The mass of the skater without arms is 67kg; her height is 1.6m; the mass of eac

Question

The mass of the skater without arms is 67kg; her height is 1.6m; the mass of each of her arm is 3.6kg; the length of each arm is 0.9m. When her arms are outstretched, she is rotating at 1.3rev/s. In your calculations, when her arms are stretched out, approximate her body (without arm) to a long cylinder of height 1.6 m and radius 0.22m. Assume also that arm start at her center (not at the end of her body). When her arms are pulled towards her body, approximate her body (with arms) to a long cylinder of height 1.6 m and radius 0.22m.

a. What is her approximate moment of inertia (in Kg m^2) when her arms are outstretched?

b. What is her angular momentum (in Kg m^2/s) when her arms are outstretched?

c. What is her kinetic energy (in J) when her arms are outstretched?

Assuming no losses duw to friction or air resistance.

d. What is her new (with her arms pulled towards her body) moment of inertia (in Kg m^2)

e. What is the new (with her arms pulled towards her body) angular speed (in rev/s)?

Explanation / Answer

Part A)

I = .5MR2 + 2(1/3)ML2

I = (.5)(67)(.22)2 + 2(1/3)(3.6)(.9)2

I = 3.57 kg m2

Part B)

L = Iw

w = 1.3(2pi) = 8.168 rad/s

L = (3.57)(8.168)

L = 29.1 kg m2/s

Part C)

KE = .5Iw2

KE = (.5)(3.57)(8.168)2

KE = 119 J

Part D)

I = .5(67)(.22)2

I = 1.62 kg m2

Part E)

Iw = Iw

(3.57)(8.168) = (1.62)(w)

w = 18.0 rad/s

That is 2.86 rev/s

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