34. A student sits on a freely rotating stool holding two dumb- bells, each of m

Question

34. A student sits on a freely rotating stool holding two dumb- bells, each of mass 3.00 kg (Fig. PI 1.34). When his arms are extended horizontally (Fig. P11.34a), the dumbbells are 1.00 m from the axis of rotation and the student rotates with an angular speed of 0.750 rad/s. The moment of iner- tia of the student plus stool is 3.00 kg·m' and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.300 m from the rotation axis (Fig. P11.34b). (a) Find the new angular speed of the stu- dent. (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward. Figure P11.34

Explanation / Answer

moment of inertia of student+stool I = 3 kg m^2

initial angular momentum Li = (I + 2*m*r1^2)*wi

m = mass of dumbell

r1 = 1 m

initial angular speed wi = 0.75 rad/s

after the dumbells are pulled inward

final angular momentum Lf = (I + 2*m*r2^2)*wf

r2 = 0.3 m

from conservation of angular momentum

Lf = Li

(3 + 2*3*0.3^2)*wf = (3 + 2*3*1^2)*0.75

final angular speed wf = 1.91 rad/s

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part (b)

kinetic energy before pulling the dumbells KEi = (1/2)*(I + (2*m*r1^2))*wi

KEbefore = (3 + (2*3*1^2))*0.75^2 = 5.06 J

kinetic energy after pulling the dumbells KEf = (1/2)*(I + (2*m*rw^2))*wf

KEafter = (3 + (2*3*0.3^2))*1.91^2 = 12.9 J

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