377_Physics ProblemsTechnical Physics

377_Physics - 379 Chapter 12 P12.70(1(2 ph = I p p = MvCM h vCM If the ball rolls without slipping R = vCM So h = P12.71(a 2 I I I = = = R 5 p MvCM

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Chapter 12 379 P12.70 (1) ph I = ω (2) pM v = CM If the ball rolls without slipping, Rv = CM So, h I p I Mv I MR R == = = ωω CM 2 5 p h v CM FIG. P12.70 P12.71 (a) If the acceleration is a , we have Pm a x = and PnF yg +− = 0 . Taking the origin at the center of gravity, the torque equation gives PL d Phn d yx −+ − = af 0. Solving these equations, we find P F L d ah g y g =− F H G I K J . h P CG d H L F y n g FIG. P12.71 (b) If P y = 0, then d ah g = 200 150 980 0306 .. . . ±ms ±m m 2 2 ej . (c) Using the given data, P x 306 N and P y = 553 N. Thus, Pi j + 306 553 ±± N. *P12.72 When the cyclist is on the point of tipping over forward, the normal force on the rear wheel is zero. Parallel to the plane we have fm g m a 1 −= sin θ . Perpendicular to the plane, nm g 1 0 cos . Torque about the center of mass: mg f n 01 0 5 0 6 50 11 a f a f −+ = m m . Combining by substitution, mg n 1 f 1 FIG. P12.72 ma f mg n mg mg mg
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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 .

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