Jill ( 25 kg) runs at 4 m/s in a straight line, and jumps tangentially onto the

Question

Jill (25 kg) runs at 4 m/s in a straight line, and jumps tangentially onto the rim of an initially stationary merry-go-round, which you may approximate as a solid disk (of mass 12.5 kg and an outer radius 2 m) mounted on a frictionless pivot.

a.) After Jill and the merry-go-round achieve a common rotation, what is the final:
angular speed of the system? rad/s
linear speed of the rim/Jill? m/s

b.) Suppose it took 15 ms for Jill and the merry-go-round to achieve a common speed. During this time, what is size of the (average) torque experienced:

by the merry-go-round? N

Explanation / Answer

inertia of merry-go-round = 0.5*12.5*2^2 = 25 kg m^2
inertia of jill = 25*2^2 = 100
so.. final inertia of system = 100 + 25 = 125 kg m^2

so.. consercing angular momentum
m*v*r = I_total * w
so.. 25 * 4 * 2 = 125 * w
so.. angular speed = w = 1.6 rad/sec

linear speed = w*2 = 1.6*2 = 3.2 m/sec

b)
torque by merry go round = change in angular momentum / time = 25*1.6 / 0.015 = 2666.666667 N m

torque by Jill = same = 2666.6667 Nm

forces are friction force between Jill and mery-go-round

c) ration = 0.5*125*1.6^2 / (0.5*25*4^2) = 0.8

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