Question

A man, holding spheres of 10 kg each in his hands, is standing on a rotating table completing 1 cycle in 2 s. His arms are extended and each sphere is at a distance of 3 m from axis of rotation. If the person through away spheres, what will be new angular velocity of the table? Moment of inertia of man with table is 15 kgm².

Answer 1

Moment of inertia of man with extended arms,

I₁ = I_man + mr² + mr²

or I₁ = 15 + 2 x 10 x (3)² = 15 + 10 x 9

= 195 kgm²

^{\(\omega = \frac{2\pi n}{t} = \frac{2\pi\times1}{2} = \pi \ rad.s^{-1}\)} 

After throwing the spheres from hands, its moment of inertia

I₂ = 15 kgm²

ω₂ = 2πn²

From principle of conservation of angular momentum,

L₁ = L₂

or I₁ω₁ = I₂ω₂

∴ 195 x π = 15 x 2π x n²

\(\therefore\ n^2 = \frac{195}{30} = 6.5\ c.s^{-1}\)