# Is normal force always 0 at the top of a circle during uniform circular motion?

I'm getting lots of mixed answers from throughout the internet. Some answers say that during vertical uniform circular motion, the normal force is always equal to 0 because only gravity needs to do work as the centripetal force at that point. Other answers say that for example if a car were driving towards the top of the hill, if the normal force gets smaller then the car would fly off the hill. I would like some clarity on the matter.

Note: I'm a hs freshman self-studying physics and I havent taken an actual physics class yet, so I can't ask my teacher.

Uniform circular motion encompasses many different situations, including some situations where there is no surface involved and thus nothing to exert a normal force (like a yoyo in space, or an object in orbit), as well as other situations where there is a surface to exert a normal force, as well as yet more situations where there is a surface but it's not exerting a normal force. So based only on the information you've provided, it's impossible to say.

This may have something to do with why you're finding inconsistent information.

1) during a rotation or any kind of dynamic motion, we cant assume a value of normal force, normal force strictly depends on the forces experienced by the body at certain instant.

2) if the net force experienced by body without considering normal force cancels out i.e $$Fnet$$ = $$0$$, then we can assume normal force is $$0$$, but only if we know value of forces.

3) however in most cases we don't know the values of forces , even if we know the type or direction of force the body experience for example

when car moves towards top of hill we , know that it the body experience centrifugal force( iam talking in frame of car , non inertial frame, however result would be same in inertial as well considering centripetal force instead) and its own weight is pointing downwards, now imagine if the weight of body becomes too high that it exceeds centrifugal force $$\cfrac{mv^2}{R}$$ which is pointing outwards then there would be a $$Fnet$$ downwards , hence now to prevent body going inside the surface of hill we need to consider normal force $$N$$(pointing up) such that

$$N$$ $$+$$ $$\cfrac{mv^2}{R}$$ $$=$$ $$mg$$

the above equation in inertial frame can be viewed as that $$mg$$ $$-$$ $$N$$ is net force downwards which creates necessary centripetal force for the car to move over top of hill. however remember if forces other than normal force balance each other we cant consider normal force, but that is only possible when we know the actual value of forces, it is always better to assume a normal force ( don't assume its value yet, create equations for equilibrium of forces to find normal force) to see if forces are balanced or not

I'm assuming you're talking about a car moving in a circular loop track.

At the topmost position, if you draw the free body diagram, you'll see that the normal force and the force due to gravity ($$mg$$) are in opposite radial directions. So, they together provide the centripetal force.

$$mg - N = \frac {mv^2}{R}$$

The faster the car travels, the less the normal force becomes.

As David Z pointed out, it really depends on the case you're looking at. However, what's important to know is that the vector sum of radial forces provides the centripetal force in all cases, whether a ball on a string, car in a loop, and so on.