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 9.0 kg . When outstretched, they span 1.6 m ; when wrapped, they form a cylinder of radius 24 cm . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40kgm2.

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

Explanation / Answer

given are

m1=9kg

r1=1.9m

r2=0.25m

I=0,.40kgm^2

W=0.50rev/sec

First we will calculate skater’s initial moment of inertia

I1=0.40+(1/12)* 9*1.9^2

=3.10kgm^2

Final moment of inertial will be

I2=0.40+(8)*(0.25)^2

=0.9kgm^2

I1w1=I2w2

W2=i1w1/i2

=0.50*(3.10/0.9)

=1.72rev/sec--answer

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