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?m2and may be considered constant. Each dumbbell has a mass of 5.05kg and may be considered a point mass. Neglect friction.

A) What is the initial angular momentum of the system?

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

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

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

Explanation / Answer

A)

inital mmoment of inertia , I = 5.05 +2 * 5.05 *0.605^2

I = 8.75 Kg.m^2

Inital angular momentum = 8.75 * 3.05

Inital angular momentum = 26.7 Kg.m^2/s

B)

Now, as the anglular momenum is costant ,

If*wf = 26.7

(5.05 +2 * 5.05 *0.205^2) * wf = 26.7

wf = 4.87 rad/s

the final angular momentum is 4.87 rad/s

C)

KInetic energy = 0.5 * 8.75 * 3.05^2

Kinetic energy = 40.7 J

D)

final kinetic energy = 0.5 * (5.05 +2 * 5.05 *0.205^2)*4.87^2

final kinetic energy =64.9 J

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