The outstretched hands and arms of a figure skater preparing for a spin can be c

Q: The outstretched hands and arms of a figure skater preparing for a spin can be c

initial moment of inertia of body I1 = 0.4 + 9*1.8^2/12 = 2.83 kg.m^2 final moment of inertai a body I2 = 0.4 + 9*0.26^2 = 1.0084 kg.m^2 here initial angular momentum = final angular momentum I1*w1 = I2*w2 2.83*0.3 = 1.0084*w2 ==> w2 = 2.83*0.3/1.0084 ==> w2 = 0.8418 rev/s or 5.287 rad/s

ID: 2264780 • Letter: T

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. 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 1.8 m. when wrapped they form a cylinder of radius 26 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to .40 kg*m^2. If his original angular speed is 0.30 rev/sec, what is his final angular speed?

Explanation / Answer

initial moment of inertia of body

I1 = 0.4 + 9*1.8^2/12 = 2.83 kg.m^2

final moment of inertai a body

I2 = 0.4 + 9*0.26^2 = 1.0084 kg.m^2

here initial angular momentum = final angular momentum

I1*w1 = I2*w2

2.83*0.3 = 1.0084*w2

==> w2 = 2.83*0.3/1.0084

==> w2 = 0.8418 rev/s or 5.287 rad/s

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The outstretched hands and arms of a figure skater preparing for a spin can be c

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The outstretched hands and arms of a figure skater preparing for a spin can be c

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