Chapter 15457Additional ProblemsP15.51Let Frepresent the tension in the rod.(a)At the pivot, FMgMgMg=+=2A fraction of the rod’s weight MgyLFHGIKJas well as theweight of the ball pulls down on point P. Thus, thetension in the rod at point PisgyLMgMgyL=FHGIKJ+=+FHGIKJ1.M P pivotL yFIG. P15.51(b)Relative to the pivot, IIIML MLML+=rodball1343222For the physical pendulum, TImgd=2πwhere mM=2and dis the distance from thepivot to the center of mass of the rod and ball combination. Therefore,dMMLLL=++=234chand TMLMgLgL==2243243234af.For L=200. m,T432200980268...m±mss2.P15.52(a)Total energy =1212100
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .