# What is causing the normal force in circular motion? In the picture, at point 2 (the bottom of the ramp) the normal force of the object has a greater magnitude than weight. I understand that the normal force has to be greater than the weight since the acceleration points towards the center of the circle and the net force is in the same direction as the acceleration. However, that doesn't explain what causes the normal force to be greater than weight.

If the object was placed at point 2 without the prior motion (sliding down the ramp) in the diagram, the normal force would equal weight. But why does sliding down the ramp increase the magnitude of the normal force at point 2? Here's another way to phrase it: since normal force is the reactionary force of the force the object exerts on the surface, why is the force of the object on surface greater than the weight at point two?

• It's the difference in the angle $\cos(\theta)$. Feb 17 '15 at 1:58
• Based on my understanding of your comment, the angle explains the difference of the magnitudes of the normal force at various points along the ramp but doesn't explain why at point 2 the normal force is greater than w (weight). Feb 17 '15 at 2:01
• @Powdaq but, who told you that the body becomes heavier than its weight? Feb 17 '15 at 2:06
• @Powdaq the object gets an angular momentum that tends to be conserved, but where from can arise an additional force than gravity? Feb 17 '15 at 2:09
• @Sofia does momentum cause the normal force to be greater than the weight at point 2, as opposed to normal force=weight if the object is placed at that exact location? Feb 17 '15 at 2:13

Consider it in this way:

Suppose the normal force is not large enough, i.e.

$$N<mg+m\frac{v^2}{R}$$

the object will not have enough centripetal force to continue its circular motion, so it will try to increase the radius and leave the current circular orbit. In this way, it's "pushing" against the track, which will eventually increase the normal force until it balance the gravitational force and provide enough centripetal force.

Point 2 exerts more force than the force of gravity (a.k.a. weight of the skater) because there is an acceleration term. Remember, F=ma, or more accurately ΣF = ma. In this case ΣF is composed of the normal force and the force of gravity. Near the bottom of the curve, we are still trying to change the skater's velocity, which means there is an acceleration.

I don't think the book is drawn very well. Once the skater reaches flat land (after the ramp), he stops accelerating, so ΣF = 0 and you get what you expect: the normal force is equivalent in magnitude to that of the force of gravity. Because it is a smooth curve, it is clear to see that "near" the bottom of the curve, acceleration should be "nearly" zero. The maximum normal force (the one which causes us to see a normal force greater than gravity) should occur somewhere in between the top and bottom of the ramp.

Consider a roller coaster, which is the same idea but a bit more accessible to non-skaters. You get taken way up to the top of the hill, then dropped, with nothing but normal forces and gravity (just like the skater). During the drop, you feel weightless. At the bottom as you screech towards the first turn, you feel exactly 1g (normal weight). However, at some point during the curve from dropping to straight-and-level, you feel crushed into your seat. That crushing feeling is your body's intuitive analog of a normal force that is much much higher than gravity as it frantically tries to bend the path. It doesn't occur at the bottom of the hill; it occurs somewhere in the middle of the curve.

Also intuitive, take a bowling ball, put it on a table. It has no trouble with those normal forces. Now drop that same bowling ball off of a parking garage (check for pedestrians first). For a split moment, the normal force of the ground pushing on the ball is substantially higher than the force of gravity, as the ball-ground pair work together to try desperately to apply the decelerations needed to stop the ball before it goes through the concrete. Those normal forces are sufficient to crack the ball clean in half!