What is the minimum height that a mass can be released from rest and still make it around the loop-the-loop without falling off? Assume that the mass is a uniform sphere that rolls without slipping throughout.

For no friction, we found the minimum height to be h = 5r/2. With rolling, is the minimum height:

- greater than 5h/2
- less than 5h/2
- equal to 5h/2

From the loop-the-loop example with a freely-sliding mass, the minimum
speed at the top of the loop for the mass to remain on the track is
v^{2} = gr.

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}.

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

New feature, the kinetic
energy includes rotational motion:

K_{f} = ½ m v^{2}
+ ½ I ω^{2} = ½ m v^{2}
+ ½ (2 m r^{2}/5) ω^{2} =
7 m v^{2}/10

Energy conservation: mgh = mg(2r) + 7 m v^{2}/10

Using v^{2} = gr for the minimum velocity, we obtain:

mgh = mg(2r) + 7
mgr/10

Finally, h =
2 r + 7 r/10 = 27 r/10.