Arun
Last Activity: 6 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).
