Where a block stops on a loop-the-loop

1. Oct 10, 2012

kinof

1. The problem statement, all variables and given/known data

A small block of mass m slides along a frictionless loop-the-loop track. The block is released from rest at point P. What is the resultant force acting on it at point Q? At what height above the bottom of the loop should the block be released so that it is on the verge of losing contact with the track at the top of the loop?

2. Relevant equations

Wnc = ΔK + ΔU
Wc = -ΔU

3. The attempt at a solution

Point P is at the top of the loop 5R distance from the ground, and point Q is a point on the loop R above the ground and R from the center of the loop where the normal force is parallel to the surface. (Sorry if the image is a bit difficult).

I have found the force at point Q, so I am stuck on the second part.

This is how I have set it up.

From point P to point S (where N = 0) and where H is the height at which N = 0:

v^2 at point S is gR, so

0 = ΔK + ΔU = (1/2)mgR + (mgH - 5mgR)

(1/2)R + H = 5R

H = 4.5 R.

However, the correct answer is 2.5R. I don't understand, because ΔE = 0 the entire time because of the absence of friction, so can't I start at whatever point I want and so long as I have what the velocity is where N = 0, I can find the distance from the ground where that is?

Thanks.

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