a merry-go-round has a I = 319 kg*m^2 on its own, and is now spinning at 1.82 rad/s with a 34.7 kg kid standing 0.500 m from the axis.

Question

the kid then walks outward until he reaches 1.25 m from the axis. what is the new angular velocity of the system?

[?] rad/s

Answer 1

Use Conservation of Angular Momentum and assume the absence of external forces. The angular momentum of this system = I*w where I = moment of inertia of merry-go-round + child. The initial moment of inertia = 319+34.7*.5²= 327.7 kg-m². The initial angular velocity is given at 1.82 rad/sec thus the Angular Momentum = 327.7*1.82= 596.4 kg-m²/sec. Now, when the child moves to a radius of 1.25 m, the moment of inertia becomes 319+34.7*1.25²= 373.2 kg-m² but the angular momentum of the system is unchanged. Thus the angular velocity changes to be 596.4/373.2= 1.6 rad/sec.