Question

A person holding 2 spheres of mass 2 kg each in his hands is standing on a rotating table with angular velocity of 10 rad s. His arms are extendedand each sphere is at a distance of 1 m from axis of rotation. If moment of inertia of the person about axis of rotation be 1 kgm 2, what will be rotational energy of the system? If the person closes his arms such that each sphere becomes at a distance of 0.3 m from axis of rotation, what will be rotational energy now? Describe the change in this energy.

Answer 1

I₀ = 1 kgm²; ω₁ = 10 rad.s^-1

I₁ = I₀ + mr₁² + mr₁² = I₀ + 2mr₁²

= 1 + 2 x 2 x (1)² = 1 + 4

= 5 kgm²

 K₁ = \(\frac{1}{2}\)I₁ω₁² = \(\frac{1}{2}\) x 5 x 10 x 10

or K₁ = 250 J

I₂ = I₀ x 2mr₂² = 1 + 2 x 2 x (0.3)² = 1 + 0.36

= 1.36 kgm²

From conservation of angular momentum

I₂ω₂ = I₁ω₁

or \(\omega_2 = \frac{I_1\omega_1}{I_2} = \frac{5\times10}{1.36} = \frac{50}{1.36}\ rad.s^{-1}\)

\(\therefore\ K_2 = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}\times1.36\times\left(\frac{50}{1.36}\right)^2\) 

or K₂ = 919 J