A loop-the-loop track consists of a long incline that leads into a circular loop of radius r. If a mass is released from rest somewhere along the incline, what is the minimum height it can be released from and still make it around the loop without falling off? Assume the mass slides along the track with no friction.

Analyze the forces on the mass at the top of the loop. The free-body diagram has two forces, N and mg, both acting down.

ΣFr

=

mar

=

m v2 r

The acceleration is down, so make that the positive direction.

N + mg

=

m v2 r

So, at the top,

N

=

m v2 r

- mg

How slow can the object be going at the top and not fall off?

The limit is where N goes to zero, so at the speed corresponding to:

mg

=

m v2 r

Re-arrange this to read:   mgr = mv²

Multiply by ½ to get:

½ mgr = ½ mv²

This tells us that the minimum kinetic energy at the top is ½ mgr.

Applying conservation of energy, where the initial position is the release point and the final position is the top of the loop:

U_i + K_i = U_f + K_f

K_i = 0 and, setting the zero level for potential energy at the bottom of the loop, U_f = mg(2r). This gives:

mgh = mg(2r) + ½ mgr

So,

h

=

2r

+

r 2

=

5r 2

A mass released from lower than this will fall off the loop. A mass released from higher up will have a non-zero normal force at the top, and will be happy to loop the loop.

Note that we ignored any losses of energy. Some energy would be lost in a real system, meaning that we probably have to release the mass a little further up the track to make sure it loops the loop.