# Loop the loop

A steel ball rolls down an incline into a loop-the-loop of radius R. What is the minimum speed the ball must have at the top of the loop in order to stay on the track? (the ball has a radius r and starts at a height of h)
This is a similar image to the one in my book:
http://img98.imageshack.us/img98/1876/1m40209ji.th.jpg [Broken]

I'm not sure how to solve this question. But I think it has something to do with conservation of energy. I tried doing that and got:
v = square root of (v02 - 20/7 gR)
but the answer is the square root of (gR), which is must simpler than my answer...so I must be doing something wrong.

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Fermat
Homework Helper
It has to do with centripetal force, rather than energy conservation.
The ball has to not fall off the top of the track, which means it must maintain a minimum amount of circular motion.
It is centripetal force that provides this circular motion.
When travelling around a track, it is the normal reaction of the track against the ball which provides the centripetal force.
When the motion is in a vertical plane, this reaction is modified by any radial component of the ball's weight.
At the top of the track, the radial component of the ball's weight is mg.
If there is no reaction force from the track then mg is the minimum amount of centriptal force.
If mg maintains circular motion (no falling off the top of the track) then mg = mw²r

Fermat said:
If there is no reaction force from the track then mg is the minimum amount of centriptal force.
This 2nd last sentence I don't understand.

Doc Al
Mentor
There are two forces acting on the ball at the top of the loop: gravity and the normal force (the track pushing on the ball). These forces add to produce the centripetal force on the ball. The faster the ball goes, the greater will be the normal force. To find the minimum speed such that the ball barely maintains contact with the track, set that normal force to zero. (Then apply Newton's 2nd law.)

I think I'm beginning to understand.
But why does the normal force have to be greater if the speed is greater? Is it because there needs to be a greater force to keep the ball in the loop? (Rather than fly off tangentially)

so I use Fcentripetal = mg, and get the answer v = square root of (gR)

Doc Al
Mentor
endeavor said:
Is it because there needs to be a greater force to keep the ball in the loop? (Rather than fly off tangentially)
Exactly.

so I use Fcentripetal = mg, and get the answer v = square root of (gR)
Right.