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About 'origin' of the centrifugal force

  1. May 14, 2009 #1
    Just want to make sure my concepts about this fictitious force is correct...

    Assuming Joe is spinning around on a stationary point, while holding a string tightened onto stone. Therefore, he is providing a centripetal force to the stone through the string, so that it rotates around him.

    Suddenly Joe decides to release the string. As a result, the stone flies out in the tangential direction to its orbit at that instant. However, since he is still spinning, the stone appears to to be flying out in a direction perpendicular to its actual one. Hence, a centrifugal force appears to be acting on the stone.

    Is the above description correct?
  2. jcsd
  3. May 14, 2009 #2


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    Once he releases the stone, there is no centripetal force involved, except for some Magnus Effect if the stone is spinning fast enough. In the more generalized definiation of centrifugal simply meaning outwards, then while twirling the string exerts a centripetal force on the stone, coesitant with the stone reacting to the centripetal acceleration with a "reacitve centrifugal force".
  4. May 14, 2009 #3

    D H

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    Not quite. Rather than looking at what happens when Joe lets go of the string, it is better to look at things while Joe is still hold on to the string.

    Suppose that before Joe starts spinning he simply holds the end of the string with the rock hanging down. Joe is applying an upward force to the rock, and yet the rock is stationary. Joe knows (a) he is applying a force to the rock via the string, (b) that the rock is stationary, and (c) Newton's first law. From Newton's first law, the net force on the rock must be zero. Joe can deduce that there must exist some downward force that is exactly opposite the upward force that Joe knows he is applying to the rock. That downward force is gravity.

    Now Joe starts spinning. The rock lifts from vertical a bit toward the horizontal and, from Joe's perspective, becomes stationary again. Joe applies Newton's first law again. There must be some outward force on the rock. (Joe knows vectors.) Joe spins a bit faster. The rock lifts a bit more toward horizontal, the tension in the string increases, and the rock once again becomes stationary. This mystery force must be proportional to his rotation rate. (Joe knows data regression.) Now he lets the string out a bit. The rock lifts even closer to horizontal, the tension in the string increases, and the rock once again becomes stationary. This mystery force must also be proportional to the length of the string.

    So far, Joe has deduced this mystery force solely on the basis of Newton's first law. Our Joe is a bright boy. He decides to see what happens with Newton's second law. He lets go of the string. At first the rock flies straight out. But as soon as it gains speed it curves to the side. The static explanation of this mystery force doesn't quite jibe with what Joe is seeing -- unless there is yet another mystery force involved. While still spinning, Joe takes a rock out of his pocket and throws it. It curves away from him rather than flying in a straight line. There is yet another mystery force. (This new mystery force is of course the Coriolis force.)

    These forces appear because Joe has applied Newton's laws in a regime where they are not truly applicable.
  5. May 14, 2009 #4

    D H

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    Jeff, you aren't helping.

    annatar, ignore what Jeff said. In particular, forget about the concept of a reactive centrifugal force, which is a flawed concept. (Wikipedia has an entry on it. You can read it. Just remember that Wikipedia has entries on lots of garbage.)
  6. May 14, 2009 #5
    By this line: "There must be some outward force on the rock. (Joe knows vectors.)" Do you mean...
    This outward (i.e. centrifugal?) force along the string can be resolved into vector components, one of which is pointing upwards. Hence the stone lifts more from the vertical ?

    So what could an observer say about this outward force?
    He shouldn't be able to find it, right?
    Is it that he can only sees Joe is exerting a larger centripetal force, for example by spinning faster?

    Last but not least, how is the concept of reactive centrifugal force flawed?
    I've seen the wikipedia article prior to making this post...
  7. May 14, 2009 #6
    The centrifugal force only exists if you co-move with the rotating stone; eg if you are in a rotation frame of reference.

    If you choose Joe as your frame of reference, there is no centrifugal force. The only (horizontal) force on the rock is the tension of the rope, which acts as the centripetal force towards the center of rotation.

    If you choose the rock as your frame of reference, then the centrifugal force appears in the equations to still allow Newton's laws to hold (which really only hold in an inertial frame of reference).
  8. May 14, 2009 #7

    D H

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    The centrifugal force is

    [tex]\boldsymbol F_{\text{centrifugal}} =
    - m \boldsymbol{\omega}\times (\boldsymbol{\omega} \times \boldsymbol r)[/tex]

    The centrifugal force therefore has zero upward component. The upward component of the force that counterbalances gravity comes solely from the tension in the string. The string is not perfectly horizontal, even if Joe is spinning very, very fast.

    Assuming the observer is not rotating, that is correct. The outward force "observed" by Joe is specific to Joe, and exists only in Joe's mind as a result of trying to apply Newton's laws to a regime where Newton's laws aren't really valid.

    Correct. Non-rotating observers do not see the rock as stationary. They see the rock accelerating inward toward Joe. Something is making that rock curve inward. The tension in the string and gravity are all that are needed to explain the motion of the rock from the perspective of a non-rotating observer (Note: This ignores the fact that our non-rotating observer is rotating -- at one revolution per sidereal day. Ignoring the Earth's rotation is often, but not always, a reasonable assumption.)

    Some examples:
    • Does the Earth exert a reactive centrifugal force on the Sun? The Sun is orbiting the Sun-Earth center of mass. Calling this inward pull on the Sun centrifugal is confusing at best. Now consider a binary star system, where the two stars have nearly the same mass. To an inertial observer, the two stars are clearly revolving about their common center of mass. The pull each star exerts on the other is very clearly inward. There is no outward force.
    • Now consider a robotic probe that has approached a small asteroid. The operators of the probe want the probe to make a fly-around of the asteroid so they can see the whole thing. Suppose that just letting gravity do its thing would make the fly-around take too long. If the vehicle has some extra fuel, the operators can use that to augment the asteroid's puny gravity. The vehicle can reduce the fly-around time by 29% by firing the thrusters outward equal the asteroid's gravitational force (the net force is now double that of the asteroid's gravitational force). So, what is this so-called reactive centrifugal force in this case? It's split between two objects, the exhaust and the asteroid. While the exhaust is pushed outward, the asteroid is being pulled inward.

    This is an old concept that is only being kept alive by diehards. Not quite quacks, but close to it. Modern physics texts (and by modern, I mean 50 years old, maybe older, as
    "modern") simply don't introduce the concept, period.
  9. May 14, 2009 #8


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    Hi annatar! :smile:

    Short answer: yes … a non-inertial observer has to invent non-physical forces (such as centrifugal and Coriolis force) so that he can apply good ol' Newton's first law :wink:

    For a longer answer, see the https://www.physicsforums.com/library.php?do=view_item&itemid=84") …
    Last edited by a moderator: Apr 24, 2017
  10. May 14, 2009 #9


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    Reactive forces are real (Netwons third law), the only controversy is calling that reactive force "centrifugal".

    If Joe is rotating while holding the string, like a guy flying a control line model, Joe is in a rotating frame of reference, in which case there is a centrifugal force. As another example, Joe could be on the edge of a fast turning carosel facing outwards. I don't think it matters that Joe is rotating at the axis of a rotational frame of reference.
  11. May 14, 2009 #10
    I see now that Joe is the one rotating. I was under the impression that he simply stood still and rotated the stone (like 'winding up' a lasso).

    I think my point however is still valid if we don't choose Joe, but rather Joe's position as the frame of reference. I'm not sure if that is relevant to this thread however :p
  12. May 14, 2009 #11
    Thanks you for the replies!

    So in short words:

    Joe knows he is exerting a centripetal force onto the stone.
    However since the stone is not flying towards him already, a outward force must be balancing the centripetal one.

    Is this the case? :smile:
  13. May 15, 2009 #12


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    If he regards the stone as stationary (in other words, if he is using a rotating frame), then yes …

    the stone is stationary, there is a centripetal force on it, so by Newton's first law there must be another force, outward.

    He further knows that the force is proportional to the mass, but is otherwise the same on all objects (the "inertial" mass, so same-on-all-objects forces are sometimes called "inertial forces"), and is proportional to the distance from him. :wink:
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