164_Physics ProblemsTechnical Physics

164_Physics ProblemsTechnical Physics - 166 Circular Motion...

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166 Circular Motion and Other Applications of Newton’s Laws P6.20 (a) a v r c = 2 r v a c == = 2 2 13 0 2980 862 . . . ms ±ms m 2 bg ej (b) Let n be the force exerted by the rail. Newton’s law gives FIG. P6.20 Mg n Mv r += 2 nM v r gM g g M g =− F H G I K J = 2 2 , downward (c) a v r c = 2 a c 13 0 20 0 845 2 . . . m 2 If the force exerted by the rail is n 1 then g Mv r Ma c 1 2 = ag c 1 which is < 0 since a c = . m s 2 Thus, the normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars. To be safe we must require n 1 to be positive. Then c > . We need v r g 2 > or vr g >= 20 0 9 80 .. m m s 2 af , v > 14 0 . ms . Section 6.3 Motion in Accelerated Frames
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