Cylindrical Components

Introduction

Cylindrical coordinates are used when an object is rotating around a fixed point or axis. An example of this would be a car going around a bend. The motion of the particle is described in terms of radial distance $r$, angular position $\theta$, and, when necessary, a vertical position $z$.

Polar Coordinates

The radial coordinate ($r$) extends from the origin to the particle. It is notated as $\overrightarrow{u_{r}}$.The transverse coordinate ($\theta)$ is the counterclockwise angle from a reference line and the r-axis. It is notated as $\overrightarrow{u_{\theta}}$. The position vector is $\overrightarrow{r} = r\overrightarrow{u_{r}}$.

Velocity Components

Radial velocity: $v_{r} = \dot{r}$

Transverse velocity: $v_{\theta} = r\dot{\theta}$

Therefore, the velocity vector is $v = \dot{r}\overrightarrow{u_{r}} +$$r\dot{\theta}\overrightarrow{u_{\theta}}$, with a magnitude of $v = \sqrt{v_r^2+v_\theta^2}$

For 3-dimensional motion, $v = \sqrt{v_r^2 + v_\theta^2 + \dot{z}^2} = \sqrt{(\dot{r})^2 + (r\dot{\theta})^2 + (\dot{z})^2}$

Acceleration Components

Radial acceleration: $a_{r} = \ddot{r} - r\dot{\theta}\ ^{2}$

Transverse acceleration: $a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta}$

For 3-dimensional motion, $a = \sqrt{a_r^2 + a_\theta^2 + \ddot{z}^2} = \sqrt{(\ddot{r} - r\dot{\theta}^2)^2 + (r\ddot{\theta} + 2\dot{r}\dot{\theta})^2 + (\ddot{z})^2}$

Chain Rule Relationships

When $r$ is given as a function of $\theta$ instead of time, chain rule is required. If $r = f(\theta)$, then $\dot{r} = \frac{dr}{d\theta}\dot{\theta}$. For instance, if $r = sin(\theta)$ and $\theta = 8t$, then $\dot{r} = \dot{\theta}cos(\theta) = 8cos(\theta)$. Alternatively, you can substitute $\theta$ immediately. In this example, $r = sin(\theta) = sin(8t)$, and you can take the derivatives as normal with chain rule.

Special Cases

If constant radius, then $r = constant$, $\dot{r} = 0$, and $\ddot{r} = 0$.

If constant angular velocity, then $\dot{\theta} = \omega = \frac{d\theta}{dt}$

Example:
Find the magnitude of the block’s velocity and acceleration when t = 1s. The block moves along the sling with a speed of $\dot{r} = (4t)\ m/s$ and the platform rotates at a constant rate of 6 rad/s.

1) Find $r,\ \dot{r},\ \ddot{r},\ \dot{\theta},\ \ddot{\theta}\ \ at\ t = 1$

Integrate $\dot{r}$ to get $r$: $r = \int_{}^{}4t\ dt = 2t^{2}$

Differentiate $\dot{r}$ to get $\ddot{r}$: $\ddot{r} = 4$

Evaluate everything at $t = 1$:

$r$	$\dot{r}$	$\ddot{r}$	$\dot{\theta}$	$\ddot{\theta}$
$2(1)^{2} = 2\ m$	$4(1) = 4\ m/s$	$4\ m/s^{2}$	$6\ rad/s$	$0$

2) Velocity

Radial velocity: $v_{r} = \dot{r} = 4\ m/s$

Transverse velocity: $v_{\theta} = r\dot{\theta} = (2\ m)(6\ rad/s) = 12\ m/s$

$v = \sqrt{v_r^2 + v_\theta^2} = \sqrt{(4\,m/s)^2 + (12\,m/s)^2} = 12.65\,m/s$

3) Acceleration

Radial acceleration: $a_{r} = \ddot{r} - r\dot{\theta}\ ^{2} = 4\ m/s^{2} - (2\ m)(6\ rad/s)^{2} = \ - 68\ m/s^{2}$

Transverse acceleration: $a_{\theta} = r\ddot{\theta} + 2\dot{r}\dot{\theta} = (2\ m)(0) + 2(4\ m/s)(6\ rad/s) = 48\ m/s^{2}$

$a = \sqrt{a_r^2 + a_\theta^2} = \sqrt{(-68\,m/s^2)^2 + (48\,m/s^2)^2} = 83.28\,m/s^2$