A student on a piano stool rotates freely with an angular speed of 2.95 rev/sec.

Question

A student on a piano stool rotates freely with an angular speed of 2.95 rev/sec. The student holds a 1.15 kg mass in each outstretched arm, 0.759 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.03 kg m^2, a value that remains constant. As the student pulls his arms inward, his angular speed increases to 3.50 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points? Calculate the initial and final kinetic energy

Explanation / Answer

Given that theinitial angular velocity is ?1 = 2.95rev/s mass ofeach object is m = 1.15 kg The initialdistance between the axis of rotation and the each mass is d =0.759 m Moment of inertiaof the student and the stool is I1 = 5.03kg.m2 Final angularvelocity is ?2 = 3.50 rev/s --------------------------------------------------------------------------------------- From theconservation of angular momentum initial angular momentum = final angular momentum ( I1 + md^2 +md^2 )?1 = ( I1 +mx^2+mx^2)?2 ( I1 + 2md^2 )?1 = ( I1 +2mx^2)?2 (I1 + 2mx^2) = ( I1 + 2md^2 )?1 /?2 2mx^2 = ( I1 + 2md^2 )?1 / ?2 - I1 x^2= ( I1 + 2md^2 )?1 /2m*?2 - I1 /2m then we get x =0.141889 m The initial kinetic energy is K1 = (1/2)( I1 + md^2 +md^2)?1^2 =27.6521J The final kinetic energy is K2 = (1/2)( I1 + mx^2 +mx^2)?2^2 =17.0957J

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