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

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:

I2 = (2/5) I1

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

Hence, for the two situations, we can write:

I2ω2 = I1 ω1

ω2 = (I1/I2) ω1

= [ I1 / (2/5)I1 ] × 40 = (5/2) × 40 = 100 rev/min

Answer 2

Solution

Ans.)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:

I2 = (2/5) I1

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

Hence, for the two situations, we can write:

I2ω2 = I1 ω1

ω2 = (I1/I2) ω1

= [ I1 / (2/5)I1 ] × 40 = (5/2) × 40 = 100 rev/min