Physics
posted by Christina .
The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thinwalled hollow cylinder. His hands and arms have a combined mass of 8.50 kg. When outstretched, they span 1.80 m; when wrapped, they form a cylinder of radius 25.0 cm (.25m).
The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.350 kg*m^2 If the skater's original angular speed is 0.450 rev/s, what is his final angular speed?

Use conservation of angular momentum. Assume the outstretched arms have a moment of inertia contribution of
Iarms2 = (1/3) M(arms) L^2, where L = 1.80m m.
[(Ibody + Iarms1)w1] = [(Ibody + Iarms2)w2]
Solve for w2
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