Dot Physics

It’s odd that I have talked about these forces so much. First, I talked about how centrifugal forces were not real and the difference between centrifugal and centripetal forces. Then I talked about how sometimes, fake forces are good. Finally, I talked about the origin of the words centrifugal and centripetal. (note: “talked about” means wrote a blog post)

In thinking about centripetal forces, I realized that I could come up with a situation in which the centrifugal force is the centripetal force. This is great. I can end all the confusion between centrifugal and centripetal by making a case where they are the same. So, here is the situation:

Suppose I am on a merry go round standing near the edge and I am swinging a rock around in a circle by a string. The rock is making vertical circles with the rock at one point in between me and the axis of rotation for the merry-go-round. Here is a picture. (I was going to model this in vpython, but I would rather just make a sketch)


I didn’t draw my person in there, but hopefully you get the idea. The arrow pointing up on the axis of rotation of the merry-go-round is the angular velocity vector for the platform. The point that I am interested in is when the ball is in the position shown. If I look at this from the frame of the rotating merry-go-round (or platform as I call it), there can be several forces acting on the ball.

Real forces on the ball

  • Gravity. This is acting straight down.
  • Tension in the rope. Ropes can only pull. So this force could be pulling to the right (towards the center of the circle in my frame). I am going to look at the case where this tension is zero though.

Faux Forces

  • Centrifugal force. This is a force proportional to the the angular velocity of the platform times the distance to the ball from the axis squared. The direction of this forces is always directly away from the axis of rotation. In the case of the ball, this will be to the right (at that instant).
  • Coriolis force. This is a force on objects that depends on its velocity in the frame. If the object is moving in the same direction as the rotational velocity vector, there is no coriolis force (so for the instant I chose, this is the zero vector).

Centrifugal force

Technically, the centrifugal force would be written as:


Where capitol omega is the angular velocity vector of the merry-go-round. If you don’t understand cross products, you can find the magnitude in this case as:


Coriolis Force

The coriolis force is:


At the instant shown above, the velocity of the rock is parallel to the angular velocity of the merry-go-round. This makes the cross product the zero vector. So, there is no coriolis force.

Now let me draw a free body diagram for the rock as seen from the rotating frame (including fake forces).


So, I assume it is possible to make the ball rotate at an angular speed (in the frame of the merry-go-round) such that at the instant shown, the tension has a magnitude of zero. This would mean that the rope is just about to go slack. At such an instant in such a case, the force pushing towards the center of the circle (thus the centripetal force) would be the centrifugal force. Would this make the world explode? Or maybe it would unless endless amounts of free energy. Needless, this would be a case where the centripetal and centrifugal force would be the same thing.

As I was writing this, I was thinking about the view from an inertial frame. In an inertial frame (such as outside the merry-go-round), there would be no centrifugal force. Also, both observers should agree on the tension in the rope (it could have a big digital tension read out). So, how would the external person see the ball moving in a circle with no centripetal force? I guess he wouldn’t. From his view point, the ball would not be going in a circle at that point. This should be verifiable by running this situation in vpython or something. Maybe I will put this on the classical mechanics final as a take home computational problem.

One final note, I made the ball go in a vertical circle (in the frame of the merry-go-round) so that I wouldn’t have to deal with coriolis force.


  1. #1 Jaime
    January 1, 2010

    Without working out the details my bet is that, when you have zero tension on the string, the curve followed by the particle has no (i.e. infinite) curvature. Or maybe some more complicated relationship, but you can definitely work everything out using Frenet’s trihedron and formulas.

  2. #2 tesseral
    January 3, 2010

    Isn’t centrifugal force real in the sense of Newton’s Third Law (equal and opposite reaction)?

  3. #3 Rhett Allain
    January 3, 2010


    I don’t think you could consider centrifugal force Newton’s Third Law.

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