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 0.605m from the axis of rotation of the stool. She is given an angular velocity of 3.05rad/s , after which she pulls the dumbbells in until they are only 0.205m distant from the axis. The woman's moment of inertia about the axis of rotation is 5.05kg*m^2 and may be considered constant. Each dumbbell has a mass of 5.05kg and may be considered a point mass. Neglect friction. Part A What is the initial angular momentum of the system? Part B What is the angular velocity of the system after the dumbbells are pulled in toward the axis? Part C Compute the kinetic energy of the system before the dumbbells are pulled in.

Explanation / Answer

a)

L = Itot

Itot = Iwrmax + 2mr2

L = [4.90kgm2 + 2 * 4.85 * (0.595)2] 3.05

= 25.41kg m2/s

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b)

from cons. of ang momentum

Ii wi = If wf

wf = Ii wi / If

= 25.41 / (4.90 + (2.42* 0.1952))

= 5.09 rad/s

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c)

KE = 1/2 Ii * wi2

= 0.5 * 8.33 * (3.05)2

= 38.7 J

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d)

KE = 1/2 If * wf2

= 0.5 * 4.99 * 5.092

= 64.64 J

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