Question

A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction.

Answer 1

Suppose, initial moment of inertia of the child is I1 Then final moment of inertia,

`I_2 = 2/3 I_l`

Also `v_1 = 40 " rev min"^(-I)`

By using the principle of conversion of angular, momentum, we get

`I_1omega_1 = I_2omega_2  or I_1(2piv_1) = I_2(2piv_2)`

or `v_2 = (I_1v_1)/I_2 = (I_1xx40)/(2/5 xx I_1) = 100 rev min^(-1)`

Answer 2

100 rev/min

Initial angular velocity, ω₁= 40 rev/min

Final angular velocity = ω₂

The moment of inertia of the boy with stretched hands = I₁

The moment of inertia of the boy with folded hands = I₂

The two moments of inertia are related as:

`I_2 = 2/5 I_1`

Since no external force acts on the boy, the angular momentum L is a constant.

Hence, for the two situations, we can write:

`I_2omega_2 = I_1omega_1`

`omega_2=I_1/I_2 omega_1`

`= I_1/(2/5 I_1) xx 40 = 5/2 xx  40`

= 100 rev/min