Question

Three children are riding on the edge of a merry-go-round that is 105 kg, has a 1.60-m radius, and is spinning at 22.0 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

Using angular momentum conservation:

Li = Lf

Ii*wi = If*wf

Ii = initial moment of inertia = I_m + I1 + I2 + I3

Ii = 0.5*m*r^2 + m1*r^2 + m2*r^2 + m3*r^2

Ii = r^2*(0.5*m + m1 + m2 + m3)

Ii = 1.60^2*(0.5*105 + 22 + 28 + 33)

Ii = 346.88 kg-m^2

when 2nd children moves to the center, moment of inertia will be

If = final moment of inertia = I_m + I1 + I2 + I3

here I2 = 0, since for 2nd child, r = 0

If = 0.5*m*r^2 + m1*r^2 + 0 + m3*r^2

If = r^2*(0.5*m + m1 + m3)

If = 1.60^2*(0.5*105 + 22 + 33)

If = 275.2 kg-m^2

wi = initial angular velocity = 22 rpm

wf = ?

wf = wi*(Ii/If)

wf = 22*(346.88/275.2)

wf = 27.73 rpm

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