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 0.650m from the axis of rotation of the stool. She is given an angular velocity of 2.85rad/s , after which she pulls the dumbbells in until they are only 0.240m distant from the axis. The woman's moment of inertia about the axis of rotation is 4.75kg?m2 and may be considered constant. Each dumbbell has a mass of 5.50kg and may be considered a point mass. Neglect friction.

1. What is the initial angular momentum of the system?

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

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

Explanation / Answer

1)

initial total inertia = 4.75 + 2*5.5*0.65^2 = 9.3975 kg m^2

so.. intial angular momentum = inertai * angular speed = 9.3975 * 2.85 = 26.782875

2)

let the final angular speed be w ...

final inertia = 4.75 + 2*5.5*0.24^2 = 5.3836 kg m^2

so.. final momnetun = 5.3836 * w = inital ( for conserving momentum )

so.. 5.3836 * w = 26.782875

so.. final angular speed = w = 4.9749 rad/sec

3)

kinetic energy before dumbells pulled in= 0.5 * inital inertia * inital angular speed^2

= 0.5 * 9.3975 * 2.85^2 = 38.165596875 J

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