# Loop the loop

clarineterr

## Homework Statement

A cyllinder of radius rc starts a height h above a loop the loop on a track. If the radius of the loop is rl, how high does h need to be. The cyllinder does not slip. Express the answer in terms of the radius of the loop.

## Homework Equations

Conservation of energy

## The Attempt at a Solution

The cyllinder has potential energy mgh at the beginning and 2mgrl at the top of the loop. It also has kinetic energy 1/2mv^2 +1/2Iw^2 where v=$$\sqrt{rg}$$ is the critical velocity. The moment of inertia is I = 1/2mrc^2. I can't figure out how to get rid of rc in the answer.

I get h = rl/2 + rc^2/4rl + 2rl

## Answers and Replies

Staff Emeritus
Science Advisor
Gold Member
how high does h need to be.

To do what, exactly? Can you clarify what the problem is asking, and what it means when it says that the cylinder starts at height h "above" the loop?

Science Advisor
Homework Helper
Hi clarineterr!

(try using the X2 and X2 tags just above the Reply box )
The cyllinder has potential energy mgh at the beginning and 2mgrl at the top of the loop. It also has kinetic energy 1/2mv^2 +1/2Iw^2 where v=$$\sqrt{rg}$$ is the critical velocity. The moment of inertia is I = 1/2mrc^2. I can't figure out how to get rid of rc in the answer.

I get h = rl/2 + rc^2/4rl + 2rl

I can't see how you got that

can you give us some detail?

(and it isn't 2mgrl at the top … don't you need to subtract some rc?)

clarineterr
Im sorry. How high must h be for the cyllinder to be able to go around the loop.

At the top of the loop the only centripetal force is gravity if the cyllinder just makes it around the loop so v=$$\sqrt{grl}$$

then I used mgh = 1/2mv^2 + 1/2Iw^2 + 2mgrl and if you v/r for w, the moment of inertia and the velocity above mg cancels out.