A loop-the-loop track consists of an incline that leads into a circular loop of radius r. What is the minimum height that a mass can be released from rest and still make it around the loop without falling off? Neglect friction.

First, we need to know the minimum speed at the top of the loop for the mass to remain on the track. We do this by DID TASC.

Step 1: Draw diagram and coordinate system.

Step 2: Isolate the system.

Step 3: Draw all forces.

Step 4: Take components.

Step 5: Apply Newton's second law.

Step 6: Solve.

Step 7: Check.

Step 3: At the top of the loop, the two forces are N and mg, both acting down.

Step 5:
ΣF_{r} = N + mg =
mv^{2}/r → N = mv^{2}/r - mg.

Step 6:
Minimum speed is given by N=0
→ v^{2} = gr.

Multiply by ½ m to get: ½
mgr = ½ mv^{2}.

The minimum kinetic energy at the top is ½ mgr.

Now determine the initial kinetic energy needed by D0EL.

Step 1: Define/draw diagram and coordinate system.

Step 2: Choose a consistent zero.

Step 3: Energy conservation.

Step 4: Losses.

Step 2: The zero level for potential energy is the bottom of the loop.

Step 3: U_{i} + K_{i} = U_{f} + K_{f}.

K_{i} = 0, U_{i} = mgh,
while K_{f} = ½ mgr, U_{f} = mg(2r).

Energy conservation: mgh = ½ mgr + mg(2r)

Finally, h = 2r + r/2 = 5r/2.

A mass released from lower than h = 5r/2 will fall off the loop. A mass
released from higher up will have a non-zero normal force at the top, and
will loop the loop.