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

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

Initial angular speed, w1 = 0.4 rev/s Initial moment of inertia of the system, I1 = 0.35 + m*L^2/12 = 0.35 + 7.5*1.8^2/12 = 2.375 kg.m^2 final moment of inertia of the system, I2 = 0.35 + m*R^2 = 0.35 + 7.5*0.25^2 = 0.819 kg.m^2 Apply conservation of angular momentum I1*w1 = I2*w2 ==> w2 = w1*(I1/I2) = 0.4*2.375/0.819 = 1.16 rev/s

ID: 1395965 • Letter: T

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. (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 25.0 cm .

The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.350 kgm2. If the skater's original angular speed is 0.400 rev/s , what is his final angular speed?

Explanation / Answer

Initial angular speed, w1 = 0.4 rev/s

Initial moment of inertia of the system, I1 = 0.35 + m*L^2/12

= 0.35 + 7.5*1.8^2/12

= 2.375 kg.m^2

final moment of inertia of the system, I2 = 0.35 + m*R^2

= 0.35 + 7.5*0.25^2

= 0.819 kg.m^2

Apply conservation of angular momentum

I1*w1 = I2*w2

==> w2 = w1*(I1/I2)

= 0.4*2.375/0.819

= 1.16 rev/s <<<<<<<<<----------Answer

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

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

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