Question

A skater has a moment of inertia of 105.0 kg.m^2 when his arms are outstretched and a moment of inertia of 70.0 kg.m^2 when his arms are tucked in close to his chest. If he starts to spin at an angular speed of 80.0 rpm (revolutions per minute) with his arms outstretched, what will his angular speed be when they are tucked in?

Answer 1

120 rpm

Step-by-step explanation:

I1 = 105 kgm^2, I2 = 70 kgm^2

f1 = 80 rpm, f2 = ?

Let the angular speed be f2 when his arms are tucked.

If no external torque is applied, then the angular momentum remains constant.

L1 = L2

I1 w1 = I2 w2

I1 x 2 x 3.14 x f1 = I2 x 2 x 3.14 x f2

105 x 80 = 70 x f2

f2 = 120 rpm