A cart slides down a frictionless inclined track to a circular loop of radiusR =

Question

A cart slides down a frictionless inclined track to a circular loop of radiusR = 8.0 m. In order for the cart to negotiate the loop safely, the normal force acting on the cart at the top of the loop, due to the track, must be at least equal to the cart's weight. (Note: This is different from the conditions needed to just negotiate the loop.)

|vmin|=

h=

|v|=

|a|=

Explanation / Answer

a) By mv^2/r = mg =>v^2 = rg =>v = sqrt[9 x 9.8] = 9.39 m/s (b) PE(initial) = PE(final) + KE(final) =>mgH = 1/2mv^2 + mgh =>9.8 x H = 1/2 x (9.39)^2 + 9.8 x 18 [ as h = 2R] =>H = 22.50m (c) We have to know the position of point (c). (d) By work energy theorem:- W = delta E =>F x d = PE(initial) =>m x a x d = m x g x H =>a = gH/d = [9.8 x 22.50]/25 =>a = 8.82 m/s^2

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