A frictionless track A B C D E ends in a circular loop of radius R , from the given figure. A body slides down the track from point A which is at a height h = 5   cm . The maximum value of R for the body to successfully complete the loop is

Step-by-step solution

H is sliding on a smooth plane, now using conservation of energy at the highest point A , there is potential energy m g H . Now, at a point C the normal force is, N = m g + v 2 R Here, due to circular motion, a centripetal force is also present, so according to the law of conservation of energy, N = m g + v 2 R = 0 v = g R . . . i Now, according to the law of conservation of energy, m g H = m g 2 R + 1 2 m v 2 m g H = m g 2 R + 1 2 m g R H = 2 R + 1 2 R H = 5 R 2 = 5 R = 2   cm