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 0.585 m from the axis of rotation of the stool. She is given an angular velocity of 2.85 rad/s , after which she pulls the dumbbells in until they are only 0.165 m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.60 kgm2 and may be considered constant. Each dumbbell has a mass of 5.50 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

The initial moment of inertia of the system will be

Ii = I(woman) + I(dumbell)

Ii = 4.6 + 2 x 5.5 x 0.585^2 = 8.36 kg-m^2

Li = I(i) w

Li = 8.36 x 2.85 = 23.83 kg-m^2/s

Hence, Li = 23.83 kg-m^2/s

B)Moment of inertia after the dumbells are pulled in is:

I(f) = I(woman) + I(dumbell)

I(f) = 4.6 + 2 x 5.5 x 0.165^2 = 4.89 kg-m^2/s

from conservation of angular momentum

L(i) = L(f)

Ii wi = If wf

wf = Ii wi/If = 8.36 x 2.85/4.89 = 4.87 rad/s

Hence, wf = 4.87 rad/s

C)KEi = 1/2 Ii wi^2

KEi = 0.5 x 8.36 x 2.85^2 = 33.95 J

Hence, KEi = 33.95 J

D)KEf = 1/2 If wf^2

KEf = 0.5 x 4.89 x 4.87^2 = 57.99 J

Hence, KE = 57.99 J

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