Originally Posted by bitingduck

Not quite.

You stay on the track due to the friction between your tire and the track, while gravity is trying to pull you down off the track. The force holding you up is the friction coefficient between your tire and the track times the force you're applying to the track, which has two components-- a downward component from gravity and an outward component from centripetal/centrifugal force. The downward component is always the same for a given angle (given round tires). The outward component is your mass times the acceleration, which is (V^2)/R. Higher up on the track, the radius of the turn (the R) is larger because you're farther from the center of the circle (assuming a circle for convenience of description), so to get the same frictional force to keep you from sliding down you have to have a higher speed.

If you draw the free body diagram and do the math (just a few sines and cosines) you'll see it's correct - it's a standard freshman physics problem but it's usually phrased in terms of a car on a banked track and at what speed will they slide off the top. It's similar to skidding a car on the flat when you take a tight turn too fast (you just have to set the banking to zero degrees).

If you solve the whole problem you'll also see that the mass cancels out - the minimum speed is independent of your mass, which is why a big rider and small rider can both creep along together on relief in a madison or in a match sprint, and the one with less skill will slide first.