Question

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?

Answer 1

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