The outstretched hands and arms of a figure skater preparing for a final spin ca

Question

The outstretched hands and arms of a figure skater preparing for a final spin can be modeled as a thin rod rotating about an axis through its center. As the skater's hands and arms are brought in close to his body, he can be considered as a hollow cylinder. His hands and arms have a combined mass of 8.0kg. When outstretched, they span 1.8m: when wrapped, they form a cylinder of radius 25cm. The final moment of inertia about the rotation axis is constant at 0.40kg-m^2. If his original angular speed is 0.40rev/s, what is his final angular speed?

Explanation / Answer

oustretched: I = Io + mL^2/12 = 0.40 + 8kg * (1.8)^2 / 12 = 2.56 kg·m^2
brought in: I = 0.4 + 8*(0.25)^2 = 0.9 kg·m^2

Conserve momentum: initial L = final L .
2.56 * 0.4 = 0.9 * w
w = 1.14 rev/s

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