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.

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

Explanation / Answer

kinetic enrgy of the system = momnet of inertia of the system in final state * w in final state

final w of the system when pulled in =(4.75 + 2*5.5* 0.65 ^2)*2.85/ ( 4.75+2*5.5 *0.24^2 ) =4.974

kintic enrgy of the system = 1/2* ( 4.75+2*5.5 *0.24^2 ) *4.974^2 =66.59J

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