If his original angular speed is 0.40 rev/s, what is his final angular speed?

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. 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 8.0 kg. When outstretched, they span 1.8 m; when wrapped, they form a cylinder of radius 25 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40kg.m/s^2. If his original angular speed is 0.40 rev/s, what is his final angular speed?

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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=2.16 kgm^2

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

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Total Moment of inertia of the outstretched arms and body = I1 + I2 = 2.16+0.4 = 2.56 kgm^2

initial angular speed =wi=2pi* 0.40=0.8pi rad/s=2.5133 rad/s

initial angular momentum Li=Iwi=2.56*2.5133=6.434 kgm^2/s

Let final angolar speed=wf=?

radius of hollow cylinder=r=0.25 m

moment of inertia of hollow cylinder=mr^2=8*(0.25)^2=0.5 kgm^2

final total momentof inertia =0.4+0.5=0.9 kgm^2

final angular momentum=Lf=I!wf=0.9wf

Lf = Li

0.9wf = 6.434

wf =6.434/0.9 =7.1489 rad/s

final angular speed of skater is 7.1489 rad/s
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Just substitute the figures in the formulae given in the link below.

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