a woman stands at the center of a platform. the woman and the platform rotate with an angular speed of 5.00 rad/s. friction is negligible. her arms are outstretched, and she is holding a dumbbell in each hand. in this position the total moment of inertia of the rotating system is 5.40 kgm^2. by pulling in her arms, the moment of inertia is reduced to 3.80 kg m^2. find her new angular speed.

asked by Remie

5 years ago

1,525 views

1

0

1 answer

conservation of momentum applies
I*w=I'*w'
w'=(5.4/3.9)5 rad/sec

answered by bobpursley

1

0

Ask a New Question or answer this question.

Question

a woman stands at the center of a platform. the woman and the platform rotate with an angular speed of 5.00 rad/s. friction is negligible. her arms are outstretched, and she is holding a dumbbell in each hand. in this position the total moment of inertia of the rotating system is 5.40 kgm^2. by pulling in her arms, the moment of inertia is reduced to 3.80 kg m^2. find her new angular speed.

1 answer

conservation of momentum applies
I*w=I'*w'
w'=(5.4/3.9)5 rad/sec

Ask a New Question or answer this question.

bobpursley · Accepted Answer

conservation of momentum applies I*w=I'*w' w'=(5.4/3.9)5 rad/sec