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 23 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

Moment of inertia of outstreched hands and arms = I1 = ml^2/12 = 9*1.7^2 / 12 = 2.1675 kg-m2

MOI of the remaining body = I2 = 0.4 kg-m2

Total MOI =I1+I2 = 2.1675 + 0.4 = 2.5675 kg-m2

initial angular speed = Wo = 2piN = 2*3.14*0.5 = 3.14 rad/s

initial angular momentum = I*Wo = 2.5675*3.14 = 8.066 kg-m2/s

MOI of the hollow cylinder = mr^2 = 9*0.23^2 = 0.4761kg-m2/s
final MOI = 0.4+0.4761 = 0.8761 kg-m2/s
Final angular momentum = I*Wf = 0.8761Wf

by the Conservation of angular momentum , we have
Initial angular momentum = Final Angular momentum
8.066 = 0.8761Wf
Wf = 9.206 rad/s
Wf = 9.206/2pi rev/s = 1.465 rev/s

so, the final angular velocity is 1.465 rev/s

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