The situation of vertical circular motion is fairly common. Examples include:

• roller coasters
• water buckets
• cars traveling on hilly roads

The analysis of all these problems is similar, although there's a difference between cars on hilly roads and the other two examples. Let's start with the roller coaster / water bucket example.

As usual, begin with a free-body diagram.
Follow this up with an appropriate choice of coordinate system.

At rest, the free-body diagram is simple, with an upward normal force and a downward force of gravity. These are the only two forces in the system even when circular motion is going on. The force of gravity has a constant magnitude and direction. The normal force, however, changes both magnitude and direction.

In the radial direction (toward the center) there is a net force as long as the object is moving along the circle. This object also slows down on the way up, and speeds up on the way down, so there is an acceleration in the tangential direction, too.

Focus on particular points.

At the bottom, applying Newton's Second Law gives (in the radial direction):

SFr

=

mar

=

m v2 r

N - mg

=

m v2 r

So, at the bottom,

N

=

m v2 r

+ mg

What do we get at the top?

SFr

=

mar

=

m v2 r

N + mg

=

m v2 r

So, at the top,

N

=

m v2 r

- mg

How slow can the object be going at the top and not fall off?

The limit is where N goes to zero, so at the speed corresponding to:

mg

=

m v2 r

[Note that the m's cancel.]

At the point halfway up or down:

SFr

=

mar

=

m v2 r

N

=

m v2 r

What is mg doing at these points?

It provides the tangential acceleration that slows the object down (on the way up) or speeds it up (on the way down).

Now consider the case of a car going over a hilly road. What is the equation for the normal force when the car is at the bottom of a valley traveling at a speed v? Assume the car is traveling on a circular arc at that point.

1. N = mg
2. N = mv²/r
3. N = mg + mv²/r
4. N = mg - mv²/r
5. N = mv²/r - mg

At the bottom:

N

=

mg +

m v2 r

1. N = mg
2. N = mv²/r
3. N = mg + mv²/r
4. N = mg - mv²/r
5. N = mv²/r - mg

At the top:

N

=

mg -

m v2 r

Remember that the normal force is equal to your apparent weight, so this is why you feel heavier at the bottom and lighter at the top.

Finally, note that the key difference between the car on the road and the roller coaster/water bucket situation is that the normal is up at both the bottom and the top for the car.