Question

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 5.4 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

To solve this problem we will apply the concepts related to the conservation of angular momentum, which is defined as the equality between the initial state and the final state of the angular momentum. Mathematically this is defined as:

`S_1 = S_2`

`I_1 \omega_1 = I_2 \omega_2`

Here,

I = Moment of inertia at each state

`\omega`= Angular velocity at each state

Replacing with our values

`(9kg\cdot m^2)(5.4rev/s) = (3kg\cdot m^2)(\omega^2)`

`\omega_2 = ((9kg\cdot m^2)(5.4rev/s))/( (3kg\cdot m^2))`

`\omega_2 = 16.2rev/s`

Therefore the angular speed is 16.2rev/s