# Answer on Mechanics | Relativity Question for Jason Ranak

Question #5134

A skater has a moment of inertia of 150 kgm2 when his arms are outstretched and a moment of inertia of 60 kgm2 when his arms are tucked in close to his chest. If he spins at an angular speed of 2.0 rps (revolutions per second) with his arms outstretched, what will his angular speed be when they are tucked in?

Expert's answer

This is a question about the conservation of angular momentum (L) L = I*w where I is the moment of inertia and w is the angular velocity. In the absence of external forces or torques, this quantity is conserved, just as linear momentum (m*v) is conserved.

In the original configuration w = 2*pi*2 rad/s, so initially L1 = 150*2*pi*2

In the final configuration L2 = 60*2*pi*n2 where n2 is the final number of revs per sec.

Equating L1 to L2 gives 150*2*pi*2 = 60*2*pi*n2

n2 = 2*150/60 = 5rps If you want an answer in radians/s, you multiply this by 2*pi.

This is a formal solution, but you can take a short cut: since I*n is a constant, I1*n1 = I2*n2, leading to n2 = I1*n1/I2 - a simple problem in proportion. Subscripts 1 & 2 refer to the initial and final conditions.

In the original configuration w = 2*pi*2 rad/s, so initially L1 = 150*2*pi*2

In the final configuration L2 = 60*2*pi*n2 where n2 is the final number of revs per sec.

Equating L1 to L2 gives 150*2*pi*2 = 60*2*pi*n2

n2 = 2*150/60 = 5rps If you want an answer in radians/s, you multiply this by 2*pi.

This is a formal solution, but you can take a short cut: since I*n is a constant, I1*n1 = I2*n2, leading to n2 = I1*n1/I2 - a simple problem in proportion. Subscripts 1 & 2 refer to the initial and final conditions.

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