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I am trying to understand why polar coordinates accelerations in the theta and r directions cannot be interchanged for the normal and tangential components of acceleration. I am currently doing a dynamics course where we will have to choose which of these to use for specific problems.

My understanding is that these are in the same directions (normal acceleration with r's acceleration and the tangential acceleration with the acceleration in the theta direction.

Please could the explanation include why these cannot be interchanged.

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    $\begingroup$ Could you add a sketch and equations for what you mean by "acceleration in the theta direction"? $\endgroup$ – user1583209 May 28 '18 at 14:54
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If I understand your question right, you are placing a coordinate system at the periphery of a circular path, with the y-axis always pointing tangentially. In this setup, you are comparing polar with Cartesian/rectangular coordinates.

The polar and rectangular coordinate (basis) vectors may appear to point in the same direction at first glance, but they are not of the same dimension.

One is

$$\text{Cartesian:}\quad\Big(\text{(tangential) acceleration}\;\;,\;\text{(perpendicular) acceleration}\Big)$$

while the other is

$$\text{Polar:}\quad\Big(\text{(radial) acceleration}\;\;,\;\text{turning angle}\Big)$$

In other words,

  • metres-per-second-squared by metres-per-second-squared and
  • metres-per-second-squared by degrees (or radians if you will).

The angular coordinate in a polar coordinate set can never equal a Cartesian (rectangular) coordinate, simply due to its different dimension (different unit).

Thinking correctly of the angular coordinate as a number-of-degrees also gives you the impression of a curved axis and basis vector rather than a straight vector. It doesn't say "how much" in a straight direction, but "how much" around. Such curved basis vector is clearly not a tangent to the curve - rather, it is (it defines) the curve.

Now, since the angular coordinate in one system can't equal the tangential coordinate of another system, in a polar coordinate system the radial coordinate is the only one left to carry the entire "size"/magnitude of the acceleration.

In Cartesian (rectangular) coordinates we are used to the magnitude of a vector being shared between both coordinates, and we find the magnitude as a mix (through Pythagoras' relation). So clearly, none of the coordinates in a Cartesian coordinate system equals the radial coordinate in a polar system - although the dimension (the unit) fits, they carry different information. They are not the same thing.

Conclusion: There is no overlap between the coordinates of polar and rectangular systems. No coordinate in one equals that of the other. None of them fit to the other.

The way to think of these two system is thus

  • intuitively go-a-bit-out-and-a-bit-up for Cartesian coordinates while
  • less intuitively go-the-full-amount-out-and-turn for polar coordinates,

and you can only convert from one to the other through the well-known trigonometric sine, cosine and tangens relations in a right-angled triangle.

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  • $\begingroup$ I believe that the OP is asking about coordinate directions ($\hat{n},\hat{tang},\hat{r},\hat{\theta}$) rather than coordinate values. $\endgroup$ – Bill N May 28 '18 at 15:20
  • $\begingroup$ @BillN I believe I am already covering that in this answer in non-mathematical fashion. $\endgroup$ – Steeven May 28 '18 at 15:21
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    $\begingroup$ Ah! I see that now. $\endgroup$ – Bill N May 28 '18 at 15:25
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Whether you're talking about acceleration components, velocity comonents, or position components, you must consider vector components in general. There's nothing special about an acceleration vector when examining vector descriptions.

The coordinates $r$ and $\theta$ are defined based on some origin which is constant. Also, the zero points in that system are defined by some person and usually are time-independent (unless one is doing an accelerated-reference-frame problem).

The normal-tangential coordinate description is body-centered and is not time-independent. In fact, the tangential direction is defined to be parallel to the instantaneous velocity of the particle, and the normal direction is perpendicular to the velocity, with a choice of left-handed or right-handed perpendicularity. (Some problems even lend themselves to non-perpendicular systems, especially general relativity. But, I diverge ...)

Let's consider a system in which the $r$ and $\theta$ components are functions of time. For example, $$r=2\theta \text{ and } \theta=0.5t.$$

In polar coordinates, the position of the particle is $$\vec{r}=r(t) \hat{r}(t)=1.0 t\ \hat{r}(t).$$ What is $\hat{r}(t)$? If we shift to a Cartesian description with the same origin, $$\hat{r}=\hat{i}\cos\theta+\hat{j}\sin\theta=\hat{i}\cos (0.5 t)+\hat{j}\sin(0.5 t)$$

Now, the tangential direction will be parallel to the instantanceous velocity, $\vec{v}=\dfrac{d\vec{r}}{dt}$: $$\vec{v}=\frac{d\vec{r}}{dt}=\hat{r}+t\frac{d\hat{r}}{dt}$$ If the tangential direction is always parallel to $\hat{\theta}$, then is will always be perpendicular to $\hat{r}$, so we can examine $\vec{v}\cdot\hat{r}$ to see if it is zero: $$\vec{v}\cdot\hat{r}=\hat{r}\cdot\hat{r}+t\frac{d\hat{r}}{dt}\cdot\hat{r}$$ $$=1+0.5t(-\hat{i}\sin(0.5t)+\hat{j}\cos(0.5t))\cdot(\hat{i}\cos(0.5t)+\hat{j}\sin(0.5t))=1\ne 0.$$

So, by example we have at least one situation in which the tangential component is not the $\theta$ component.

You should try tackling the more general case for yourself in which r is not time-independent.

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