# Loop de loop problem

1. Jun 22, 2009

### the_obs

Just a quick question, concerning the (probably common) loop de loop problem.
At the top, why can we say that the centripetal force has to be at least equal to the force of gravity, so that the car not fall (excluding any friction)?

Both forces point towards the bottom (or the centre of the circle, more exactly), so shouldn't the centripetal force be opposite that of gravity?

2. Jun 22, 2009

### Staff: Mentor

Note that the term "centripetal force" refers to the net force acting towards the center. The actual forces that comprise the centripetal force are gravity and the normal force. Since both of those forces act toward the center, the centripetal force must be at least equal to gravity.

At the bottom of the loop things are different, because the centripetal force must act upward while gravity acts downward.

Make sense?

3. Jun 22, 2009

### the_obs

Ohhhh right, I get it.

Thank you very much! I didn't really grasp the concept of centripetal force as a net force, which actually makes sense, seeing as it's calculated by sum of forces =ma, a = v^2 /r, thus sum of forces = mv^2/r...

Thanks!

4. Jun 22, 2009

### tiny-tim

Hi the_obs!

but I think calling it centripetal force is bad and confusing …

there are only two forces on the car, the gravitational force and the normal force …

by good ol' Newton's second law, their sum equals the mass times the centripetal acceleration

there is no centripetal force.

5. Jun 22, 2009

### Staff: Mentor

I agree with the overall point, tiny-tim, but to say that there's no centripetal force is a bit extreme. :tongue2: That's why I refer to centripetal force as a net force, not as an actual individual force that would appear on a free body diagram. But good point!

6. Jun 24, 2009

### unscientific

My physics teacher always say, never include centripetal force into free-body diagram, cause its the NET force. It only exists when a some force exerted on a body PROVIDES for the REQUIREMENT of centripetal force.

Take earth orbiting the sun, the gravitational force exerted on earth by sun PROVIDES for the centripetal force.

7. Jun 24, 2009

### rcgldr

In order for the car to stay on the track at the top of the loop, the minimum normal force is zero. The car is in free fall during the instant it is at the top of the loop. There is no force between the track of the loop and the tires of the car (or vice versa). In this special case, the only force at the top of the loop is gravity, and being the only force, centripetal force = gravitational force.

If the car were moving slower it would fall. If the car were moving faster, then there would a be a normal force exerted by the track onto the car, coexisting with an equal and opposing force exerted by the car onto the track at all times.

The force the car exerts onto the track is the sum of the gravitational and reactive (to centripetal acceleration) forces.