Question

12) A 24.5-kg child (momentum of inertia is mR2 )is standing on the outer edge of a horizontal merry-go-round (treat the merry-go-around as solid cylinder, which has moment of inertia of 1/2mR2) about a vertical axis through its center and a radius of 2.50 m. The entire system (including the child) is initially rotating at 0.200 rev/s a) Find the angular velocity if the child moves to a new position 1.25 m from the center of the merry-go-round. b) If the merry-go-round plus the child system need to be stopped in 10.0 seconds, how much torque is needed.

Answer 1

a] By conservation of angular momentum,

(mr^2 + mR^2/2)*w2 = (mR^2 + mR^2/2)*w1

[m*1.25^2 + m*2.50^2/2]w2 = [m*2.50^2 + m*2.50^2/2]*0.200

w2 = [m*2.50^2 + m*2.50^2/2]*0.200 / [m*1.25^2 + m*2.50^2/2]

= 0.40 rev/s answer

b] torque needed = angular momentum/time

=  [m*2.50^2 + m*2.50^2/2]*0.200*2pi /10

=  [24.5*2.50^2 + 24.5*2.50^2/2]*0.200*2pi /10

= 28.86 Nm