Question

6 kids with a mass of 40kg each, climb a merry-go-round w/ radius of 1.5 meters & moment of inertia of 685kg*m^2 and are on the outer edge of the ride. It then gets an angular velocity of 6 revs per minute and the kids walk closer to the middle and stop when they are 0.75 meters from axis of rotation. What is the new angular velocity of the ride?

Answer 1

Before the kids start walking,

Moment of inertia of the merry-go-round, I_mog = 685 kg-m²

MOI of one child while they are the the periphery, I_c = mR² = 40*1.5² = 90 kg-m²

MOi of 6 children = 6*I_c

Total MOI, I_i = I_mog + 6*I_c = 685 + 6*90 = 1225 kg-m²

angular velocity, w_i = 6rpm = 0.6283 rad/s

After the kids start walking,

Moment of inertia of the merry-go-round, I_mog = 685 kg-m²

MOI of one child, I_c'= m(0.75)² = 40*0.75² = 30 kg-m²

MOI of 6 children = 6*I_c'

Total MOI, I_f = I_mog + 6*I_c' = 685 + 6*30 = 865 kg-m²

angular velocity, w_f

Applying conservation of momentum

I_i*w_i = I_f*w_f

1225*0.6283 = 865*w_f

w_f = 0.8897 rad/s = 8.496 rev per minute