Question

A student on a piano stool rotates freely with an angular speed of 3.11 rev/s. The student holds a 1.37-kg mass in each outstretched arm, 0.787 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.35 kg.m2, a value that remains constant. As the student pulls his arms inward, his angular speed increases to 3.48 rev/s. How far are the masses from the axis of rotation at this time, considering the masses to be points? Submit Answer Tries 0/8 Calculate the initial kinetic energy of the system. Submit Answer Tries 0/8 Calculate the final kinetic energy of the system. Submit Answer Tries 0/8

Answer 1

Iten that |- given 3 w 3.11 reva m 1.37 Kg m r 0,787 m 12 I = wa . 5.35 kg.m2 3.48 rev/s a) pet d in the final position position of the arms form the axis of rotation then apply the law of consenna. -tron of angular non

2 2 T d = ܢܐ ay (2mo?t. I) a we (2nd ? HI) 3.112X1.37 x 0.787'+5:35 ) = (3-48)(2x 1.94d?+5.25) 27.086 (2.7402 + 5.35) 7. 7833 -5.35 = 2.7462 0.9423m b) initial kinetic energy, Ei= f I, w,2 Ź (2x 1.37 *0.7872+3:48) (2.11*2*) 2 =987.405 Ez = Ź Iz W22 |(2X1.37 x3.5782 (3-48x2x)? Ez 1958.62 J c ) 0.942375.35) 2 2