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 7.0kg . When outstretched, they span 1.8m ; when wrapped, they form a cylinder of radius 26cm . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40kg?m 2 .

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

Explanation / Answer

The outstretched hands and arms are treated as a thin rod rotating about an axis through the center and perpendicular to its length.

Moment of inertia of the outstretched hands and arms=I1=ml^2/12=1.89 kgm^2

Moment of inertia of the remaining body =I2=0.40 kgm^2

Total Moment of inertia of the outstretched arms and body = I1 + I2 = 1.89+0.4 = 2.29 kgm^2

initial angular speed =wi=2pi* 0.50=1 pi rad/s=3.14 rad/s

initial angular momentum Li=Iwi=2.29*3.14=7.1904 kgm^2/s

Let final angolar speed=wf=?

radius of hollow cylinder=r=0.26 m

moment of inertia of hollow cylinder=mr^2=7*(0.26)^2=0.4732 kgm^2

final total momentof inertia =0.4+0.4732=0.8732 kgm^2

final angular momentum=Lf=I!wf=0.8732wf

Lf = Li

0.8732wf = 7.1904

wf =7.1904/0.8732 =8.2345 rad/s

final angular speed of skater is 8.2345 rad/s

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