Question

Three children are riding on the edge of a merry-go-round that is 105 kg, has a 1.70-m radius, and is spinning at 24.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? Ignore friction, and assume that the merry-go-round can be treated as a solid disk and the children as points.

Answer #1

Moment of inerta of the merry-go-round = 1/2 Mr^{2}, M =
105 kg, r = 1.7 m

Moment of inertia of all the three children = m1r^{2} +
m2r^{2} + m3r^{2}

m1 = 22 kg, m2 = 28 kg, m3 = 33 kg

Total initial moment of inetia, Ii = [M/2 + m1 + m2 + m3]
r^{2}

= [105/2 + 22 + 28 + 33] x (1.7)^{2}

= 391.595 kg m^{2}.

Initial angular velocity, i = 24 rpm

Final moment of inertia

When the child with mass 28 kg is movend to the center, he is not
contributing to the net moment of inertia since r = 0 at the
center.

If = = [105/2 + 22 + 33] x (1.7)^{2}

= 310.675 kg m^{2}.

Take f is the final
angular momentum.

Since there is no external torque, angular momentum is
conserved.

Ii x i = If x
f

f = [Ii x
i] / If

= [391.595 x 24] / 310.675

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