Chapter 12379P12.70(1)phI=ω(2)pMv=CMIf the ball rolls without slipping, Rv=CMSo, hIpIMvIMRR====ωωCM25phvCMFIG. P12.70P12.71(a)If the acceleration is a, we have Pmax=andPnFyg+− =0 . Taking the origin at the center ofgravity, the torque equation givesPL d Phndyx−+ − =af0.Solving these equations, we findPFLdahgyg=−FHGIKJ.hPCGdHLFyngFIG. P12.71(b)If Py=0, then dahg=2001509800306....±ms±mm22ej.(c)Using the given data, Px306 N and Py=553 N.Thus, Pij+306553±±N.*P12.72When the cyclist is on the point of tipping over forward,the normal force on the rear wheel is zero. Parallel to theplane we have fmgma1−=sinθ. Perpendicular to theplane, nmg10cos. Torque about the center of mass:mgfn0105065011afaf−+=m m.Combining by substitution,mgn1f1FIG. P12.72mafmgnmgmgmg
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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 .