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Conservation of Angular Momentum |
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| Nov20-10, 10:34 AM | #1 |
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Conservation of Angular Momentum
1. The problem statement, all variables and given/known data
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). 2. Relevant equations L(final) = L(initial) L = Iω The change in theta = -60 degrees. 3. 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? |
| Nov24-10, 08:24 AM | #2 |
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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.
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