Conservation of Angular Momentum of ballerina

[Nontrad]

Homework Statement

I am new to Physics Forums and was wondering if anyone would be willing to help me with this problem.

A ballerina begins a tour jete with an angular speed ω(initial) and a rotational inertia consiting of two parts: I(leg) = 1.44 kg*m^2 for her leg extended outward at angle theta=90.0 degrees to her body and I(trunk)= 0.660 kg*m^2 for the rest of her body (primarily her truck). Near her maximum height she holds both legs at angle theta=30.0 degrees to her body and has angular speed ω(final). Assuming that I(trunk) has not changed, what is the ratio ω(final)/ω(initial).

Homework Equations

L(final) = L(initial)
L = Iω
The change in theta = -60 degrees.

The Attempt at a Solution

L(final leg) + L(final trunk) = L(initial leg) + L(initial trunk)
so,
(ω(final))(I(final leg) + 0.660 kg*m^2) = (ω(initial))(2.1 kg*m^2)

I have 3 unknowns and do not know what to do with the angles. Can anyone help me with the next thought?

 

[Nontrad]

It took me a while, but I figured this one out. It actually wasn't as difficult as I thought. I still don't know why only one leg is considered in the initial equation while both legs are considered in the final equation.