Question

A 40.0-kg child running at 3.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg • m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?

Answer 1

Answer:

0.26087 rad/s

Explanation:

mass of the child (m) = 40 kg

velocity (v) = 3 m/s

distance (r) = 1.5 m

moment of inertia (I) = 600 kg.m^{2}

rotational momentum of the child = Iω

where

• moment of inertia of the child (I) = $mr^{2}$ = 40 x 1.5 x 1.5 = 90 kg/m^{2}

• angular velocity (ω) = velocity / distance = 3 / 1.5 = 2 rad/s

rotational momentum of the child = Iω = 90 x 2 = 180 kg$m^{2}$/s

from the conservation of momentum the initial momentum of the child must be the same as the final momentum of the child

initial momentum of the child = final momentum of the child

180 = (90 + 600) ω

180 = 690 ω

ω = 180 / 690 = 0.26087 rad/s