Chapter 10297*P10.29We consider the cam as the superposition of the original solid disk and a disk of negative mass cutfrom it. With half the radius, the cut-away part has one-quarter the face area and one-quarter thevolume and one-quarter the mass M0of the original solid cylinder:MMM0014−=043=.By the parallel-axis theorem, the original cylinder had moment of inertiaIMRMRMRCM+FHGIKJ=+=02020202212434.The negative-mass portion has R=−FHGIKJFHGIKJ12142320202. The whole cam hasRM RMRMR===34322332233243232402020222and KIMRMR===121223242348222ωωω.Section 10.6TorqueP10.30Resolve the 100 N force into components perpendicularto and parallel to the rod, asFparN N=°=10057 054 5afcos..andFperpN=10057 083 9sin..The torque of Fparis zero since its line of action passesthrough the pivot point.FIG. P10.30
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