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In general, the acceleration vector for planar motion in polar coordinates is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

Although you don't say it, I assume you are asking about circular motion where $\dot r=\ddot r=0$, so that the centripetal acceleration becomes $$a_r=r\dot\theta^2$$ and the tangential acceleration becomes$^*$ $$a_\theta=r\ddot\theta$$

Therefore, one could argue that these values are related through the $r$ variable such that $$\frac{a_r}{\dot\theta^2}=\frac{a_\theta}{\ddot\theta}$$

Of course, if there is no tangential acceleration for our circular motion we get an undefined value mathematically, but this makes sense physically. You can have any centripetal acceleration for uniform circular motion.

Note that the angular acceleration is just $\ddot\theta$, so if you want to bring that in too then that is a pretty simple thing to do.

In terms of how useful this relation is for doing physics problems, I'm not so sure. Both accelerations are easy to determine if you already know $\dot\theta$ and $\ddot\theta$, but I digress.

You might be more familiar with notation $a_c=r\omega^2=v^2/r$ and $a_\theta=r\alpha$. I could have started here, but I wanted to give a more general approach starting from "the beginning" with general planar motion in polar coordinates.

In general, the acceleration vector for planar motion in polar coordinates is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

For circular motion we have $\dot r=\ddot r=0$, so that the centripetal acceleration becomes $$a_r=r\dot\theta^2$$ and the tangential acceleration becomes$^*$ $$a_\theta=r\ddot\theta$$

Therefore, one could argue that these values are related through the $r$ variable such that $$\frac{a_r}{\dot\theta^2}=\frac{a_\theta}{\ddot\theta}$$

I should say that this relation is more "descriptive" than "constraining" in the sense that it doesn't necessarily mean that the centripetal acceleration determines the tangential acceleration or vice versa. It really only says that these two ratios have to be equal, as $a_c$ depends on $\dot\theta^2$ in the same way $a_\theta$ depends on $\ddot\theta$. I know this probably isn't as insightful as you were hoping for, but it's what we have.

Therefore, in terms of how useful this relation is for doing physics problems, I'm not so sure. Both accelerations are easy to determine if you already know $\dot\theta$ and $\ddot\theta$, but I digress.

Note that the angular acceleration is just $\ddot\theta$, so if you want to bring that in too then that is a pretty simple thing to do.

Also note that if there is no tangential acceleration for our circular motion we get an undefined value mathematically, but this makes sense physically. You can have any centripetal acceleration for uniform circular motion.

You might be more familiar with notation $a_c=r\omega^2=v^2/r$ and $a_\theta=r\alpha$. I could have started here, but I wanted to give a more general approach starting from "the beginning" with general planar motion in polar coordinates. This is usually my starting point for investigating questions like these.

In general, the acceleration vector for planar motion in polar coordinates is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

Although you don't say it, I assume you are asking about circular motion where $\dot r=\ddot r=0$, so that the centripetal acceleration becomes $$a_r=r\dot\theta^2$$ and the tangential acceleration becomes $$a_\theta=r\ddot\theta$$

Therefore, one could argue that these values are related through the $r$ variable such that $$\frac{a_r}{\dot\theta^2}=\frac{a_\theta}{\ddot\theta}$$

Of course, if there is no tangential acceleration for our circular motion we get an undefined value mathematically, but this makes sense physically. You can have any centripetal acceleration for uniform circular motion.

Note that the angular acceleration is just $\ddot\theta$, so if you want to bring that in too then that is a pretty simple thing to do.

In terms of how useful this relation is for doing physics problems, I'm not so sure. Both accelerations are easy to determine if you already know $\dot\theta$ and $\ddot\theta$, but I digress.

In general, the acceleration vector for planar motion in polar coordinates is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

Although you don't say it, I assume you are asking about circular motion where $\dot r=\ddot r=0$, so that the centripetal acceleration becomes $$a_r=r\dot\theta^2$$ and the tangential acceleration becomes$^*$ $$a_\theta=r\ddot\theta$$

Therefore, one could argue that these values are related through the $r$ variable such that $$\frac{a_r}{\dot\theta^2}=\frac{a_\theta}{\ddot\theta}$$

Of course, if there is no tangential acceleration for our circular motion we get an undefined value mathematically, but this makes sense physically. You can have any centripetal acceleration for uniform circular motion.

Note that the angular acceleration is just $\ddot\theta$, so if you want to bring that in too then that is a pretty simple thing to do.

In terms of how useful this relation is for doing physics problems, I'm not so sure. Both accelerations are easy to determine if you already know $\dot\theta$ and $\ddot\theta$, but I digress.

You might be more familiar with notation $a_c=r\omega^2=v^2/r$ and $a_\theta=r\alpha$. I could have started here, but I wanted to give a more general approach starting from "the beginning" with general planar motion in polar coordinates.

In general, the acceleration vector for planar motion in polar coordinates is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

Although you don't say it, I assume you are asking about circular motion where $\dot r=\ddot r=0$, so that the centripetal acceleration becomes $$a_r=r\dot\theta^2$$ and the tangential acceleration becomes $$a_\theta=r\ddot\theta$$

Therefore, one could argue that these values are related through the $r$ variable such that $$\frac{a_r}{\dot\theta^2}=\frac{a_\theta}{\ddot\theta}$$

Of course, if there is no tangential acceleration for our circular motion we get an undefined value mathematically, but this makes sense physically. You can have any centripetal acceleration for uniform circular motion.

Note that the angular acceleration is just $\ddot\theta$, so if you want to bring that in too then that is a pretty simple thing to do.

In general, the acceleration vector for planar motion in polar coordinates is given by $$\mathbf a=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

Although you don't say it, I assume you are asking about circular motion where $\dot r=\ddot r=0$, so that the centripetal acceleration becomes $$a_r=r\dot\theta^2$$ and the tangential acceleration becomes $$a_\theta=r\ddot\theta$$

Therefore, one could argue that these values are related through the $r$ variable such that $$\frac{a_r}{\dot\theta^2}=\frac{a_\theta}{\ddot\theta}$$

Of course, if there is no tangential acceleration for our circular motion we get an undefined value mathematically, but this makes sense physically. You can have any centripetal acceleration for uniform circular motion.

Note that the angular acceleration is just $\ddot\theta$, so if you want to bring that in too then that is a pretty simple thing to do.

In terms of how useful this relation is for doing physics problems, I'm not so sure. Both accelerations are easy to determine if you already know $\dot\theta$ and $\ddot\theta$, but I digress.