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 0.615 m from the axis of rotation of the stool. She is given an angular velocity of 3.45 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 5.40 kgm2 and may be considered constant. Each dumbbell has a mass of 4.65 kg 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.

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

Explanation / Answer

Here ,

a)

initial angular momentum = initial moment of inertia * angular velocity

initial angular momentum = (5.4 + 0.615^2 * 2 * 4.65) * 3.45

initial angular momentum = 30.8 Kg.m/s

the initial angular momentum is 30.8 Kg.m/s

part b)

let the final angular speed is wf

as the angular momentum of the system is conserved

Ii * wi = If *wf

(5.4 + 0.615^2 * 2 * 4.65) * 3.45 = (5.4 + 0.155^2 * 2 * 4.65) * wf

wf = 5.47 rad/s

the final angular velocity is 5.47 rad/s

c)

initial kinetic energy = 0.5 * Ii * w^2

initial kinetic energy = 0.5 * (5.4 + 0.615^2 * 2 * 4.65) * 3.45^2

initial kinetic energy = 53.1 J

the initial kientic energy is 53.1 J

d)

final kinetic energy = 0.5 * If *wf^2

final kinetic energy = 0.5 * (5.4 + 0.155^2 * 2 * 4.65) * 5.47^2

final kinetic energy = 84.13 J

final kinetic energy is 84.13 J

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