Intuition for formula of tangential component of acceleration in general curvillinear motion

Question

In certain problems of plane motion, the position of the particle P is defined by its polar coordinates $r$ and $\theta$. It is then convenient to resolve the velocity and acceleration of the particle into radial and tangential components.

Let $\hat{r}$ denote unit vector along radial direction and $\hat{\theta}$ denote unit vector along tangential direction.

Now, by writing velocity $v$ = $r\dot{\hat{r}} + \dot{r}\hat{r}$ and using some calculus it is not hard to show that acceleration a can be expressed as, $$a = (\ddot{r}-r{\omega}^2)\hat{r} + (\alpha r + 2\dot{r}\omega)\hat{\theta}$$ where $\omega = \dot{\theta}$ and $\alpha = \ddot{\theta}$, are the angular velocity and acceleration.

Now, the math is easy but I am trying to make intuitive sense of this formula above (I don't want to memorize it). The first part of the formula makes complete sense. It is like saying, subtract from the centripetal acceleration the part which is used in increasing the radial velocity, makes sense to me.

But what about the second part? Sure, the $\alpha r$ part is due to the contribution from angular acceleration and in case of circular motion, this would have been the only part present but where does the second term (radial velocity * angular velocity * 2!!!?) pop up from? Is there any significance to it or is it a purely mathematical construction. I am having a hard time understanding it intuitively.

Answer 1

Here is a relatively simple thought experiment. Imagine some rotating object, say a ball on a string that you're swinging. Let the angular acceleration be zero during the entire experiment. You hold the string such that the ball is at some radius $r_0$. Then, you let some more string out, so that the radius increases to $r_1$. Since the angular acceleration is zero, the angular velocity did not change, so the quantity $r\omega$ has increased, meaning the velocity of the ball in the $\hat{\theta}$ direction is greater now than at the smaller radius. This means there was an acceleration in the $\hat{\theta}$ direction; that's the $\dot{r}\omega$ term.

In other words, that term is due to the fact that a changing radius means that the velocity in the tangential direction changes, if angular acceleration is zero.

The reasom for the 2 is much deeper; for that, please refer to Farcher's comment on your question.

Answer 2

After reading on many different opinions and answers I have obtained the right intuition. I am sharing it here just so that it may help anyone who stumbles on it again.

Disclaimer : I am gonna build this answer directly over flevin's thought experiment. Please read it if you haven't already.

I am gonna completely ignore the radial component as the intuition for it is fairly obvious. In the discussion that follows I will exclude it.

Now, the tangential component of acceleration is best seen as composed of not 2 but three different terms (stay with me for a sec, I will clarify) -

$$a_t = (\alpha r) + (\omega \dot{r}) + (\omega \dot{r})$$ Of these 3 terms, the first term is fairly obvious and intuitive. So let's drop it by saying it that the body carries no angular acceleration (taking hint from @flevinBombastus' thought experiment). Now the expression becomes -

$$a_t (\text{at constant} \space \omega) = 0 + (\omega \dot{r}) + (\omega \dot{r})$$

Now, as @flevinBombastus' rightly argued in his thought experiment, as the radial distance of particle changes from origin, it's tangential velocity must also change. Intuitively (hell, also rigorously) for $\Delta r$ change in radius, tangential velocity changes by $\omega \Delta r$. Thus, we need an tangential acceleration of $\omega \dot{r}$ to bring about this change. This explains the second term in our expression. Thanks to @flevinBombastus' for the thought experiment which gave me this wonderful idea.

But wait, it seems like we accounted for everything already, so from where does that last $\omega \dot{r}$ pop up from? That's the tricky part, but absolutely not non-intuitive. Here's the big idea -

Let's ask ourselves, what's the difference between a purely uniform circular motion versus a motion in which we slowly reel out string as described in the flevin's thought experiment? Answer : It's the radial velocity. It is absent in case of circular motion but obviously present in the present case under study. So, our particle has this radial velocity and if you think about it, the radial velocity is a rotating vector. But, and here is the essence of argument, if radial velocity is a rotating vector it implies that we need another tangential acceleration to change it's direction! Now, it can be shown by explicit calculation that the acceleration required to bring about this rotation is equal in magnitude to $\omega \dot{r}$ !!!!

That means, that means ... that the third term of our expression is also unveiled.

TL;DR, there is no factor of 2 in the "coriolis" acceleration. It is actually made up of 2 terms arising from entirely different contexts - one to bring about the change in tangential velocity arising due to radial movement of particle and the other to bring about rotation of radial velocity vector. It just so happens, I like to say, that by coincidence the magnitude of both turned out to be same.