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
This is the end of the preview. Sign up
access the rest of the document.