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 2.90 kg (see figure below). When his arms are extended horizontally (figure a), the dumbbells are 0.90 m from the axis of rotation and the student rotates with an angular speed of 0.741 rad/s. The moment of inertia of the student plus stool is 2.51 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 (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.

Kbefore = J Kafter = J

Explanation / Answer

Mass of each dumbbell m = 2.90 kg
Distance of the dumbell from axis of rotation when his arms are extended horizontally r = 0.90 m Angular speed ' = 0.741 rad/s The moment of inertia of the student plus stool I = 2.51 kg · m2 Distance of the dumbell from axis of rotation after the student pulls the dumbbells inward horizontally r ' = 0.305 m Moment of inertia of the student when his arms streached I ' = I + 2mr 2                                                                                            = 2.51 + 4.698                                                                                            = 7.208 kg m 2 Moment of inertia of the student when his arms closed inward I " = I + 2mr ' 2                                                                                            = 2.51 + 0.5395                                                                                            = 3.049 kg m 2 From law of ocnservation of angular momentum . I ' ' = I"" From this the new angular speed of the student " = I''/ I"                                                                              = 1.751rad/s

(b) The kinetic energy of the rotating system before he pulls the dumbbells inward K = ( 1/ 2)I'' 2                                                                                                                                 = 1.9788 J The kinetic energy of the rotating system after he pulls the dumbbells inward K ' = ( 1/ 2)I"" 2                                                                                                                           =4.674 J                                                                                                                                 = 1.9788 J From law of ocnservation of angular momentum . I ' ' = I"" From this the new angular speed of the student " = I''/ I"                                                                              = 1.751rad/s

(b) The kinetic energy of the rotating system before he pulls the dumbbells inward K = ( 1/ 2)I'' 2                                                                                                                                 = 1.9788 J The kinetic energy of the rotating system after he pulls the dumbbells inward K ' = ( 1/ 2)I"" 2                                                                                                                           =4.674 J                                                                                                                                 = 1.9788 J

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