Problem 11.74 A student on a piano stool rotates freely with an angular speed of

Question

Problem 11.74 A student on a piano stool rotates freely with an angular speed of 3.05 rev/s The student holds a 1.50 kg mass in each outstretched arm, 0.749 m. from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.43 kg m2 a value that remains constant. Part A As the student pulls his arms inward, his angular speed increases to 3.80 rev/s How far are the masses from the axis of rotation at this time, considering the masses to be points? In Submit My Answers Give Up Part B Calculate the initial kinetic energy of the system. Submit My Answers Give Up

Explanation / Answer

Given

the totating stool with initial moment of inertia is I1 = 5.43 kgm2 , w1 = 3.05 rev/s = 3.05*2pi rad/s when he

   holds the masses of 1.5 kg in the outstrectched arm of length l= 0.749 m

when he pulled the his arms inward the new angular speed increased to w2 = 3.08 rev/s = 3.08*2pi

Part A

From conservation of angular momentum

       I1W1 = I2W2 ---------------(1)

       I1W1 = (I1+2mr^2)W2

       5.43*3.05= (5.43+2*1.5*r^2)(3.08)

       r = 0.133 m = 13.3 cm from the axis of rotation

PArt B

   initial kinetic energy of the system is k.e1 = 0.5*I1*W1^2 = 0.5*5.43*(3.05*2pi)^2 J = 997.0782650607125 J

PArt c

   final kinetic energy of the system is k.e2 = 0.5*I2*W2^2

   here from equation (1) I2 = I1*W1 /W2

                       = 0.5*(I1*W1/W2)W2^2 J

                       = 0.5*I1*W1*W2
                       = 0.5*5.43*3.05*2pi*3.08*2pi J

                       = 1006.8856 J

final kinetic energy of the system is k.e2 = 1006.8856 J

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