A student sits on a freely rotating stool holding two dumbbells, each of mass 3.
ID: 1489900 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 3.05 kg (see figure below). When his arms are extended horizontally (figure a), the dumbbells are 0.94 m from the axis of rotation and the student rotates with an angular speed of 0.748 rad/s. The moment of inertia of the student plus stool is 2.69 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.299 m from the rotation axis (figure b).(a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.
Explanation / Answer
moment of inertia of total system=moment of inertia of man+stool+dumbell
initial moment of inertia of both dumbell=2mr^2=2*3.05*0.94^2=5.38996 kgm^2
total initial moment of inertia=I1=2.69+5.38996=8.07996 kgm^2
final moment of inertia of dumbell=2mr^2=2*3.05*0.299^2=0.5453 kgm^2
final total moment of inertia=I2=2.69+0.5453=3.235 kgm^2
w1=0.748 rad/s
conserving angular momentum
I1*w1=I2*w2
w2=I1*w1/I2=1.8682 rad/s
(b) initial kinetic energy=0.5*I1*w1^2=2.26 J
final kinetic energy=0.5*I2*w2^2=5.645 J
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