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

Question

10.43

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 7.0kg . When outstretched, they span 1.6m ; when wrapped, they form a cylinder of radius 26cm . 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 0.50rev/s , what is his final angular speed?

Express your answer using two significant figures.

Explanation / Answer

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

I (initial) = Ibody + I arm = 0.400 kg-m^2 + 1/12*m*l^2 = 0.400 kg-m^2 + 1/12*7.00kg*(1.60m)^2 = 1.893 kg-m^2

I (final) = 0.400kg-m^2 + m*r^2 = 0.400 + 7.00*0.26^2 = 0.873 kg-m^2

Now1.893*0.500 = 0.873*omega(f)

So omega(f) = 1.893*0.500 /0.873

= 1.084 rev/s

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