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 9.0kg . When outstretched, they span 1.7m ; when wrapped, they form a cylinder of radius 24cm . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40kg?m2.

Part A

If his original angular speed is 0.50rev/s , what is his final angular speed?

Express your answer using two significant figures.

Explanation / Answer

The Initial and final moment of inertia of skater is

Ii = Ic + ( m r12 / 12 )

= 0.40 + ( 9 * 1.7 * 1.7 / 12)

= 2.567 kg-m2

If = Ic + ( m r22 / 12 )

= 0.40 + ( 9 * 0.24 * 0.24 / 12)

= 0.4432 kg-m2

According to conservation of angular momentum

Li = Lf

wi Ii = wf If

wf = wi Ii / If

= 0.50 * 2.567 / 0.4432

= 2.895 rev/s

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