Question

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg?m2 is rotating freely with an angular speed of 3.0 rad/s. Two people, each having a mass of 60 kg are standing right outside the edge of the merry-go-round. One person radially steps on the edge merry-go-round with negligible speed and the angular speed changes to ?1. A few seconds later, the second person radially steps on the merry-go-round with negligible speed but at distance of 4.0 m from the axis of rotation (so there are now two people on the merry-go-round) and the angular speed changes to ?2. Determine ?1-?2.

Answer #1

Initial angular momentum Lo = I_merrygo*wo

after the first person steps

angular momentum L1 = (Imerrygo + m*R^2)*w1

from conservation of angular momentum

total angular momentum is conserved

L1 = Lo

(Imerrygo + m*R^2)*w1 = (Imerrygo)*wo

( 2000 + (60*5^2))*w1 = 2000*3

w1 = 1.714 rad/s

==================================

after the second person steps

angular momentum L2 = (Imerrygo + m*R^2 + m*r^2)*w2

r = 4 m

R = 5 m

from conservation of angular momentum

total angular momentum is conserved

L1 = Lo

(Imerrygo + m*R^2 + m*r^2 )*w2 = (Imerrygo)*wo

( 2000 + (60*5^2) +(60*4^2) )*w2 = 2000*3

w2 = 1.34 rad/s

==================================

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