Question

The 160 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 13 lb. Model the torso as a solid 134-lb cylinder 13 in. in diameter. Treat the man with arms retracted as a solid160 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.)

Answer 1

initial rotational speed, wi = 1 rev/s = 2pi rad/s

when arms extended :

moment of inertia of rod about its centre of mass = m L^2 / 12

= 13 x 27^2 / 12 = 789.75 lb in^2

moment about centrlien,

I = 789.75 + ( 13 (27/2 + 13/2)^2 ) = 5989.75 lb in^2

Ii = 2 x 5989.75 + ( 134 x 13^2 / 8 ) = 14810.25 lb in^2

When arms retracted:

If = 160 x 13^2 / 8 = 3380 lb in^2

Using angular momentum conservation,

Ii wi = If ef

14810.25 x 1 rev/s = 3380 x wf

wf = 4.38 rev /s OR 27.53 rad/s