Loop-the-loop Revisited

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:

  1. greater than 5h/2
  2. less than 5h/2
  3. 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   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.