- #1

NoHeart

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a diver rotates at 1 revolution per second in the stretched out position. when the diver tucks her head in and bends her legs, assume her length is shortened by 1/2. what is her rate of rotation in this position?

my choices are 4 Hz or 5 Hz, you'd think having only 2 choices would make this easier, but alas, i am dumbfounded.

i have tried thinking of the diver in the stretched out position as a rod, with rotational inertia 1/12ML^2

the other position would be like a solid sphere, with rotational inertia 2/5MR^2

1 revolution per second is 1 Hz, or 2pi r/s

the angular momentum (which may be irrelevant) of the stretched out position is

L= 1/12M(2r)^2 * 6.28 radians/1 revolution * 1 revolution/1second

this leaves me with L=2.093Mr^2 rad/sec

L of ball position is

2/5Mr^2 * 6.28 radians/?sec= 2.51Mr^2 radians/?sec

?=0.39Mr^2 rad/sec

slowly going insane, i see that rotational accleration = net torque/rotational inertia

this also leads me nowhere

any help, hints, or a slap across the face would be greatly appreciated

(p.s.- anyone else think online physics is the STUPIDEST idea EVER?)