Forces on a horizontal loop of a roller coaster?

Question

What would a free-body diagram look like of the net and normal forces (actually all the forces) on a horizontal loop of a roller coaster which is perfectly parallel to the ground. I think the normal force is pointing directly to the inside of the circle and the force of gravity (mg) is pointing directly down to the ground, which would make the normal force at a right angle or perpendicular to the gravitational force. What would be the equation for Fnet then?

Thank you for your help!

Answer 1

I think you're right. In my mind, I generalized your model somewhat ... suppose you have a tall hollow cylinder standing upright, so that the cross-section of the interior wall is a circle in the horizontal plane.

Now suppose you have a jet-powered sliding block that you can accelerate up to speed v around that circle some distance above the ground. Ignore friction, and when you attain the speed v, cut the power.

Now look at the forces acting on the free body. You have momentum tending to perpetuate straight line motion (as a function of v), and you have the normal force (a function of both m and v) directed toward the central axis of the cylinder, producing acceleration (the turning in the horizontal plane with no change in speed). In addition, as you point out, the gravitational force acts vertically.

The net force is just the vector sum (or you can do it with sines and cosines). There will be a downward trend because there's no vertical force to counteract gravity. You do have considerable angular momentum, so the faster you go, the longer the friction-free sliding block will stay up. (Similar to why a bicycle does not tip over.)

Anyway, that's how I'd approach the problem.

(BTW, an angular momentum approach might also do the job. That vector is vertical, pointing upward with proper orientation, with motion of the block around the horizontal circle following the right-hand rule. Just a thought, and you'd still have to consider gravity.)

[Edit: An afterthought ... in many ways, this problem is the same as firing a cannonball horizontally from the top of a cliff. The gravitational force is constant, and the distance it will travel is a function of the initial velocity. The faster it starts, the farther it will go before hitting the ground. In your problem, you can divide the horizontal distance (or better yet, the Pythagorean distance) by the circumference of the circular track. Of course, you want a force diagram, and you still need to calculate that. End edit.]