The Figure shows a skater doing a pirouette. The mass of the skater without arms

Question

The Figure shows a skater doing a pirouette. The mass of the skater without arms is 60kg; her height is 1.6m; the mass of each of her arms is 3.8kg; the length of each arm is 1.1m. At the instant shown, with her arms outstretched, she is rotating at 1.5 rev/s. In your calculations, when her arms are stretched out, approximate her body (without arms) to a long cylinder of height 1.6m and radius 0.24m. 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 1.6m and radius 0.24m.

(a) What is her approximate moment of inertia (in Kg m2) when her arms are outstretched?

(b) What is her angular momentum (in Kg m2/s) when her arms are outstretched?

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

Assuming no losses due to friction or air resistance.

(d) What is her new (with her arms pulled towards her body) moment of inertia (in Kg m2)?

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

Explanation / Answer

(a) I1 = Ibody + Iarms = MbR2/2 + 2[MaL2/3] = (60 * 0.242 / 2) + (2 * 3.8 * 1.12 / 3) = 4.79 kg-m2

(b) L1 = I11 = 4.79 * (1.5 * 2) = 45.14 kg-m2/s

(c) KE1 = I112/2 = 4.79 * (1.5 * 2)2 / 2 = 212.74 J

(d) I2 = (Mb + 2Ma)R2/2 = [60 + (2 * 3.8)] * 0.242 / 2 = 1.95 kg-m2

(e) Angular momentum is conserved.

I11 = I22

=> 2 = I11/I2 = 4.79 * 1.5 / 1.95 = 3.7 rev/s

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