Take the top of the loop. The cart is upside down here, and the rail is above it.
The centre of the curve is straight below. Gravity points down, and the push of the rail points down as well. Both point at the centre.
So this time they add: . In the valley and on the hill the two forces were subtracted; here alone they are added.
Solved, . If the speed is too low the right side is negative, and there is no such thing as a negative normal force. The story ends there.
A rail can only push; it cannot hold the cart back. This is a cart with no wheels gripping from above. So the condition is .
Solving gives , the very equation that appeared as the condition for floating at the top of a hill.
The edge is , where . At that point gravity alone is holding the cart to the circle. The figure lowers the speed and stops where the arrow disappears.
Below this the cart leaves the rail and falls. Whether it makes it round is settled at this one point at the top and nowhere else.
We have a condition on the speed, but what can be settled before the ride is the height. So has to be translated into .
The equation from the previous lecture does it. Without friction, . The top of the loop is at a height of , so the speed there satisfies .
The path makes no difference, so the shape in between may be anything at all. Steep slope or gentle, only the descent from to is turned into speed.
Put this into . From here it is only a matter of solving.
Divide both sides of by to get , which tidies into .
has gone, and the mass was never in it. What is left is the ratio of a height to a radius. Fix the size of the loop and the height required is fixed at times it.
The figure varies the radius, and the line for the required height always sits at times it. That a larger loop needs a higher start is the whole content of this one inequality.
A real coaster starts from very much higher than this. It has to allow for what friction takes away, and for the fact that riding at the very edge of would be miserable.