Finding the Angle of Release for a Pellet on Top of a Cylinder

[Bling]

I need a little help with this problem.

There is a small 0.2kg pellet sitting on the top of a cylinder of radius 0.8m.

What is the value of the angle at which the pellet loses contact with the surface of the cylinder?

I just need a little help on how to get started with this one. I was thinking about finding the point where the normal and tangential acceleration of the pellet were equal, but wasn't sure if this was the right path.

Any other additional help would be greatly accepted.

 

[Doc Al]

Originally posted by Bling
I just need a little help on how to get started with this one. I was thinking about finding the point where the normal and tangential acceleration of the pellet were equal, but wasn't sure if this was the right path.

First, realize that the pellet is centripetally accelerated when it follows the cylinder. Then consider that the pellet leaves the cylinder when the normal force equals zero.

 

[Bling]

I know the value of the normal acceleration is V^2/R, but how is this used to find the angle?

 

[Bling]

Also, I know how to find the velocity of the pellet at a given angle. Does this lead to the answer?

 

[NateTG]

You know what the forces are:
The normal force and gravity. If you draw an FBD, you can determine the magnitude of the maximal centipetal acceleration due to gravity as a function of the angle of the pellet on the cylinder, and thus the height.

Clearly the pellet leaves when the two are equal.