A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. What is the ratio of the normal force to the gravitational force?

Question

Freddy

Physics

26 May, 13:30

A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. What is the ratio of the normal force to the gravitational force?

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Peter Avila · Answer 1

Critical speed is assumed to be that tangential velocity v where g = v^2/R and g = 9.81 m/sec^2 and R is the radius of rotation. The normal force N = mg - F; and F = mv^2/R = mg when at critical speed, as I defined it. Thus N = mg - mg = 0 at v. Now, when v = 2v, we have n = mg - m4v^2/R = mg - 4mg so that n/mg = - 3 QED. n is the normal force and mg is the gravitational force. The minus sign means the car is going flying unless it is somehow anchored.