I AM SORRY I JUST HAVE ONE OF THE MOST HORRIBLE TEACHERS, AND I AM TEACHING MYSELF FOR I GET NOTHING IN CLASS.

Question

On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.565m from the axis of rotation of the stool. She is given an angular velocity of 2.80rad/s , after which she pulls the dumbbells in until they are only 0.200m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.90kg⋅m2 and may be considered constant. Each dumbbell has a mass of 4.95kg and may be considered a point mass. Neglect friction.
Part A
What is the initial angular momentum of the system?

ANSWER IN kg⋅m2/s in the direction of the motion

Part B
What is the angular velocity of the system after the dumbbells are pulled in toward the axis?
ω =

rad/s

Part C
Compute the kinetic energy of the system before the dumbbells are pulled in.
K = J

Part D
Compute the kinetic energy of the system after the dumbbells are pulled in.K = J

Answer 1

I initial = 4.90 + .565^2 (4.95*2)
= 4.90 + .319 (9.9)
= 4.90 + 3.16
= 8.06

Ifinal = 4.90 + .2^2(9.9)
= 4.90 + .396
= 5.30 oh my, much less

angular momentum initial and final
= I omega1 = 8.06*2.8
= 22.6 (answer A)
= 5.3*omega2
so
omega 2 = 22.6/5.3 = 4.26 rad/s

Now I think you can do
(1/2) I omega^2 for both

You will find the Ke is higher after she pulls the dumb bells in. She has to put work in to pull the weights in because they are held out by centripetal acceleration.

Answer 2

Thanks! you are a life saver