A merry-go-round with a a radius of R = 1.95 m and moment of inertia I = 192 kg-m2 is spinning with an initial angular speed of ω = 1.41 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 63 kg and velocity v = 4.7 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round. What is the angular speed of the merry-go-round after the person lets go?

Question

Evan · Accepted Answer

This is a conservation of angular momentum question. Momentum will stay the same before and after.
L_initial,merry-go-round: I*ω=(192)(1.41)=193.41
L_initial,person: m*v*r=(63)(4.7)(1.95)=577.395
L_final: (I_merry-go-round+m*r^2)*ω=(192+239.56)(ω)
193.41+577.395=431.56ω
ω=1.79
Hopefully I didn't make any math errors, but this is how to solve it.

Damon · Answer

There is no change in the angular velocity because the person is moving at the rim speed before and after and therefore the total angular momentum does not change before and after. If he jumped on from a dead stop, the rotation would slow down but he is already at the right speed.