A student sits on a freely rotating stool holding two dumbbells, each of mass 2.

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 J

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