A student sits on a freely rotating stool holding two dumbbells, each of mass 2.
ID: 2013020 • Letter: A
Question
A student sits on a freely rotating stool holding two dumbbells, each of mass 2.94 kg. When his arms are extended horizontally the dumbbells are 0.99 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 2.70 kg · m2 and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position 0.291 m from the rotation axis.(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
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) + 2.70 kg . m2 intiallywe get r = 0.99 m Ii = 2 (2.94 kg) (0.99 m)2 + 2.70 kg .m2 = 8.463 kg. m2 finally r =0.291 m If = 2 (2.94 kg) (0.291 m)2 + 2.70 kg .m2 =3.198 kg . m2 according to the law of conservation of momentum weget Iff = Iii solvefor f f =(8.463/3.198)*0.753 = 1.992 rad /s (b) KEi = (1 / 2) Iii2 = .0.5* 8.463 * (0.753)^2 = 2.399 J KEf = (1 / 2) Iff2 = 0.5 * 3.198*(1.992)^2 = 6.345 J f =(8.463/3.198)*0.753 = 1.992 rad /s (b) KEi = (1 / 2) Iii2 = .0.5* 8.463 * (0.753)^2 = 2.399 J KEf = (1 / 2) Iff2 = 0.5 * 3.198*(1.992)^2 = 6.345 JRelated Questions
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