If a trapeze artist rotates 1 each second while sailing through the air, and contracts to reduce her rotational inertia to 0.40 of what it was, how many rotations per second will result?

Explanation:

A trapeze is rotating with 1 rotation per second .

Thus its angular velocity ω = 2π n

here n is the number of rotations per second

Thus ω = 2π b because n = 1 in this case

Suppose the moment of inertia of his is = I

Then angular momentum L₁ = I  ω = 2 I π

In the second case , the moment of inertia becomes = 0.4 I

Let his angular velocity is  ω₀

Thus angular momentum L₂ = 0.4 I  ω₀

Because no external torque is applied , therefore angular momentum will remain constant .

Thus L₁ = L₂

Therefore  2 I π = 0.4 I x 2 n₀ π

here n₀ is the number of rotations per second

n₀ = = = 2.5