius, subsequently the centripetal acceleration of factor B is better than factor A 4) factor B with the aid of fact the tangential velocity of factor B is better than factor A so the linear or tangential acceleration of factor B would be better.