The 175 lb ice skater with arms extended horizontally spins about a vertical axi

Question

The 175 lb ice skater with arms extended horizontally spins about a vertical axis with a rotational speed of 1 rev/sec. Estimate his rotational speed if he fully retracts his arms, bringing his hands very close to the centerline of his body. As a reasonable approximation, model the extended arms as uniform slender rods, each of which is 27 in. long and weighs 17 lb. Model the torso as a solid 141-lb cylinder 13 in. in diameter. Treat the man with arms retracted as a solid 175 lb cylinder of 13 in. diameter. Neglect friction at the skate-ice interface. (Hint: You'll need to use the parallel-axis theorem to figure out the moment of inertia of the extended arms.)

. rev/s

Explanation / Answer

here,

mass of skater , m = 175 lb

initial rotational speed , w0 = 1 rev/s

let the final rotational speed be w

using conservation of angular momentum

Iinitial * w0 = Ifinal * w

( 2 * 1/3 * 17 * 27^2 + 0.5 * 141 * (13/2)^2 ) * 1 = ( 0.5 * 175 * (13/2)^2) * w

solving for w

w = 3.04 rev/s

the final angular speed is 3.04 rev/s

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