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. (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 8.00 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 0.400 kgm2. If the skater's original angular speed is 0.400 rev/s , what is his final angular speed?

Explanation / Answer

use the sum of Moments = 0

P sin 85 + 3 cos 10 = P cos 85 * 3 sin 10 + 500 * 9.8 * 12 cos 10 + 1000* 9.81* 24* cos10

P * 2.898 = 289769.88

P = 99.89 kN

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

I (initial) = Ibody + I arm

I initial = 0.4 + 1/12*m*l^2

Iinitial = 0.4 + 1/12* 7.5 *(1.80)^2

I initial = 1.65 kg-m^2

I (final) = 0.4 + m*r^2

Ifinal = 0.4 + 8.00*0.25^2 = 0.9 kg-m^2

So Now

1.65 *0.400 = 0.9*W f

So Wf = 1.65 *0.4/0.9

Wf = 0.734 rev/s

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