Acceleration is the rate of change of velocity. In the uniform circular motion the acceleration is produced due to change of direction of the velocity(the magnitude remains same). The direction is changing at a uniform rate depending upon the speed of the body. Then why is it said that acceleration is variable? What is that thing that is variable?
[Physics] Why is acceleration variable in uniform circular motion
accelerationkinematicsnewtonian-mechanicsvectors
Related Solutions
Radial acceleration $\vec a_{rad}$ takes care of turning (when pulling perpendicular to the velocity vector $\vec v$, it can only turn it, not increase it), and tangential acceleration $\vec a_{tan}$ takes care of speeding up (when pulling parallel to $\vec v$, it can only increase it, not turn it).
- A car speeding up while driving straight, has a $\vec a_{tan}$ but no $\vec a_{rad}$.
- A car turning but not speeding up has a $\vec a_{rad}$ only and no $\vec a_{tan}$.
- A car speeding up while turning has both a $\vec a_{rad}$ and a $\vec a_{tan}$.
A uniform circular motion is case 2. The word "uniform" means constant speed.
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.
Best Answer
Acceleration is a vector quantity. It has both magnitude and direction. Just as you said in the question, the direction of acceleration is changing in uniform circular motion. Therefore acceleration is variable, as a vector quantity is said to be variable if either its magnitude or its direction is changing.