A student sits on a rotating stool holding two 2.7-kg objects. When his arms are
ID: 1461597 • Letter: A
Question
A student sits on a rotating stool holding two 2.7-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.32 m from the rotation axis. (a) Find the new angular speed of the student.(b) Find the kinetic energy of the student before and after the objects are pulled in.
Explanation / Answer
the Initial moment of inertia of the system(student + stool+ the two masses) is
Ii = 3.0 + 2 ( 2.7*1.02 ) = 8.4 kgm2
the Final moment of inertia is
If = 3.0 + 2 ( 2.7*0.322 )= 3.55 kgm2
from conservation of angular momentum
Ii i = If f
8.4 * 0.75 = 3.55 *
= 1.77 rad/s
initial kinetic energy of the system is
KE = (1/2) I 2
= (1/2) * 8.4 * 0.752
= 2.36J
final kinetic energy of the sytem is
KE = (1/2) I 2
= (1/2) * 3.55 * 1.772
= 5.56J
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