On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbel

Question

On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.650 m from the axis of rotation of the stool. She is given an angular velocity of 2.55 rad/s , after which she pulls the dumbbells in until they are only 0.180 m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.65 kgm2 and may be considered constant. Each dumbbell has a mass of 5.25 kg and may be considered a point mass. Neglect friction.

What is the initial angular momentum of the system?

What is the angular velocity of the system after the dumbbells are pulled in toward the axis?

Compute the kinetic energy of the system before the dumbbells are pulled in.

Compute the kinetic energy of the system after the dumbbells are pulled in.

Explanation / Answer

Iw = moment of inertia of woman = 4.65

m = mass of each dumbbell = 5.25 kg

ri = initial distance = 0.65 m

rf = final distance = 0.180 m

wi = initial angular velocity = 2.55 rad/s

Ii = initial total moment of inertia = Iw + 2 m ri2 = 4.65 + 2 (5.25)(0.65)2 = 9.1 kgm2

If = Final total moment of inertia = Iw + 2 m rf2 = 4.65 + 2 (5.25)(0.18)2 = 5 kgm2

Li =initial angular momentum = Ii wi = 9.1 x 2.55 = 23.2

using conservation of angular momentum

Li = Lf

23.2 = If wf

23.2 = 5 wf

wf = 4.64 rad/s

KEi = (0.5) Ii w2i = (0.5) (9.1) (2.55)2 = 29.6 J

KEf = (0.5) If w2f = (0.5) (5) (4.64)2 = 53.82 J

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