A student on a piano stool rotates freely with an angular speed of 2.85 rev/s .

Question

A student on a piano stool rotates freely with an angular speed of 2.85 rev/s . The student holds a 1.00kg mass in each outstretched arm, 0.739m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.43 kg*m^2 , a value that remains constant. A. As the student pulls his arms inward, his angular speed increases to 3.25 . How far are the masses from the axis of rotation at this time, considering the masses to be points? d = _____ M B. Ki = ______ J C. Kf = ______J

Explanation / Answer

Best Answer - Chosen by Voters The combined moment of inertia of the student and the stool, ignoring the two masses, is 4.89 kg · m2, moment of inertia due to 2 masses = 2 * 1.41 * 0.759^2 = 1.625 kg - m^2 Total moment of inertia = 4.89 + 1.625 = 6.515 kg-m^2 K.E. = 0.5 x moment of inertia x angular speed ^2 = 0.5*6.515*(3.10 * 2 * 3.1416)^2 = 1235.861 N-m Now for the same K.E. the new angular speed is 3.54 rev/s, we must calculate new moment of inertia. new moment of inertia = (1235.861) / (0.5 * ( 3.54 * 2 * 3.1416)^2) = 4.996 kg -m2 4.996 = 4.89 + ( 2 x 1.41 x d^2) d^2 = 0.356 d = 0.194 m are the masses from the axis of rotation at this time, ...................ans

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