Particle Kinematics in Cylindrical Coordinates

I recently had two different students ask me two different, but related questions. One had asked about the derivation of the equations he used in his dynamics I class for acceleration when using cylindrical coordinates, and another had asked me about Coriolis acceleration. Fortunately, both can be explained with the exact same answer, so let’s answer them both with the minimum amount of work required!

Let’s start with a little drawing illustrating what we’re dealing with here:

Cool! We are dealing with a coordinate system characterized by a radial distance, R, azimuth, $\theta$ , and axial distance, z. As we can see from the above diagram we have a right handed coordinate system defined by the unit vectors

$1) \displaystyle \qquad\bar{e_{R}} \; \times \; \bar{e_{\theta}} = \bar{e_{z}}$

From the above diagram we can relate these cylindrical coordinate system unit vectors back to traditional Cartesian coordinate system unit vectors with the following relationships.

$2) \displaystyle \qquad \bar{e_{R}} = ( \cos \theta) \hat{i} + ( \sin \theta ) \hat{j}$

$3) \displaystyle \qquad \bar{e_{\theta}} = -( \sin \theta ) \hat{i} + ( \cos \theta ) \hat{j}$

$4)\displaystyle \qquad \bar{e_{z}} = \hat{k}$

Now, let’s write the equation for our position vector, r.

$5) \displaystyle \qquad \bar{r} = R \bar{e_{R}} + z \bar{e_{z}}$

Note, in equation 5, from equation 3, the radial unit vector, $\bar{e_{R}}$ implicitly depends on $\theta$ , and theta may depend on time.

Velocity Derivation

If we wish to obtain the generic form of velocity in cylindrical coordinates all we must do is differentiate equation 5 with respect to time, but remember that the radial unit vector must be treated as a variable since it implicitly depends on $\theta$ . This dictates that we must use the chain rule to differentiate the first term of equation 5.

$6) \displaystyle \qquad \bar{v} = \dot{\bar{r}} = \dot{R} \bar{e_{R}} + R \frac{d \bar{e_{R}}}{dt} + \dot{z} \bar{e_{z}}$

Now, to differentiate the radial unit vector with respect to time we must employ the chain rule.

$7) \displaystyle \qquad \frac{d \bar{e_{R}}}{dt} = \frac{d \bar{e_{R}}}{d \theta} \; \frac{d \theta}{dt} = \frac{d \bar{e_{R}}}{d \theta} \cdot \dot{\theta} = \dot{ \theta} \bar{e_{\theta}}$

Woah, woah, woah, how the hell did I do that last part and change unit vectors you may ask. Well, let’s differentiate equation 2 with respect to $\theta$ .

$8) \displaystyle \qquad \frac{d \bar{e_{R}}}{d \theta} = -(\sin \theta) \hat{i} + (\cos \theta) \hat{j} = \bar{e_{ \theta}}$

That works out nicely, and it also informs us that the second velocity term below is the result of the radial unit vector changing as a function of time.

So, condensing everything from equations 6, 7, and 8 we obtain the general equation for velocity in cylindrical coordinates.

$9) \displaystyle \qquad \bar{v} = \dot{R} \bar{e_{R}} + R \dot{\theta} \bar{e_{ \theta}} + \dot{z} \bar{e_{z}}$

Let’s revisit the differentiation performed for the radial unit vector with respect to $\theta$ , and do the same thing for the azimuth unit vector.

$10) \displaystyle \qquad \frac{d \bar{e_{ \theta}}}{dt} = \frac{d \bar{e_{ \theta}}}{d \theta}\; \frac{d \theta}{dt} = \frac{d \bar{e_{ \theta}}}{d \theta}\; \dot{ \theta}= -\dot{ \theta} \bar{e_{R}}$

$11) \displaystyle \qquad \frac{d \bar{e_{\theta}}}{d_{ \theta}} = -(\cos \theta) \hat{i} - (\sin \theta) \hat{j} = - \bar{e_{R}}$

Let’s look at equation 9 for a moment and discuss the contributions from the terms. The first and last are easy to understand; the second basically says that the azimuthal velocity results from changing the azimuth angle, with the magnitude growing in proportion to the radial distance.

Acceleration Derivation

Now, this is where things start to make us look smarter than we actually are because of all the cool symbols we get to use. Moving on to acceleration, we obviously just continue differentiating!

$12a) \displaystyle \qquad \bar{a} = \frac{d}{dt} [ \dot{R} \bar{e_{R}}] + \frac{d}{dt} [ R \dot{ \theta} \bar{e_{\theta}} ] + \frac{d}{dt} [ \dot{z} \bar{e_{z}}]$

Using the relations in 7, 8, 10, and 11 we end up with

$12b) \displaystyle \qquad \bar{a}= \ddot{R} \bar{e_{R}} + \dot{R} \dot{ \theta} \bar{e_{ \theta}} + \dot{R} \dot{ \theta} \bar{e_{ \theta}} + R \ddot{ \theta} \bar{e_{ \theta}} - R \dot{ \theta}^2 \bar{e_{R}} + \ddot{z} \bar{e_{z}}$

Collecting like terms we finally obtain the general equation for acceleration in cylindrical coordinates.

$13) \displaystyle \qquad \bar{a} = ( \ddot{R} - R \dot{ \theta}^2) \bar{e_{R}} + (R \ddot{ \theta} + 2 \dot{R} \dot{ \theta}) \bar{e_{ \theta}} + \ddot{z} \bar{e_{z}}$

The Coriolis acceleration is then the sum of the second and third quantities in equation 12b). I left this intermediate step in there to show that, although the second and third terms are the same quantity, they originate from different effects. The second term is the result of the radial direction not being constant, as evidenced by equation 7, whereas the third term is the rate at which the azimuthal velocity, $R \dot{ \theta}$ , will change if R is non-constant.

Cylindrical Coordinate System Kinetics

To further elucidate the Coriolis acceleration, and in turn the Coriolis force or much more buzz worthy Coriolis Effect, let’s take a look at the scalar form of Newton’s second law in each principal direction.

$14a) \displaystyle \qquad \sum F_{R} \equiv \sum \bar{F} \cdot \bar{e_{R}} = m ( \ddot{R} - R \dot{ \theta}^2)$

$14b) \displaystyle \qquad \sum F_{\theta} \equiv \sum \bar{F} \cdot \bar{e_{\theta}} = m ( R \ddot{\theta} + 2 \dot{R} \dot{ \theta})$

$14c) \displaystyle \qquad \sum F_{z} \equiv \sum \bar{F} \cdot \bar{e_{z}} = m \ddot{z}$

It is worth mentioning here that the famous centrifugal force appears in equation 14a as the negative quantity, and the much less famous Euler force appears as the first term in equation 14b.

So what can we say about Coriolis acceleration from equation 14b? Well, as explained previously, it is the result of 1) the radial direction being non-constant, as is the case of a rotating coordinate system (like, say, the Earth), and 2) the magnitude of the position vector changing in that rotating coordinate frame. Equation 14b indicates that this results in a force acting perpendicular to the radial direction. The force resulting from this acceleration term contributes to many of Earth’s largest atmospheric circulation systems such as trade winds and hurricanes.

Another way to describe the Coriolis acceleration without the use of vector calculus is to say that at a given rate of rotation of an observer, the magnitude of this acceleration on an object is proportional to the velocity of the object as well as the sine of the angle between the direction of travel of the object and the axis of rotation.

In a different coordinate system this may be easier to see, as we end up with the equation

$15) \displaystyle \qquad \bar{a_{c}} = 2 \bar{v} \times \bar{\omega}$

where $\bar{a_{c}}$ is the Coriolis acceleration vector, $\bar{v}$ is the velocity vector of the particle in the rotating coordinate system, and $\bar{\omega}$ is the angular velocity vector of the rotating coordinate system.

$\blacksquare$

Velocity Derivation

Acceleration Derivation

Cylindrical Coordinate System Kinetics

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