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A small body starts sliding down an inclined groove passing through into a half circle of radius h/2. find the speed of the body when it reaches the highest point of trajectory. ans:2/3*root(h/3)

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2 years ago

```							Suppose the particle leaves the groove at P (x,y) relative to the centre of the loop. x = Rsinθ and y = Rcosθ where R = h/2 and θ is measured anticlockwise from the vertical. The top of the loop is level with the initial position of the particle, so the velocity of the particle at P is v where by conservation of energy : v^2 = 2g(R-R cosθ) = gh(1 - cosθ). The condition for leaving the groove is that the normal reaction N becomes less than zero. N is given by N + mgcosθ = mv^2/R = 2mv^2/h. At point P where N = 0 we have mgcosθ = (2m/h)gh(1 - cosθ) cosθ = 2(1 - cosθ) cosθ = 2/3. After leaving contact with the groove at P, the horizontal component of velocity remains constant and has value vcosθ where. v^2 = gh(1 - cosθ) = gh(1/3) v = √(gh/3) vcosθ = (2/3)√(gh/3). So the speed of the particle at the highest point of its trajectory (where the vertical component is zero) is (2/3)√(gh/3).
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2 years ago
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