Question

A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 255 kg · m2 and is rotating at 8.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round

Answer 1

I_merry_go_round = 255 kg-m2

R = 1.8 m

Mass of child m = 24 kg

So, moment of inertia of child I_child = m*R^2 = 77.76 kg-m2

Initial angular speed w_initial = 8 rev/min

By conservation of angular momentum (due to frictionless axle),

I_merry_go_round * w_initial = (I_merry_go_round + I_child) * w_final

=> 255 * 8 = (255 + 77.76) * w_final

So, w_final = 6.13 rev/min

Answer 2

here angular momentum is conserved.
I1*w1 = I2*w1

255*8 = (255+ 24*1.8^2)w2

w2 = 255*8/(255+24*1.8^2)

= 6.13 rev/min

= 6.13*2*pi/60

= 0.641 rad/s