A skater is initially spinning at a rate of 11.1 rad/s with a rotational inertia of 2.54 kg·m2 when her arms are extended. What is her angular velocity after she pulls her arms in and reduces her rotational inertia to 1.65 kg·m2?
2006-10-27 13:26:24 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
angular momentum is constant. So L = Iw is the same before and after she pulls her arms.
Before: L = I(wi) =(2.54 kg·m²)(11.1 rad/s) = 28.194 kg·m²/s
After: L = I(wf)
wf = L/I = [28.194 kg·m²/s]/[1.65 kg·m²) = 17.1 rad/s
2006-10-27 13:33:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Angular momentum is conserved. Angular momentum is I*w, and if I is reduced, w increases proportionally.
2006-10-27 13:30:51 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
11.1 *2.54/1.65 rad/sec
2006-10-27 13:29:32 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
http://ruina.tam.cornell.edu/Book/RuinaPratapProblems.pdf
2006-10-27 13:27:41 · answer #4 · answered by god knows and sees else Yahoo 6 · 0⤊ 0⤋