1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Centripetal force confusion

  1. Jan 4, 2012 #1
    This might be silly question for some people, yet better to ask than dont know the answer :(.
    so the thing is , i know that centripetal force holds a thing to keep running on a path,
    but when i think of earth and all those planet which surround it, the centripetal force is gravity right ? and why those planets dont collide with earth ?
    Thanks
     
  2. jcsd
  3. Jan 4, 2012 #2

    Ken G

    User Avatar
    Gold Member

    Centripetal force is just the name for the F in F = ma when "a" is centripetal acceleration, i.e., when the motion is in a circle at given speed. Since F is the sum of whatever real forces add up to the F in that equation, so is centripetal force. So if a planet orbits the Sun in a circle, then it has a centripetal (net) force on it. If only one force is doing it, then that one force is the centripetal force-- as is typical for gravity. The planet does not fall into the Sun, even though it has a net force on it, because the force is needed to keep it going in a circle-- a different (stronger) force would be needed to get it to fall in.
     
  4. Jan 5, 2012 #3
    Hi,
    centripetal force on the planet due to sun's gravitation is balanced by centrifugal force due to planets rotation. That is why planets are stable in their respective orbits.

    PS: are you from CNU daejeon, south korea?
     
  5. Jan 5, 2012 #4
    No, im from melbourne, australia :)
    oh ok , so if i think of a car travels in a circular path, the centripetal force is its gravity and the force which balances it is its normal force right ? but when the car goes too fast, thus, the force in tangential direction is so large that the centripetal force cant hold it anymore then the car off the ground ? am i wrong here ?
     
  6. Jan 5, 2012 #5

    Doc Al

    User Avatar

    Staff: Mentor

    If you view the planet's motion from an inertial frame (the rest frame of the sun), then there would be no need to introduce 'fictitious' forces such as centrifugal force. In an inertial frame, the planet is accelerating towards the sun, thus there must be a net force on it.

    (And the centrifugal force, should you care to use it, is due to viewing the motion from an accelerating frame moving with the planet. It's not due to the planet's rotation.)
     
  7. Jan 5, 2012 #6

    Doc Al

    User Avatar

    Staff: Mentor

    When you drive over a curved surface like a hill or bump, you are accelerating centripetally (downward, in this case). The only vertical forces acting on the car are gravity (acting down) and the normal force of the road (acting up). As long as the net force required (the 'centripetal force', which depends on speed and the radius of the hill) is less than or equal to the weight of the car, you can maintain contact with the road. But go too fast and you will not be able to generate the required centripetal force to maintain contact--you'll come off the ground.
     
  8. Jan 6, 2012 #7

    Ken G

    User Avatar
    Gold Member

    I think you may be confusing the vertical and horizontal directions. For a car going in a circle, gravity and the normal force balance, but that is in the vertical direction. The centripetal force needed to get it to go around the circle is in the horizontal direction-- toward the center of the circle. It is generally provided by friction from the tires. The only connection between that horizontal centripetal force and the vertical forces is that "static friction" (the friction when the wheels are rolling not skidding) is often limited by some order-unity coefficient times the normal force. You wouldn't even need this at all if you knew the wheels were rolling, but if asked when the car skids out, you need to find the maximum static frictional force possible, and ask if it is enough to provide the centripetal force to get the car to stay on the circular road.
    Generally speaking, it is simplest to work in an inertial frame, like Doc Al was talking about (say the frame of the road). In that frame, there is only one tangential (horizontal) force, and that's the friction-- which is also called the centripetal force if the car is holding the turn. In that frame, we would not frame a skid by saying that some new tangential force appeared and overpowered the centripetal force, we would simply say the maximum possible frictional force is insufficient to provide the necessary centripetal force to keep the car on the circular road. In that case, the car skids, and there is no centripetal force at all any more-- the car is now skidding in a straight line.
     
  9. Jan 6, 2012 #8

    Doc Al

    User Avatar

    Staff: Mentor

    We need to distinguish two situations:

    (1) A car going in a horizontal circle, which Ken G described. (And which is probably what you had in mind.)

    (2) A car going over a bump or hill, which I described. (I may have misinterpreted what you had in mind, which was probably the first situation.)

    Similar considerations apply, but obviously the forces involved--and their directions--are different.
     
  10. Jan 6, 2012 #9
    Actually what i had in mind is a car going over a hill.
    and i understand that with enough velocity the car will be off the ground
    but the thing that still confuses me is :
    lets say when the car is speeding up, the normal force experienced by the car is getting smaller and smaller. and if it's so, why it's still on the track ? ( what i have in mind is : to be on track, gravity=normal force, action reaction pair)...
     
  11. Jan 6, 2012 #10

    Doc Al

    User Avatar

    Staff: Mentor

    That's true. The faster the car goes over the hill, the less the normal force pressing it against the road
    As long as the normal force is non-zero, there's still contact between car and road.
    No.

    (1) To be on the track just means that there's a non-zero normal force.

    (2) Gravity and the normal force are not 'action/reaction' pairs! There are two 'action/reaction' pairs in this problem:
    - The earth exerts a gravitational force down on the car (the car's weight) and the car exerts an equal and opposite gravitational force up on the earth.
    - The car presses down on the earth and the earth presses up with an equal and opposite force (the normal force) on the car.

    (3) Gravity does not equal the normal force! If you were to stop the car on the top of the hill, then gravity will equal the normal force. But if you're moving, the normal force will be less than the car's weight.
     
  12. Jan 6, 2012 #11

    Ken G

    User Avatar
    Gold Member

    OK, so the scenario Doc Al answered is what you have in mind. Your additional questions can be handled easily, they are relatively common misconceptions that once you iron out you'll be fine. First of all, gravity and the normal force are not an action/reaction pair. There are many situations in physics where two forces will be equal and opposite for various reasons that are special to the problem, but resist the temptation to interpret any two equal/opposite forces as action/reaction pairs! Action reaction pairs never act on the same object, and they are never two different types of forces. Gravity and the normal force both act on the car, and are two different types of forces, so are not an action/reaction pair. What this means is, they do not have to be equal and opposite! Sometimes they are, sometimes they are not, you have to write F=ma and work it out. If a car is going over a hill, it will always have the same force of gravity on it, but the normal force won't equal that, and indeed the normal force goes to zero if the car flies off the track. F=ma will tell you when that is, where F is the sum of all the real forces (including the normal force and gravity).
     
  13. Jan 6, 2012 #12
    ok. i think i got the idea. so the force which make the car still on the ground is gravity (whenever it is off the ground gravity will force it to come on the ground). But looking at the force experienced by the track/ road or whatever it is , this force is influenced by the car's velocity. The faster the car goes over the hill, the less force experienced by the track/road (which results in the less normal force). So the centripetal force in this case is surely gravity. Am i wrong ?
     
  14. Jan 6, 2012 #13
    ah sry the centripetal force depends on Fnet. not gravity
     
  15. Jan 7, 2012 #14
    wanna ask a basic question. let say there is an apple , 1. there is a force of earth on apple 2. there is force of apple on earth. these two are equal but in opposite direction. So where is the gravity from ? i mean what i have in mind is the force of earth on apple = gravity. i know i am wrong here, please correct me.:(
     
  16. Jan 7, 2012 #15

    Doc Al

    User Avatar

    Staff: Mentor

    Nothing wrong with what you are saying. The earth and the apple exert equal and opposite gravitational forces on each other. The gravitational force on the apple is its weight, mg.
     
  17. Jan 7, 2012 #16
    Does that mean that the earth is moving a bit (because of the force of apple on earth) but we cant see it ?
     
  18. Jan 7, 2012 #17

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. If you drop an apple, the apple accelerates. The same force exerted on the earth accelerates the earth, but since the mass of the earth is so huge, the acceleration is ludicrously small.
     
  19. Jul 4, 2012 #18
    I am probably reviving a dead thread, but am not a troll. I wanted to ask, since the topic is already present, how exactly does the frictional force(which is acting as the tangent to circle tight?) act as a centripetal force. Can some one show it by a diagram or something?
     
  20. Jul 5, 2012 #19

    Doc Al

    User Avatar

    Staff: Mentor

    I presume you are thinking of (for example) a car moving at constant speed in a horizontal circle? In that case the friction force acts toward the center of the circle, not tangentially.
     
  21. Jul 5, 2012 #20
    Yes, thats exactly my point! How? If i take one instant, of the car in the circular motion, the velocity will be acting at a tangent. This is because the force of the engine, is acting straight(i.e at a tangent to the circle), and hence shouldn't friction oppose this motion, acting at the opposite direction(but still at the tangent)??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Centripetal force confusion
  1. Centripetal confusion (Replies: 9)

  2. Centripetal force (Replies: 11)

  3. Centripetal force . (Replies: 6)

  4. Centripetal Force (Replies: 9)

  5. Centripetal force (Replies: 7)

Loading...