The 170 lb ice skater with arms extended horizontally spins about a vertical axi
ID: 1437246 • Letter: T
Question
The 170 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 14 lb. Model the torso as a solid 142-lb cylinder 13 in. in diameter. Treat the man with arms retracted as a solid 170 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
skater with arms extended horizontally
=> Moment of inertia = Inertia of two arms + inertia of torso
= 2 * (6.35 * 0.68582/12 + 6.35 * 0.5082) + 64.41 * 0.16512/2
= 4.653 kg-m2
When he fully retracts his arms
=> moment of inertia = 77.1107 * 0.16512/2
= 1.0509 kg-m2
Applying conservation of angular momentum
=> 4.653 * (2 * 3.14) = 1.0509 * w
=> w = 27.805 rad/sec ---------------> his rotational speed
= 4.427 rev/sec ---------------> his rotational speed
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