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:
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 v2 = 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: Ui + Ki = Uf + Kf.
Ui = mgh, Ki = 0 while Uf = mg(2r).
New feature, the kinetic
energy includes rotational motion:
Kf = ½ m v2 + ½ I ω2 = ½ m v2 + ½ (2 m r2/5) ω2 = 7 m v2/10
Energy conservation:   mgh = mg(2r) + 7 m v2/10
Using v2 = gr for the minimum velocity, we obtain:
mgh = mg(2r) + 7 mgr/10
Finally, h =
2 r + 7 r/10 = 27 r/10.