Centripetal Confusion: Circular Motion Explained

AI Thread Summary
Centripetal force is essential for maintaining circular motion, and it acts toward the center of the circular path. When a car travels on the outside of a circular hill, gravity serves as the centripetal force, but if the car's speed exceeds a certain threshold, gravity alone may not be sufficient to keep it on the circular path, causing it to leave the track. The discussion clarifies that without a force directed toward the center, an object cannot maintain a circular trajectory, whether on the inside or outside of a loop. The concept of centripetal force can manifest in various forms, such as tension or gravity, depending on the situation. Ultimately, understanding centripetal force involves recognizing that it is the result of external forces acting on the object in motion.
Muon12
Messages
34
Reaction score
0
This is a classical mechanics question requarding circular motion. I understand that centripetal force causes objects to accelerate toward the center of the circular path they travel in, but how does this apply to objects traveling on the outside of a circle? I have come to understand that if an object is traveling within a loop, (with enough v) it's natural tendency is to remain on the inner edge of that circle, since its velocity is directed tangentially, and the centripetal acceleration is directed toward the center of the circle. But my question is this: would that same object be pulled toward the center of the circle if it were traveling on the outside of the loop? Say for example, a car is driving over a hill that has a circular shape. If Fr=-mv^2/r (- in this case since its traveling on the outside), then would we consider the centripetal accel. to be inward, especially when, giving enough velocity, this car would take off and leave the circle (until gravity pulled it back down, that is)? Conceptually, how is the acceleration (ar) directed toward the hill's center when the centripetal force seems to have little effect on the car at this point?
 
Physics news on Phys.org
Frankly it is rather difficult to find your question among the verbiage.

To maintain a circular path a centripetal force is required. A car does not always stay on the outside of circular loop because there is NO centripetal force to keep it there. A centripetal force is not present the instant something is curved, the force must be provided by some mechanism. When the car is on the inside loop, the track provides the centripetal force, there is no corresponding force when the car is on the outside of the loop. So if gravity is not sufficient to provide the force the car will leave the track.

Does that help?
 
Let me add a few comments to Integral's answer.
Originally posted by Muon12
This is a classical mechanics question requarding circular motion. I understand that centripetal force causes objects to accelerate toward the center of the circular path they travel in, but how does this apply to objects traveling on the outside of a circle?
In exactly the same way. If an object moves in a circle, then there must be a force pushing it towards the center.
I have come to understand that if an object is traveling within a loop, (with enough v) it's natural tendency is to remain on the inner edge of that circle, since its velocity is directed tangentially, and the centripetal acceleration is directed toward the center of the circle.
Careful! The "natural" tendency for any object is to keep going in a straight line at a constant speed. The only reason it goes in a circle is because something is pushing it towards the center. The only way a car can go around a circular track is if a force exists to push it towards the center. The only external forces on the car are the road and it's weight. The road exerts two forces on the car: a friction force sideways and a "normal" force pushing straight out of the ground. The weight just pulls down. The only way a car can go in a circle (inside or outside a loop) is if these forces happen to point towards the center!
But my question is this: would that same object be pulled toward the center of the circle if it were traveling on the outside of the loop? Say for example, a car is driving over a hill that has a circular shape.
Gravity pulls it down (which is toward the center)! But that force is fixed; go too fast and it won't be enough to hold you to the ground.
If Fr=-mv^2/r (- in this case since its traveling on the outside), then would we consider the centripetal accel. to be inward, especially when, giving enough velocity, this car would take off and leave the circle (until gravity pulled it back down, that is)?
The acceleration is just a description of the motion. If it goes over a hill, then there is some "centripetal" acceleration. Think of acceleration as the effect of some force, which is the cause. No force, no acceleration, centripetal or otherwise.
Conceptually, how is the acceleration (ar) directed toward the hill's center when the centripetal force seems to have little effect on the car at this point?
The only "centripetal" force in this case is gravity: the weight of the car. Sure it has an effect---it pulls the car down! But if you go too fast, it won't be enough to make you go in a tight enough circle (centripetal)---you'll keep going straight: into the air! Of course, once the car is in the air, the tires can't push the ground any longer, so the car falls just like a tossed ball would.
 
Just Curious

Within the Event Horizon of a Black Hole, centrifugal force works in the opposite dsrection from common experience. This means that orbitting at a high rate of speed around the center generates a force pushing one toward the center, rather than away.

Does this mean that within the EH, a centrifugal force becomes centripedal, or is it still considered centrifugal force with a reversed vector value or something like that? Just a matter of terminology, but I'd be interested in knowing the answer.
 
Lurch,
That is the first I have heard of that phenomena. Does it connect to the concept of closed geodesics, which I believe occur inside the EH. I am sure that once you have crossed the EH you have entered the singularity, therefore you must be very careful about application of normal laws of physics.
 
Circles do NOT HAVE "inner" and "outer" edges. Centripetal force and acceleration applies to any object moving in a circle. It is not a question of "inside" or "outside".
 
Originally posted by Integral
Lurch,
That is the first I have heard of that phenomena. Does it connect to the concept of closed geodesics, which I believe occur inside the EH. I am sure that once you have crossed the EH you have entered the singularity, therefore you must be very careful about application of normal laws of physics.

Yeah, I found it in an article in SciAm about two years ago. I will include a link if I can find the article. However, I do not think it is entirely accurate to say that once one crosses the EH, one has entered the singularity. Perhaps it would be more correct to say that inside the EH, all world-lines lead to the singularity. This article was about one surprising feature found in the math that describes what happens in the space in between.
 
Well, I found http://fy.chalmers.se/~number44//PAPERS/Pop-Papers/Sci-Amer.html that makes refference to the artical, including the issue in which the article is located (March 1995).

Doesn't include any of the article, though.
 
Last edited by a moderator:
arial

Thanks for the replies. I think I understand what the term centripetal force means now. It is only a way of describing forces as they affect objects in motion and cause them to move in circular paths. What I forgot was that centripetal force can take on diffrent forms, like tension, gravity, and so on. Now, if I've got this right, you're saying that centripetal force is only as strong as the force causing it. So back to my car on a hill analogy, if that car's engine creates enough force, then it will leave the hill because gravity (the centripetal force at the time), is not strong enough to keep it on it's circular path...
 
  • #10
None of the other guys in this thread have answered, so I'm just piping in here to say that you have it exactly correct!
 
Back
Top