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377_Physics ProblemsTechnical Physics

# 377_Physics ProblemsTechnical Physics - 379 Chapter 12...

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Chapter 12 379 P12.70 (1) ph I = ω (2) p Mv = CM If the ball rolls without slipping, R v ω = CM So, h I p I Mv I MR R = = = = ω ω CM 2 5 ω p h v CM FIG. P12.70 P12.71 (a) If the acceleration is a , we have P ma x = and P n F y g + = 0 . Taking the origin at the center of gravity, the torque equation gives P L d P h nd y x + = a f 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 = = = 2 00 1 50 9 80 0 306 . . . . m s m m s m 2 2 e j a f . (c) Using the given data, P x = − 306 N and P y = 553 N . Thus, P i j = − + 306 553 ± ± e j 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 f mg ma 1 = sin θ . Perpendicular to the plane, n mg 1 0 = cos θ . Torque about the center of mass: mg f n 0 1 05 0 65 0 1 1 a f a f a f + = . . m m . Combining by substitution, mg n 1 f 1 FIG. P12.72 ma
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