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 (Figure 1) . 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 9.0 kg . When outstretched, they span 1.7 m ; 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 0.40kgm2.

If his original angular speed is 0.50 rev/s , what is his final angular speed?

Explanation / Answer

first from conservation of angular momentum (I*omega)i = (I*omega)f

I (initial) = Ibody + I arm = 0.4 kg-m^2 + 1/12*m*l^2 = 0.4 kg-m^2 + 1/12 * 9kg * (1.7m)^2 = 2.57 kg-m^2

I (final) = 0.4kg-m^2 + m*r^2 = 0.4 + 4 * 0.24^2 = 0.6304 kg-m^2

Now 2.57 * 0.5 = 0.6304*omega(f)

So omega(f) = 2.57 * 0.5 / 0.6304 = 2.03 rev/s

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