skater physics problem?

Question

A skater is initially spinning at a rate of 14.7 rad/s with a rotational inertia of 2.72 kg·m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to 1.58 kg·m2?

Answer 1

Iω^2(before)=Iω^2(after)

2.72*(14.7)^2=1.58*(ω^2)
ω^2=2.72*14.7/1.58
ω^2=25.3
ω=5.03 rad/sec

Answer 2

Angular momentum is a constant, equal to the rotational inertia times the angular velocity. Her spin rate will therefore increase by a factor of 2.72/1.58.

14.7 x 2.72/1.58 = 25.3