Question

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.1 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 4.0 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

Answer:

The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s

Explanation:

Given;

moment of inertia of a skater with arms out, $I_{arms \ out}$ = 3.1 kg.m²

moment of inertia of a skater with arms in, $I_{arms \ in}$ = 0.9 kg.m²

inward angular speed, $\omega _{in}$ = 4 rev/s

The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.

$L_{out} = L_{in}$

$I_{out} \omega_{out} = I_{in} \omega_{in}\\\\\omega_{out} = \frac{ I_{in} \omega_{in} }{I_{out} } \\\\\omega_{out} = \frac{0.9*4}{3.1} \\\\\omega_{out} = 1.161 \ rev/s$

Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s