# Conservation of energy on loop. Nearly done!

1. Oct 17, 2011

### irishbob

1. The problem statement, all variables and given/known data
A 1500-kg roller coaster car starts from rest at a height H=23.0m above the bottom of a 15.0-m-diameter loop. If friction is negligible, determine the downward force of the rails on the car when the upside-down car is at the top of the loop.

2. Relevant equations
Conservation of energy: U+K=Esys
U=mgh
K=0.5mv2
F=ma
acentripetal=v2/r

3. The attempt at a solution
Uinitial=Esys because starts from rest. Uinitial=mghinitial
at the top of the loop: Esys=U+K=mghtop+0.5mv2
meaning mghinitial=mghtop+0.5mv2
simplifying: v2=2(ghinitial-ghtop).
so:
F=macentripetal=mv2/r=
1500kg*2(9.81m/s2*23m-9.81m/s2*15)/7.5m

So that gives an answer of 31,392N. I know the answer is 16.7 kN (back of book), and I know you get there by subtracting mg. So why am I subtracting mg? I would think that would be the TOTAL force and not solely the force of the tracks on the car.

Any and all help is much appreciated!

Edit: You know, I think I get it. I solved for centripetal force, which is the total inward force toward the center of the circle. Because mg is in that direction, I subtract it and get the force from the tracks. Is this right?

Last edited: Oct 17, 2011
2. Oct 18, 2011

### Staff: Mentor

Yes, that's right. The 'centripetal force' is just the net force. To find the contribution from the tracks, you must subtract mg. It's always best to think in terms of Newton's 2nd law:
ΣF = ma
N + mg = mv2/r
N = mv2/r - mg

3. Oct 18, 2011

### PeterO

Glad you worked it out - I highlighted some wording in the question, and your pondering which re-inforces what you realised.