In non-uniform circular motion, the centripetal acceleration and tangential acceleration are perpendicular to each other, but does that mean that they don't affect each other because centripetal acceleration depends on velocity (=$v^2/r$) and tangential acceleration's direction (i.e., whether it will act along the direction of motion or opposite to the direction of motion) also depends upon the change in velocity. So are they dependent on each other or not?


3 Answers 3


Let's look at the general acceleration vector in polar coordinates$^*$:

$$\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat r+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$

If we want our object to remain on the same circle, which I'm assuming this is what you are interested in, we must have $\dot r=0$ and $\ddot r=0$. This means our acceleration must have the form: $$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$

Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force magnitude$^{**}$ of $$F_r=-mr\dot\theta^2$$ and a tangential force magnitude of $$F_\theta=mr\ddot\theta$$

Because we are confined to move around a single circle we can determine the role of each force. The radial force must only be responsible for changing the direction of the velocity, since if it could effect the speed then this means the object would have to change its $r$ coordinate, hence knocking us off the circle. Similarly, the tangential force must only be responsible for changing the speed of the particle as it moves around the circle, since if it could effect the direction of the velocity then it would do so by knocking us off the circle.

You can probably tell by now that if we want to stay on the circle, these forces are required to be "linked", in a sense. Indeed, if you take the time derivative of $F_r$ and use the requirement that $\dot r=0$, you will find that $$\frac{\text dF_r}{\text d t}=-2\dot\theta F_\theta$$ showing that the presence of a tangential force requires a change in the magnitude of the radial force in order for the particle to remain on the circle (or, on the flip side, a change in the magnitude of the radial force must be accompanied by a tangential force).

What if this condition is not met? Well, looking at the derivative of $F_r$ without the condition that $\dot r=0$, we must have $$-m\dot r\dot\theta^2\neq 0$$ This means that $\dot r\neq0$, which means we are no longer engaging in circular motion. Therefore, we need these two force components to be linked in this way to keep the motion circular.

A subtle point remains to be cleared up (as realized in comments to other answers). This does not necessarily mean that these forces are physically linked in general. This answer assumes we have an object undergoing non-uniform circular motion, and then investigates what must be true about the forces acting on the object. However, there could be situations where the radial and tangential force are not physically linked, at which point to achieve non-uniform circular motion you would have to make these forces act in such a way so that non-uniform circular motion is achieved.

$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $\dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.

$^{**}$ Note that this is what you usually encounter in your introductory physics classes as $F_{r}=mv^2/r$, since for motion along a circle of radius $r$, $\dot\theta=v/r$. The negative sign in this answer is to keep track of the direction of increasing/decreasing $r$, but if you are working a question where you only care about the magnitude of the radial force then this is irrelevant, hence why you usually don't see the negative sign.


The first answer explains the seeming link well. Though I would explicitly say that they don't depend on each other and don't affect each other. For an example with an orbiting spaceship is gravity affected by its horizontal speed or vice versa? Each may tell you how much is required of the other to maintain a circular orbit, but they are their own accelerations. Does acceleration in one direction affect acceleration in a perpendicular direction just because we want circular motion? I wouldn't say that.

  • $\begingroup$ They are their own accelerations. But if they aren't linked then you can only have uniform circular motion. As soon as you include a tangential force then you are no longer in circular motion. You need the link to have non-uniform circular motion. With your example of gravity it would be impossible to have a orbit with non-uniform circular motion. $\endgroup$ Commented Jul 11, 2019 at 12:52
  • $\begingroup$ You are saying a spaceship 'affects' gravity in order to maintain circular motion? It may be a question of word usage like one depends on the other to maintain a circle but it's more like a 'hey, I'm depending on you buddy' kind of depends than actually affecting the other. $\endgroup$
    – user234190
    Commented Jul 11, 2019 at 14:51
  • $\begingroup$ I am not saying that at all. I am saying for non-uniform circular motion that would have to be the case. Since that cannot happen with gravity, non-uniform circular motion cannot be possible for gravitational orbits. $\endgroup$ Commented Jul 11, 2019 at 14:52
  • $\begingroup$ You can if the spaceship is accelerating in a way to make it so. This would mean it is accelerating radially in ways also. $\endgroup$
    – user234190
    Commented Jul 11, 2019 at 15:04
  • $\begingroup$ By gravitational orbit we usually just mean gravity is the only radial force, but your example is fine too... So then you would need the radial acceleration supplied by your spaceship to behave just in the right way compared to the tangential acceleration. The point is you can't just pick any radial and tangential force and produce non-uniform circular motion. There definitely needs to be some link. $\endgroup$ Commented Jul 11, 2019 at 15:08

Nature knows nothing about circular motion. At any given space-time event, a particle hast only linear acceleration. Circular motion is a man-made construct. One way to appreciate this is to transform the path of an unaccelerated particle to a rotating system.

I posted this from my phone because I haven't had internet connectivity. So it was necessarily terse. I find this to be a very difficult topic which I have approached in many different ways. You can slog through all the math and follow all the symbol manipulations, but until you understand the essence of my original answer, you will not understand the physics of rotational motion.

If you don't like my original answer, you can try to sort our my notes on curves and the kinematics of non-inertial transformations beginning (as of today) on page 38 of this notebook:


An enlightening example I cribbed from Wells is on page 3 of https://drive.google.com/open?id=1WJ05eVLVYZLDdAtIZlcTr8MfmtLir9R6

Also have a look at the discussion of Coriolis Force on page 19 of the same.

My various derivations of Kepler's laws found here may also be instructive. https://drive.google.com/file/d/1OIgf7m8pIYMCN4Oru0cwvRztH619vjoo/view?usp=sharing

  • $\begingroup$ Why can't we say rotating reference frames are a man-made construct? Just because you can change reference frames to change the motion doesn't mean it doesn't exist. You can use the same reasoning to say linear motion isn't real. $\endgroup$ Commented Jul 11, 2019 at 12:48
  • $\begingroup$ Well, if circular motion wasn't "real" and just a construct of the human perception, as you say, why can't we then ask a question about it within the confines of this perception? Philosophically you could argue that we all live in a computer simulation - then this question would simply be about the rules within this simulation. This is no argument to just brush off the question. $\endgroup$
    – Steeven
    Commented Jul 11, 2019 at 16:58
  • $\begingroup$ @Steeven I grant that my answer is not rigorous. Another way I might have stated it is that, for purposes of dynamics all accelerations should be determined relative to an inertial system. Of course we can consider a rotating system to be a rest-frame endowed with pseudo-force fields. But that requires a number of special definitions. These require appeals to artificial constructs such as "rigid bodies". I didn't say, don't think in terms of rotating systems. I was suggesting that they are less natural than inertial systems. $\endgroup$ Commented Aug 18, 2019 at 2:53

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