Problem 10.29 The spinning figure skater. The outstretched hands and arms of a f

Question

Problem 10.29 The spinning figure skater. 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. (See the figure below (Figure 1) .) 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 thin-walled hollow cylinder. His hands and arms have a combined mass of 7.50 kg . When outstretched, they span 1.80 m ; when wrapped, they form a cylinder of radius 25.0cm . Part A The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.450 kg/m^2 If the skater's original angular speed is 0.450 revs/sec , what is his final angular speed?

Explanation / Answer

From conservation of angular momentum (I*omega)i = (I*omega)f I (initial) = Ibody + I arm = 0.450 kg-m^2 + 1/12*m*l^2 = 0.450 kg-m^2 + 1/12*8.00kg*(1.80m)^2 = 2.61 kg-m^2 I (final) = 0.450kg-m^2 + m*r^2 = 0.450 + 8.00*0.25^2 = 0.95 kg-m^2 Now 2.61*0.400 = 0.95*omega(f) So omega(f) = 2.61*0.400/0.95 = 1.10 rev/s

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