8. A student sits on a freely rotating stool holding two weights each of mass 3.

Question

8. A student sits on a freely rotating stool holding two weights each of mass 3.00 kg (Fig. P10.48). When his arms are extended horizontally, the weights are 1.00 m from the 1.00 m of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus Stool is 3.00 kg.m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the weights inward.

Explanation / Answer

(1)

moment of inertia of masses= 2*m*r^2

so, I1= 2*3*1^2= 6 kg.m^2

so, initial angular mementum= Inet*w= .75*(6+3)= 6.75 ................(1)

so, now, final moment of inertia of maases=0

so,  initial angular mementum= 3*wf ................... (2)

so, by conservation of angular momentum,

6.75=3*wf

so, Wf= 2.25 rad/s (answer)

(2)

initial Kinetic energy,

E1= .5*(6+3)*.75^2= 2.531 J

initial Kinetic energy,

E2= .5*(3)*2.25^2= 7.59 J

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