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 0.570 m from the axis of rotation of the stool. She is given an angular velocity of 3.20 rad/s , after which she pulls the dumbbells in until they are only 0.190 m distant from the axis. The woman?s moment of inertia about the axis of rotation is 4.85 kg. m^2 and may be considered constant. Each dumbbell has a mass of 5.25 kg and may be considered a point mass. Neglect friction. Part A What is the initial angular momentum of the system? L = kg. m^2/s in the direction of the motion 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

a) The inital angular momentum of the system will be given by the formula

           L = Iw = (mr2) (3.2rad/ s) =2 ( 5.25 )(0.570)2(3.2)Kg m2 /s = 10.91kg m2 / s

b) This could be found by using the formula of law of conservation of agnular momentum

              Li = Lf

              I1 w 1 = I2 w2

               w2 = I1 w 1 / I2 =( 10.91 Kg m2 / s ) / 10.5(0.19)2 = 28.78 rad / s

Kinetic energy of the system is

      K.E1 = 0.5 Mr2 w2 = 0.5I1w12 = 0.5 (3.4)(3.2) =5.45 J

part D

    K.E2 = 0.5 I2 w22 = 0.5(0.379)2(28.78)2 = 156.98J

   

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