Question

A person is standing on a platform rotating with an angular speed of 1 rps. With stretched
hands carrying weights. On drawing his hands close to the body, the M. 1.of the system
changes from 8kg m² to 4kg m². The increase in his kinetic energy nearly is.
1) 320 J 2) 40 J
3) 80J
4) 160 J​

Answer 1

Given info : a standing on the rotating platform with an angular speed 1 rps. Moment of inertia of system changes 8 kgm² to 4 kgm².

To find : the increase in kinetic energy nearly to

solution : from law of conservation of angular momentum,

I₁ω₁ = I₂ω₂

Here, I₁ = 8 kgm², I₂ = 4kgm² , ω₁ = 1rps = 2π rad/s

So, 8 × 2π = 4 × ω₂

⇒ω₂ = 2 × 2π rad/s = 4π rad/s

Now change in kinetic energy = final kinetic energy - initial kinetic energy

= 1/2 I₂ω₂² - 1/2 I₁ω₁²

= 1/2 × 4 × (4π)² - 1/2 × 8 × (2π)²

= 2 × 16π² - 4 × 4π²

= 16π² ≈ 160 J

Therefore the increase in kinetic energy is 160 J