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 0.560m from the axis of rotation of the stool. she is given an angular velocity of 2.75rad/s , after which she pulls the dumbbells in until they are only 0.160m distant from the axis. the woman's moment of inertia about the axis of rotation is 4.50kg*m2 and may be considered constant. Each dumbbell has a mass of 4.80kg 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) initial momnet of inertia of the woman and dumbells = 4.5 + 0.56^2*2*4.5

=7.322 kg ms^-2

so angular momentum = 7.322*2.75

=20.1355

B)conserving angular momentum,

I1=I2

or 20.1355=4.5*w

or w=4.474 rad/s

c)initial kinetic energy = 0.5*Iw^2

=0.5*7.322*2.75^2

=27.68 J

d) final kinetic energy = 0.5Iw^2

=0.5*4.5*4.474^2

=45.04 J

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