Take a particle in polar coordinate system to follow the equations: $$\theta=\omega t$$ and $$r=r_oe^{-\omega t}$$ Now, the radial acceleration will be- $$a_r=\ddot{r}-r\dot \theta^2$$ which we get as $0$. This means that the particle's radial speed is constant. But we see that $\dot r$ is $-r_o\omega e^{-\omega t}$ which is time dependent? Then how is the radial acceleration zero? I am obviously missing something simple but I'm just not getting what.
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1$\begingroup$ You know that $a_r \neq \frac{d}{dt}\dot r$ (think of a planet rotating around the Sun). What's your question then? $\endgroup$– nwolijinCommented Dec 29, 2020 at 18:01
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$\begingroup$ I don't think you get 0 for $a_r$. When you calculate $\ddot{r}$ you just get an $\omega ^2$. You don't also get a $t^2$. $\endgroup$– CGSCommented Dec 29, 2020 at 18:15
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1$\begingroup$ Yes $a_r-0$, but it would be helpful in the post to show that it comes up as 0 so people don't have to do the math on their own to confirm this before proceeding to answer. $\endgroup$– John AlexiouCommented Dec 29, 2020 at 23:53
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It is indeed correct that the radial acceleration, $a_r = 0.$ Your mistake is in thinking that $\dot{r}$ is the only contributing factor to $a_r$, whereas from the very formula you have cited for $a_r$, it is not: there is a $-r\dot{\theta}^2$ term as well.
In other words, looking at $\dot{r}$ is not sufficient to infer $a_r$. Use the correct formula for the radial acceleration, and you get $0$ once again.
