অবস্থানের পোলার ফাংশন থেকে বেগ আর ত্বরণ

When studying mechanics, three very important functions to keep in mind are the position, velocity, and acceleration functions, which are all vectors. They have both a magnitude and a direction. In addition, a unit vector is a vector with magnitude $1$. In other words, unit vectors encode for direction, not magnitude (although they have both).

One way to gain a better understanding of a vector is to break it down into its components. The vector from $(0,0)$ to $(3,5)$ on the Cartesian coordinate system has an $x$-component of $3$ and a $y$-component of $5$. We write this vector as $3 \hat{x}+5 \hat{y}$, where $\hat{x}$ is the unit vector in the $x$ direction and $\hat{y}$ is the unit vector in the $y$-direction.

The vector $x \hat{x}+ y \hat{y}$ can be thought of as going $x$ units in the $\hat{x}$ direction, and then $y$ units in the $\hat{y}$ direction.

Therefore, in the Cartesian coordinate system, if we define $\vec{r}$, $\vec{v}$, and $\vec{a}$ as the position, velocity, and acceleration vectors, we have that

$$\begin{equation} \vec{r}=x \hat{x}+ y \hat{y} \end{equation}$$ $$\begin{equation} \vec{v}=\frac{dx}{dt} \hat{x}+ \frac{dy}{dt} \hat{y} \end{equation}$$ $$\begin{equation} \vec{a}=\frac{d^2x}{dt^2} \hat{x}+ \frac{d^2y}{dt^2} \hat{y} . \end{equation}$$

The goal of this article is to rewrite the above in polar coordinates. We will be using both Cartesian and polar coordinates in order to do this. To review the definition of polar coordinates in relation to Cartesian coordinates, click here.

Since $x=r \cos \theta$ and $y=r \sin \theta$, we find that

\begin{equation} \vec{r}=x \hat{x}+ y \hat{y}=r \cos \theta \hat{x} + r \sin \theta \hat{y}=r (\cos \theta \hat{x} + \sin \theta \hat{y}). \end{equation}

Notice that the vector $\cos \theta \hat{x} + \sin \theta \hat{y}$ has magnitude $1$ because $\cos^2 \theta+\sin^2 \theta =1$. In other words, it is a unit vector. More specifically, it is the unit vector in the $r$ direction (radial), so it is denoted $\hat{r}$.

Take a moment to understand that $\hat{r}$ changes direction based on where the particle is at a given time. The $\hat{x}$ and the $\hat{y}$ vectors remain horizontal and vertical, but the $\hat{r}$ vector depends on $\theta$. Nevertheless, $\hat{r}$ always points away from the origin. This notion of a changing coordinate system is often more useful than a static one.

Thus, the position vector in polar coordinates is \begin{equation} \vec{r}=r \hat{r}. \end{equation}

The above equation has three $r$'s. Think about the differences between each one, and what the equation as a whole means.

Remember that $\vec{r}$ is the position vector, $r$ is its magnitude, and $\hat{r}$ is its direction.

Now we plow forward to find the velocity vector. It is the derivative with respect to time of the position vector. By the product rule,

$$\begin{equation} \vec{v}= \frac{dr}{dt} \hat{r}+ r \frac{d \hat{r}}{dt}. \end{equation}$$

From now on, there will be quite a few derivatives. To simplify our notation, we use Newton's dots. We define $\dot{a}$ as $\frac{da}{dt}$.

We rewrite the velocity vector as \begin{equation} \vec{v}=\dot{r} \hat{r}+r \dot{\hat{r}}. \end{equation}

We can't really do much to simplify the first term, since $\dot{r}$ depends on the motion of the particle itself. However, we can simplify the second term. We need to find the time derivative of the $\hat{r}$ vector, which is nonzero because $\hat{r}$ is changing!

We find that \begin{equation} \dot{\hat{r}}=\frac{d \hat{r}}{dt}=\frac{d}{dt}\left( \cos \theta \hat{x}+ \sin \theta \hat{y} \right ). \end{equation}

There are two things to remember when evaluating the above: that we need to use the chain rule, because $\theta$ depends on time $t$, and that $\hat{x}$ and $\hat{y}$ vectors have no time derivatives (because they don't change direction). In other words, we do need to use the product rule on each term, but the fact that $\frac{d \hat{x}}{dt}=\frac{d\hat{y}}{dt}=0$ means that $\frac{d}{dt} \left(p \hat{x} \right )=\frac{dp}{dt} \hat{x}$ and $\frac{d}{dt} \left(p \hat{y} \right )=\frac{dp}{dt} \hat{y}$ for any $p$.

Simplifying, \begin{equation} \frac{d}{dt}\left( \cos \theta \hat{x}+ \sin \theta \hat{y} \right )=(-\sin \theta) \frac{d\theta}{dt} \hat{x}+ (\cos \theta) \frac{d\theta}{dt} \hat{y}=\dot{\theta} (-\sin\theta \hat{x}+ \cos \theta \hat{y}). \end{equation}

Investigate the vector $-\sin\theta \hat{x}+ \cos \theta \hat{y}$. Its magnitude is $1$, since $\cos^2\theta+\sin^2\theta=1$. Graphing it along with $\hat{r}$ reveals that it is perpendicular to $\hat{r}$. Therefore, it is the unit vector in the $\theta$ direction, and we will denote it $\hat{\theta}$.

Combining the equations above, we find that $\dot{\hat{r}}=\dot{\theta}\hat{\theta}$. Plugging this back into the velocity vector, we find that

\begin{equation} \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}. \end{equation}

That's all for the velocity vector! We can now move forward to the acceleration vector. Since we needed to find $\dot{\hat{r}}$ in the velocity vector, you may guess that we'll also need to find $\dot{\hat{\theta}}$ to find the acceleration vector. And indeed we will, so let's get that out of the way. Note that

\begin{equation} \dot{\hat{\theta}}=\frac{d}{dt}\left( -\sin\theta \hat{x}+ \cos \theta \hat{y} \right). \end{equation}

As before, keep in mind that we need to use the chain rule. Explicitly,

\begin{equation} \frac{d}{dt}\left( -\sin\theta \hat{x}+ \cos \theta \hat{y} \right)=(- \cos \theta) \frac{d\theta}{dt} \hat{x}+(-\sin\theta) \frac{d\theta}{dt} \hat{y}=-\dot{\theta}(\cos \theta \hat{x}+\sin\theta \hat{y}). \end{equation}

Do you recognize that last vector? It is just $\hat{r}$! Therefore, we have that \begin{equation} \dot{\hat{\theta}}=-\dot{\theta} \hat{r} \end{equation} \begin{equation} \dot{\hat{r}}=\dot{\theta}\hat{\theta}. \end{equation}

That's some pretty nice symmetry.

Now, acceleration is the derivative with respect to time of velocity. Thus, using the generalized product rule for three variables $\left(\frac{d}{dt} uvw=\dot{u} vw+u\dot{v} w+ uv \dot{w} \right )$: \begin{equation} \vec{a}=\frac{d}{dt} \left( \dot{r}\hat{r}+ r \dot{\theta}\hat{\theta} \right)= \left(\ddot{r}\hat{r}+\dot{r}\dot{\hat{r}}\right)+\left(\dot{r}\dot{\theta}\hat{\theta}+r \ddot{\theta}\hat{\theta}+ r \dot{\theta}\dot{\hat{\theta}}\right). \end{equation}

We can simplify this by plugging in our expressions for $\dot{\hat{r}}$ and $\dot{\hat{\theta}}$ that we previously derived:

\begin{equation} \vec{a}=\left(\ddot{r}\hat{r}+\dot{r}\dot{\hat{r}}\right)+\left(\dot{r}\dot{\theta}\hat{\theta}+r \ddot{\theta}\hat{\theta}+ r \dot{\theta}\dot{\hat{\theta}}\right)\= \left(\ddot{r}\hat{r}+\dot{r}\left(\dot{\theta}\hat{\theta}\right)\right)+ \left(\dot{r}\dot{\theta}\hat{\theta}+r \ddot{\theta}\hat{\theta}+ r \dot{\theta}\left(-\dot{\theta}\hat{r}\right)\right). \end{equation}

We now combine the $\hat{r}$ and $\hat{\theta}$ terms to obtain:

\begin{equation} \vec{a}=\left(\ddot{r}-r\dot{\theta}^2 \right) \hat{r} + \left(2 \dot{r}\dot{\theta}+ r \ddot{\theta} \right) \hat{\theta}. \end{equation}

This cannot be simplified further, because all of the variables depend on the motion of the particle itself. We are done! Compiling our results: \begin{equation} \vec{r}=r\dot{r} \end{equation} \begin{equation} \vec{v}=\dot{r}\hat{r}+ r \dot{\theta}\hat{\theta} \end{equation} \begin{equation} \vec{a}=\left(\ddot{r}-r\dot{\theta}^2 \right) \hat{r} + \left(2 \dot{r}\dot{\theta}+ r \ddot{\theta} \right) \hat{\theta}. \end{equation}

We'll conclude with a simple example of how these equations are used in physics. If $r$ is constant, we have a particle undergoing circular motion (constant distance away from the origin). In addition, $\theta$ will change, but if it varies at a uniform rate, then the angular speed is constant. Thus, circular motion at a constant speed is characterized by $\dot{r}=\ddot{\theta}=0$. Plugging this into our expression for $\vec{a}$ above, we obtain

\begin{equation} \vec{a}_{\text{circular}}=-r\dot{\theta}^2\hat{r}. \end{equation}

Since $\hat{r}$ is in the radial (outwards) direction, $-\hat{r}$ points towards the origin. Therefore, the direction of acceleration in circular motion is towards the center, or centripetal.