In non-uniform circular motion are centripetal acceleration and tangential acceleration independent of each other?

Question

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?

Answer 1

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.

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$^*$ 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.

Answer 2

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.

Answer 3

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.

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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:

https://drive.google.com/file/d/1XOaXd5hcyh7io00bdvD2KlVJtxpS2Gy4/view?usp=sharing

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