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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards