Question

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

Answer 1

Using Angular momentum conservation on merry-go-round before and after 2nd child moves to the center:

Li = Lf

Ii*wi = If*wf

wi = initial angular speed of ride = 24.0 rpm

wf = final angular speed of ride = ? rpm

Ii = Initial angular momentum of ride = Im + I1 + I2 + I3

Im = moment of inertia of solid disk = M*R^2/2

M = mass of merry-go-round = 105 kg

R = radius of merry-go-round = 1.70 m

I = moment of inertia of child = m*R^2

m = mass of child, So

Ii = M*R^2/2 + m1*R^2 + m2*R^2 + m3*R^2

Ii = (M/2 + m1 + m2 + m3)*R^2

Ii = (105/2 + 22.0 + 28.0 + 33.0)*1.70^2/2 = 195.7975 kg-m^2

Now when 28.0 kg child moves to the center than moment of inertia due to the that child will be zero

If = M*R^2/2 + m1*R^2 + m2*0^2 + m3*R^2

If = (M/2 + m1 + m3)*R^2

If = (105/2 + 22.0 + 33.0)*1.70^2/2 = 155.3375 kg-m^2

Using these values:

wf = wi*(Ii/If)

wf = 24.0*(195.7975/155.3375)

wf = 30.25 rpm = final angular velocity

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