A child (mass - 40 kg) stands on the outer rim of a playground merry-go-round. T

Question

A child (mass - 40 kg) stands on the outer rim of a playground merry-go-round. The merry-go-round is a horizontal disk (mass = 50 kg, radius = 2.0 m) that is free to route around a vertical axis. While the disk is spinning, he walks straight toward the center of the merry-go-round, stopping when he is 1.0 from the axis. If the initial rotation speed was 1.2 rev/s. what is the final angular speed (in rev/s)? For this problem, we may consider the child as a particle, so his (initial) moment of inertia is I_= MR^2 = (40 kg)(1.0 m)^2 = 40 kg m^2, so the system's initial moment of inertia is I_i = 140 kg m^2. What are the initial and final kinetic energies of the system?

Explanation / Answer

(1)
i = 1.2 rev/s
f = ?

Using Momentum Conservation
Initial Momentum = Final Momentum
Ii* i = If * f
(Idisk + Ichild) * i = (Idisk + Ichild) * f
(1/2*M*R^2 + m*R^2)  i = (1/2*M*R^2 + m*R^2) * f

Substituing Values,
(1/2 * 50 * 2^2 + 40 * 2^2 ) * 1.2 = (1/2 * 50 * 2^2 + 40 * 1^2) * f
f = 2.23 rev/s

Final angular speed, f =  2.23 rev/s

(2)
Initial Kinetic Energy = 1/2 * I * i^2
K.Ein = 1/2 * (1/2 * 50 * 2^2 + 40 * 2^2 ) * 1.2^2 J
K.Ein = 187.2 J

Final Kinetic Energy = 1/2 * I * f^2
K.Efin = 1/2 * (1/2 * 50 * 2^2 + 40 * 1^2 ) * 2.23^2 J
K.Efin = 646.5 J

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