Question

A boy stands at the centre of a turn table with his two arms stretched. The turn table is set rotating with an angular with an angular speed of 40 r.p.m. How much is the angular speed of the boy if he folds his hands back and thereby reduces his moment of inertia to 2/5times the initial? Assume the turn table to rotate without friction.

Answer 1

Angular speed of the boy in the new situation is 100 rad / s.

By the conservation of Angular Momentum:

L = Iw = constant

Assuming w1 = Initial Angular Momentum, w2 = Final Angular Momentum, I1 = Initial Moment of Inertia, I2 = Final Moment of Inertia

I1.w1 = I2.w2
I1.40 = 0.4I1.w2
w2 = 100 rad / s.

Answer 2

Given conditions ⇒ 
Initial angular movement of the boy(ω₁) = 40 r.p.m. 
Final angular momentum of the boy = ω₂

Let the Initial moment of the Inertia be I₁.
∴ Final Moment of Inertia(I₂) = 2/5 × I₁.

Now,
We know,
I₁ × ω₁ = I₂ × ω₂
⇒ I₁ × 40 = 2/5 × I₁ × ω₂
∴ ω₂ = 40 × 5/2
∴ ω₂ = 100 

Hence, Angular speed of the boy is 100 r.p.m.


Hope it helps.