295_Physics ProblemsTechnical Physics

295_Physics ProblemsTechnical Physics - Chapter 10 *P10.29...

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Chapter 10 297 *P10.29 We consider the cam as the superposition of the original solid disk and a disk of negative mass cut from it. With half the radius, the cut-away part has one-quarter the face area and one-quarter the volume and one-quarter the mass M 0 of the original solid cylinder: MM M 00 1 4 −= 0 4 3 = . By the parallel-axis theorem, the original cylinder had moment of inertia IM R MR M R CM + F H G I K J =+ = 0 2 0 2 0 2 0 2 2 1 24 3 4 . The negative-mass portion has R =− F H G I K J F H G I K J 1 2 1 42 3 2 0 2 0 2 . The whole cam has R M R MR MR = = = 3 43 2 23 32 23 32 4 3 23 24 0 2 0 2 0 22 2 and K I MR MR == = 1 2 1 2 23 24 23 48 2 2 2 ωω ω . Section 10.6 Torque P10.30 Resolve the 100 N force into components perpendicular to and parallel to the rod, as F par N N = 100 57 0 54 5 af cos . . and F perp N = 100 57 0 83 9 sin . . The torque of F par is zero since its line of action passes through the pivot point. FIG. P10.30
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