A figure skater goes into a spin with her arms outstretched. She then pulls her arms in against her body. If she is initially rotating at 2rev / s , what is her final rotational speed? Assume that by pulling in her arms she reduces her rotational inertia from 2kg - m ^ 2 to 1.5kg - m ^ 2

Question

03-05-2022
Physics

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A figure skater goes into a spin with her arms outstretched. She then pulls her arms in against her body. If she is initially rotating at 2rev / s , what is her final rotational speed? Assume that by pulling in her arms she reduces her rotational inertia from 2kg - m ^ 2 to 1.5kg - m ^ 2

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okpalawalter8 okpalawalter8 · Answer 1 · 2022-05-13T12:19:21+02:00

For a figure skater goes into a spin with her arms outstretched. She then pulls her arms against her body. If she is initially rotating at 2rev / s is mathematically given as

Wf= 6rev / s

what is her final rotational speed?

Generally, the equation for the conservation angular momentum is mathematically given as

I1Wf=I2Wf

2kg - m ^ 2 X 2rev / s = 1.5kg - m ^ 2 X W2

Wf= 2/1.5(2)

Wf= 6rev / s

In conclusion, her final rotational speed is

Wf= 6rev / s