Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.70 kg⋅m2. if she starts out spinning at 4.8 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

Given:

I₁ = 0.70 kg-m², the moment of inertia with arms and legs in

I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.

ω₁ = 4.8 rev/s, the angular speed with arms and legs in.

That is,

ω₁ = (4.8 rev/s) * (2π rad/rev) = 30.159 rad/s

Let ω₂ = the angular speed with arms and a leg out.

Because momentum is conserved, therefore

I₂ω₂ = I₁ω₁

ω₂ = (I₁/I₂) ω₁

= (0.7/3.5) * (30.159)

= 6.032 rad/s

ω₂ = (6.032 rad/s) * (1 / (2π) rev/rad) = 0.96 rev/s

Answer: 0.96 rev/s