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 0.635 m from the axis of rotation of the stool. She is given an angular velocity of 3.50 rad/s , after which she pulls the dumbbells in until they are only 0.155 m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.70 kgm2 and may be considered constant. Each dumbbell has a mass of 5.40 kg and may be considered a point mass. Neglect friction.

Part A

What is the initial angular momentum of the system?

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Part B

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

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Part C

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

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Part D

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

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Explanation / Answer

given,

distance of the dumbbells = 0.635 m

mass of dumbbells = 5.4 kg

moment of inertia of woman = 4.7 kg.m^2

initial velocity = 3.5 rad/sec

final distance = 0.155 m

initial moment of inertia = 4.7 + 2 * 5.4 * 0.635^2

initial moment of inertia = 9.055 kg.m^2

angular momentum = moment of inertia * angular velocity

initial angular momentum = 9.055 * 3.5

initial angular momentum = 31.692 kg.m^2/sec

by conservation of momentum

initial momentum = final momentum

31.692 = (4.7 + 2 * 5.4 * 0.155^2) * final angular velocity

final angular velocity = 6.39 rad/sec

initial kinetic energy = 0.5 * moment of inertia * w^2

initial kinetic energy = 0.5 * (4.7 + 2 * 5.4 * 0.635^2) * 3.5^2

initial kinetic energy = 55.46 J

final kinetic energy = 0.5 * (4.7 + 2 * 5.4 * 0.155^2) * 6.39^2

final kinetic energy = 101.252 J

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