Answer to Question #81073 in Mechanics | Relativity for Lipsita

Question

A child of mass 50 kg is standing on the edge of a merry go round of mass 250kg and radious 3.0m which is rotating with an angular velocity of 3.0rad/s.The child then starts walking towards the centre of the merry go round .What will be the final angular velocity of the merry go round when the child reaches the centre.

Answer 1

Consider the merry-go-round as a uniform disk. Now according to conservation of angular momentum we can write:
I_1 ω_1=I_2 ω_2,
where I_1,ω_1 – moment of inertia and angular velocity of a merry-go-round with a boy on the edge correspondingly, I_2,ω_2 - moment of inertia and angular velocity of a merry-go-round with a boy in the centre respectively.
I_1=(MR^2)/2+mR^2,
where M – mass of the merry-go-round and m – boy’s mass.
I_2=(MR^2)/2,
since the boy in the centre does not affect the circular motion anymore.
Thus, express ω_2 through M,m, and R:
ω_2=ω_1 I_1/I_2 ⇔ ω_2=ω_1 (M+2m)/M=3 (250+2·50)/250=4.2 s^(-1).