A student sits on a freely rotating stool holding two dumbbells, each of mass 3.
ID: 1462194 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 3.06 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 1.07 m from the axis of rotation and the student rotates with an angular speed of 0.753 rad/s. The moment of inertia of the student plus stool is 2.67 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.307 m from the rotation axis (Figure b).
(a) Find the new angular speed of the student.
rad/s
(b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.
What is the moment of inertia of two 3.06 kg dumbbells, each a distance 1.07 from the axis of rotation? J Kafter =
You cannot assume that kinetic energy is conserved here, so you must calculate the final kinetic energy from your previous results. J
Explanation / Answer
m = mass of each dumbbell = 3.06 kg
r1 = initial distance of dumbbell = 1.07 m
initial moment of inertia of dumbbell = I1 = mr12 = 3.06 (1.07)2 = 3.504 kgm2
r2 = final distance of dumbbell = 0.307
final moment of inertia of dumbbell = I2 = mr22 = 3.06 (0.307)2 = 0.288 kgm2
Is = The moment of inertia of the student plus stool = 2.67 kg m2
Ii = initial total moment of inertia with arms strecthed out = I1 + Is = 3.504 + 2.67 = 6.17
If = final total moment of inertia = I2 + Is = 0.288 + 2.67 = 2.96
Wi = initial angular speed = 0.753 rad/s
Wf = final angular speed = ?
Using conservation of angular momentum
Ii Wi = IfWf
6.17 x 0.753 = 2.96 Wf
Wf = 1.57 rad/s
b)
initial Kinetic energy
KEi = (0.5) Ii Wi2 = (0.5) (6.17) (0.753)2 = 1.75 J
final Kinetic energy
KEf = (0.5) If Wf2 = (0.5) (2.96) (1.57)2 = 3.65 J
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