Problem: A roller coaster car is moving at 2.5 m/s when it is 40 m above the ground on a track which eventually feeds into a loop-the-loop of diameter 8 m as shown.
b) If the mass of the car and its occupants is 600 kg, what is the force that the track exerts on the cart at the top of the circular section?
Solution: Always select your reference level for potential energy as the lowest point in the system. That could be the top of the loop in this problem, but I'll use the bottom of the loop so that I have more terms in the equation to show you.
a) Using the energy balance method, and noticing that the masses divide out,
The negative sign means that the normal force was drawn in the wrong direction; it is down. That makes sense, since the cart is trying to move tangent to the circle. Apparently, the weight is insufficient to keep it moving in a circle; N supplies the additional force. (There must be flanges on the wheels so that N can act in this direction!)
Since both N and mg point down, the cart's occupants experience more "weight" than they normally feel. How much? Adding 89.1 kN and (600)(9.8), I get 94.98 kN for the cart and occupants. Equating this to mg and using 600 kg, I get a value for g of 158.3 m/s2. This is 16.2 times the normal value of 9.8; the occupants experience 16.2 g's - wow! Their next-of-kin are going to sue that amusement park company (pilots may take as much as 5 g's without blacking out, but 16.2 g's are going to make puddles in the cart's seats)!
A comment about significant digits: I have stated that we will round everything to three significant digits. If you compute N using 25.2 m/s, you'll get N = -89.4 kN. But if you use the unrounded value from the calculation of v, you'll get the value I've shown. Just another warning to avoid rounding until you get to the end of the calculations!