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235_Physics ProblemsTechnical Physics

# 235_Physics ProblemsTechnical Physics - Chapter 8 P8.51 m =...

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Chapter 8 237 P8.51 m = mass of pumpkin R = radius of silo top Fm a nm g m v R rr =⇒ = cos θ 2 When the pumpkin first loses contact with the surface, n = 0. Thus, at the point where it leaves the surface: vR g 2 = cos . FIG. P8.51 Choose U g = 0 in the 90 0 . plane. Then applying conservation of energy for the pumpkin-Earth system between the starting point and the point where the pumpkin leaves the surface gives KU KU mv mgR mgR fg fig i += + + 1 2 0 2 cos Using the result from the force analysis, this becomes 1 2 mRg mgR mgR cos cos θθ , which reduces to cos = 2 3 , and gives == ° cos . 1 23 482 bg as the angle at which the pumpkin will lose contact with the surface. P8.52 (a) Um g R A = 0 200 9 80 0 300 0 588 .. . . kg m s m J 2 ej a f (b) KUKU AABB +=+ KKUUm g R BAAB =+−= = 0588 . J (c) v K m B B = 2 20588 0200 242 . . .
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