A student sits on a freely rotating stool holding two dumbbells, each of mass 3.
ID: 2013577 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 3.08 kg. When his arms are extended horizontally, the dumbbells are 1.06 m from the axis of rotation and the student rotates with an angular speed of 0.759 rad/s. The moment of inertia of the student plus stool is 2.74 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.305 m from the rotation axis.
Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward.
Explanation / Answer
a) from the given problem the total anguar momentum of the system of thestudent the stool and the weightsabout the axis of rotation is given by Itotal= Iweights + Istudent = 2 (m r2) + 2.74 kg . m2 intiallywe get r = 1.06 m Ii = 2 (3.08 kg) (1.06 m)2 + 2.74 kg .m2 = 9.66 kg. m2 finally r =0.305 m If = 2 (3.08 kg) (0.305 m)2 + 2.74 kg .m2 =3.31 kg . m2 according to the law of conservation of momentum weget If?f = Ii?i solvefor ?f ?f =(9.66/3.31)*0.759 = 2.215 rad /s (b) KEi = (1 / 2) Ii?i2 = .0.5* 9.66 * (0.759)^2 = 2.78 J KEf = (1 / 2) If?f2 = 0.5 * 3.31*(2.215)^2 = 8.12 J
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