Loop de Loop Problem: Equal Forces?

  • Thread starter Thread starter the_obs
  • Start date Start date
  • Tags Tags
    Loop
AI Thread Summary
In the loop de loop problem, the centripetal force at the top of the loop must equal the gravitational force to prevent the car from falling, as both forces act toward the center. The concept of centripetal force is clarified as a net force resulting from the sum of gravitational and normal forces. At the top of the loop, when the car is in free fall, the only force acting is gravity, equating centripetal force to gravitational force. If the car's speed varies, the normal force will adjust accordingly; too slow results in falling, while too fast increases the normal force. Understanding centripetal force as a net force rather than an individual force is crucial for accurate physics analysis.
the_obs
Messages
2
Reaction score
0
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?
 
Physics news on Phys.org
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?
 
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!
 
Hi the_obs! :smile:
the_obs said:
I didn't really grasp the concept of centripetal force as a net force …

Maybe I'm a bit obsessive about this :rolleyes:

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. :wink:
 
I agree with the overall point, tiny-tim, but to say that there's no centripetal force is a bit extreme. :-p 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!
 
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.
 
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.
 
Back
Top