A student sits on a freely rotating stool holding two dumbbells, each of mass 3.
ID: 1466067 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 3.03 kg (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 1.04 m from the axis of rotation and the student rotates with an angular speed of 0.757 rad/s. The moment of inertia of the student plus stool is 2.60 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.306 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.
Explanation / Answer
Net moment of inertia of system = moment of inertia of studentplus stool + moment of inertia of dubells
iinitial moment of inertia Ii = (2.60 ) + ( 2 x 3.03 x 1.04^2 ) = 9.15 kg m^2
final moment of inertia = ( 2.60 ) + (2 x 3.03 x 0.306^2) = 3.17 kg m^2
a) using angular momentum conservation
Ii wi = If wf
9.15 x 0.757 = 3.17 x wf
wf = 2.19 rad/s
b) KE = I w^2 /2
KE bfore = 9.15 x 0.757^2 /2 = 2.62 J
KEafter = 3.17 x 2.19^2 /2 = 7.60 J
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