(23b) Loop-the-Loop

In the preceding section the motion of the "loop the loop" roller coaster was handled using the centrifugal force. You can also view this problem from the point of view of the outside world, using the centripetal force, but it is not as easy. At point A, on the top of the loop, both gravity and the centripetal force point downwards. So what is there that can keep riders in their seats?

Let us try solve that motion, using the concept of the centripetal force. A car going around a loop, with radius R and velocity V, is accelerating at a rate of V²/R towards the center (as long as it stays on the rails), and is therefore subject to a centripetal force mV²/R, also directed to the center. When the car is at point A, that force points downwards. Let "down" be now be taken as the positive direction along the vertical axis.

The centripetal force is provided by two sources: the weight mg of the car, directed downwards, and the reaction F_R of the rails. We have at point A

mg + F_R = + mV²/R

F_R = + mV²/R - mg

Now the car rides on rails. At point A the rails are above the car and therefore it can only push up against them. The rails then, reacting to the force, must push it down, somewhat similar to the situation in "Objects at Rest", in section #18 on Newton's second law. Thus F_R must be positive: if it were negative it would mean that the rails were pulling the car upwards, which they cannot do.

We thus require F_R > 0, that is

mV ²/R - mg > 0

mV ²/R > mg

• If all forces on the car ceased at point A, it would
     continue along a straight line to point B,
      in accordance with Newton's first law.
• If only gravity acted, it would follow a parabola
      to point C.
• For the rails to exert a positive pressure,
      they must constrain the car to a tighter curvature
      than gravity alone, forcing it to move to point D.