The outstretched hands and arms of a figure skater preparing for a spin can be c

Question

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 his hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass 7.0 kg. When outstretched, they span 1.8 cm; when wrapped, they form a cylinder of radius 24 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 kg.m^2. If his original angular speed is 0.50 rev/s, what is his final angular speed? Express your answer using two significant figures.

Explanation / Answer

From conservation of angular momentum (I*w)i = (I*w)f

I (initial) = Ibody + I arm = 0.4 + 1/12*m*l^2

= 0.4 +((1/12)*7*1.8^2) = 2.29 kg-m^2

I (final) = 0.4 + m*r^2 = 0.4 + 7*0.24^2 = 0.8032 kg-m^2

2.29*0.5 = 0.8032*wf

wf = 1.4 rev/s

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