# Velocity and acceleration of a particle in polar coordinates

I am asked to find the radial and transverse velocity and acceleration for a particle with polar coordinates $r=e^t$ and $\theta=t$

Therefore let $\underline{r}=\underline{\hat{r}}e^t$ and $\underline{\theta}=\underline{\hat{\theta}}t$ $$\therefore \dot{\underline{r}}=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{r}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{r}}$$

But$\underline{\hat{r}}=\cos\theta\underline{\hat{i}}+\sin\theta\underline{\hat{j}}$

$$\therefore\dot{\underline{r}}=\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} }{\mathrm{d} \theta}(\cos\theta\underline{\hat{i}}+\sin\theta\underline{\hat{j}})e^t+e^t\underline{\hat{r}}$$

$$=(-\sin\theta\underline{\hat{i}}+\cos\theta\underline{\hat{j}})e^t +e^t\underline{\hat{r}}$$

$$\therefore \underline{\hat{r}}=e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}$$ This is the particles velocity, the acceleration is given by: $$\ddot{\underline{r}}=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{\theta}}e^t+\underline{\hat{r}}e^t)$$

$$=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{\theta}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{\theta}}+\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{r}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{r}}$$

$$=\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} }{\mathrm{d} \theta}(-\sin\theta\underline{\hat{i}}+\cos\theta\underline{\hat{j}})e^t+e^t\underline{\hat{\theta}}+e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}$$

$$=-e^t\underline{\hat{r}}+e^t\underline{\hat{\theta}}+e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}$$

$$=2e^t\underline{\hat{\theta}}$$

This is an acceleration in the transverse direction. But if $\theta=t$ and $r=e^t$ then why do I get an acceleration in the transverse direction, and not the radial direction? Is the transverse velocity not constant (I also got $e^t\underline{\hat{\theta}}$ for transverse velocity, why is this?) Help is very much appreciated

Is the transverse velocity not constant?

You're mixing up angular velocity and transverse velocity.

$\underline{\theta}=\underline{\hat{\theta}}t$ - This is wrong. The angle $\theta$ is not a vector. Differntiating the angle with respect to time will give you angular velocity which is constant (in this case) and not a vector. [1] Angular velocity is not transverse velocity. The transverse velocity is the component of velocity along a circle centered at the origin.

The $\underline{\hat{r}}$ is the unit vector of the position vector of the particle. The rate of change of this unit vector is:

$\frac{d}{dt} (\underline{\hat{r}}) = \dot{\theta}.\underline{\hat{\theta}}$

Where $\dot{\theta} = 1$ is the angular velocity.

Angular velocity does not depend on the magnitude of the position vector of the particle. The transverse velocity depends on the magnitude of the position vector and the angular velocity.

Imagine a bicycle wheel. If you hold it up in the air and rotate it, the edge of the wheel has to cover a greater distance in one revolution than the points close to the center of the wheel. The transverse velocity of a point at the edge of the wheel is greater than the transverse velocity of a point close to the center. In contrast, the angular velocity of all points on the wheel is same at any given instant.

Since in polar coordinates we consider a circle centered at the origin, the transverse velocity is going to depend on the magnitude of the position vector of the particle. Since the magnitude of the position vector is increasing exponentially, the transverse velocity should also increase exponentially. Which is why you get $e^t \underline{\hat{\theta}}$ as the transverse velocity.

$\underline{\dot{r}}=e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}$.

$e^t\underline{\hat{\theta}}$ is due to change in direction. The circle becomes bigger and bigger and the particle has to move faster and faster so that it continues to keep a constant angular velocity.

$e^t\underline{\hat{r}}$ is due to change in the magnitude of the position vector. This is measuring the speed with the circle radius is increasing.

Why do I get an acceleration in the transverse direction, and not the radial direction?

$\ddot{\underline{r}}=(-e^t\underline{\hat{r}}+e^t\underline{\hat{\theta}})+(e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}})$

$(-e^t\underline{\hat{r}}+e^t\underline{\hat{\theta}})$ comes from differntiating the transverse velocity $e^t\underline{\hat{\theta}}$.

• The $-e^t\underline{\hat{r}}$ is the centripetal acceleration. This is continuously changing the direction of transverse velocity. Again, centripetal acceleration depends on the magnitude of the position vector for the same reasons as transverse velocity. So even this, unsurprisingly, has an exponential term.

• The $e^t\underline{\hat{\theta}}$ is increasing the magnitude of the transverse velocity. And the magnitude of transverse velocity is increasing because the circle is becoming bigger. This is one-half of the so-called Coriolis acceleration.

$(e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}})$ comes from differntiating $e^t\underline{\hat{r}}$.

• $e^t\underline{\hat{r}}$ is the radial acceleration. This is increasing the velocity of the particle in the radial direction.

• $e^t\underline{\hat{\theta}}$ is changing the direction of the radial velocity vector. This is second-half of the Coriolis acceleration.

The radial velocity is indeed increasing, but it is also turning leftwards. The centripetal acceleration and the radial acceleration cancel each other.

The trajectory of the particle:

From Wolfram Alpha "parametric plot (e^tcos t, e^t sint) t=0...1"

Note that the $\underline{\hat{\theta}}$ is changing in direction with a fixed angular velocity. The magnitude of transverse acceleration is increasing exponentially. So the acceleration is constantly chaging direction and increasing exponentially in magnitude.

Let's consider the time interval $t_0$ to $t_0 + 1$:

For all values of $t_0$ the angle covered will be the same since angular velocity is constant. But the distance covered in the interval keeps becoming greater and greater. This automatically forces radial velocity to increase. I hope you can see how this would result in the above graphs.

In polar coordinates, $v_r \neq \int a_r(t)\,dt$. This seems to be the source of your confusion. Anyway, I hope this helped. :)

[1] It is sometimes convenient to give angular velocity a direction. We assume the angular velocity points out of the page if the particle is travelling counter-clockwise and vice-versa. Angular velocity is more precisely, a pseudo-vector.