Question

A skater has outstretched arms preparing for a turn with a moment of inertia of

Answer 1

Answer:

The final angular speed is 16.1 rad/s

Explanation:

Given;

initial moment of inertia, I₁ = 2.56 kg.m²

final moment of inertia, I₂ = 0.40 kg.m²

initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s

Apply the principle of conservation of angular momentum;

I₁ω₁ = I₂ω₂

where;

ω₂ is the final angular speed

ω₂ = (I₁ω₁) / (I₂)

ω₂ = (2.56 x 2.514) / (0.4)

ω₂ = 16.1 rad/s

Therefore, the final angular speed is 16.1 rad/s