InterviewSolution
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InterviewSolution
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(4) 3.42. A small bodyof mass m can slide without friction along atrough bent which is in the form of a semi-circular arc ofradius R. At what height h will the body be at rest withrespect to the trough, if the trough rotates with uniformangular velocity about a vertical axis :(1)R(2) R-28 |
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Answer» F = mgsinθ - N When the ball loses contact, the normal force (N) = 0 => F = mgsinθ Since the ball is moving in a circular path, F = mv² / R mgsinθ = mv² / R => v² = Rgsinθ From conservation of energy, you know that: mgR = ½mv² + mgh Rg = ½Rgsinθ + gh h = R - ½Rsinθ If you draw out a force diagram you'll see that sinθ = h / R h = R - ½Rh / R 3h/2 = R h = 2R/3 So it will lose contact (2/3)rd of the way up. F = mgsinθ - N When the ball loses contact, the normal force (N) = 0 => F = mgsinθ Since the ball is moving in a circular path, F = mv² / R mgsinθ = mv² / R => v² = Rgsinθ From conservation of energy, you know that: mgR = ½mv² + mgh Rg = ½Rgsinθ + gh h = R - ½Rsinθ If you draw out a force diagram you'll see that sinθ = h / R h = R - ½Rh / R 3h/2 = R h = 2R/3 So it will lose contact (2/3)rd of the way up. F = mgsinθ - N When the ball loses contact, the normal force (N) = 0 => F = mgsinθ Since the ball is moving in a circular path, F = mv² / R mgsinθ = mv² / R => v² = Rgsinθ From conservation of energy, you know that: mgR = ½mv² + mgh Rg = ½Rgsinθ + gh h = R - ½Rsinθ If you draw out a force diagram you'll see that sinθ = h / R h = R - ½Rh / R 3h/2 = R h = 2R/3 So it will lose contact (2/3)rd of the way up. |
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