Question

tion 9 of 12 > A brave child decides to grab onto an already spinning merry-go-round. The child is initially at rest and has a mass of 25.5 kg. The child grabs and clings to a bar that is 1.45 m from the center of the merry-go-round, causing the angular velocity of the merry-go-round to abruptly drop from 37.0 rpm to 17.0 rpm. What is the moment of inertia of the merry-go-round with respect to its central axis? moment of inertia: kg.mº

Answer 1

By the principle of conservation of angular momentum, total angular momentum of the merry go round without the child must be equal to the angular momentum of merry go round with the child i.e.,

Iw1 = (I + Ichid)w2 ... (1)

I denotes the moment of inertia of the merry go round, are the initial and final angular velocities.

انا ,1ند

We consider the child to be a point mass m and grabs the merry go round at a distance r from the center.

Hence, moment of inertia of the child is

Ichild = mr2 = 25.5 x 1.452 kg.m² = 53.61 kg.m

The angular velocity of merry go round decreases from 37 rpm to 17 rpm.

Hence, using equation (i)

I X 37 = (1 +53.61) x 17

or, 371 = 171 +911.37

or, I = 911.37 kg.m² = 45.57 kg.m 20

This is the moment of inertia of the merry go round.