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# Cylindrical Coordinates

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## Introduction

This page covers cylindrical coordinates. The initial part talks about the relationships between position, velocity, and acceleration. The second section quickly reviews the many vector calculus relationships.

## Rectangular and Cylindrical Coordinates

Rectangular and cylindrical coordinate systems are related by

$\begin{eqnarray} x & = & r \cos \theta \\ y & = & r \sin \theta \\ z & = & z \end{eqnarray}$
and by

$\begin{eqnarray} r & = & \sqrt{ x^2 + y^2 } \\ \theta & = & Tan^{-1} \left( {y / x} \right) \\ z & = & z \end{eqnarray}$
Cylindrical coordinates are "polar coordinates plus a z-axis."

## Position, Velocity, Acceleration

The position of any point in a cylindrical coordinate system is written as

${\bf r} = r \; \hat{\bf r} + z \; \hat{\bf z}$
where $$\hat {\bf r} = (\cos \theta, \sin \theta, 0)$$. Note that $$\hat \theta$$ is not needed in the specification of $${\bf r}$$ because $$\theta$$, and $$\hat{\bf r} = (\cos \theta, \sin \theta, 0)$$ change as necessary to describe the position. However, it will appear in the velocity and acceleration equations because

${\partial \, \hat{\bf r} \over \partial \, t} \; = \; {\partial \over \partial \, t} (\cos \theta, \sin \theta, 0) \; = \; (-\sin \theta, \cos \theta, 0) {\partial \, \theta \over \partial \, t} \; = \; \omega \, \hat{\boldsymbol{\theta}}$
${\partial \, \hat{\boldsymbol{\theta}} \over \partial \, t} \; = \; {\partial \over \partial \, t} (-\sin \theta, \cos \theta, 0) \; = \; (-\cos \theta, -\sin \theta, 0) {\partial \, \theta \over \partial \, t} \; = \; -\omega \, \hat{\bf r}$
and finally $${\partial \, \hat{\bf z} \over \partial \, t} \; = \; 0$$ because $$\hat{\bf z}$$ does not change direction.

In summary, identities used here include

$\omega = {\partial \, \theta \over \partial \, t} \qquad \qquad \alpha = {\partial \, \omega \over \partial \, t} \qquad \qquad {\partial \, \hat{\bf r} \over \partial \, t} = \omega \, \hat{\boldsymbol{\theta}} \qquad \qquad {\partial \, \hat{\boldsymbol{\theta}} \over \partial \, t} = -\omega \, \hat{\bf r} \qquad \qquad {\partial \, \hat{\bf z} \over \partial \, t} \; = \; 0$
Returning to the position equation and differentiating with respect to time gives velocity.

${\bf v} \quad = \quad {\partial \over \partial \, t} (r \; \hat{\bf r} + z \; \hat{\bf z}) \quad = \quad (\dot r \, \hat{\bf r} + r \; \omega \; \hat{\boldsymbol{\theta}} + \dot z \, \hat{\bf z})$
This could also be written as

${\bf v} = (v_r \, \hat{\bf r} + v_\theta \hat{\boldsymbol{\theta}} + v_z \, \hat{\bf z})$
where $$v_r = \dot r, v_\theta = r \, \omega,$$ and $$v_z = \dot z$$.

Differentiating again to get acceleration...

$\begin{eqnarray} {\bf a} & = & {\partial \over \partial \, t} (\dot r \, \hat{\bf r} + r \; \omega \; \hat{\boldsymbol{\theta}} + \dot z \, \hat{\bf z}) \\ \\ & = & \ddot r \hat{\bf r} + \dot r \, \omega \; \hat{\boldsymbol{\theta}} + \dot r \, \omega \; \hat{\boldsymbol{\theta}} + r \, \alpha \, \hat{\boldsymbol{\theta}} - r \, \omega^2 \, \hat{\bf r} + \ddot z \, \hat{\bf z} \\ \\ & = & ( \ddot r - r \, \omega^2 ) \, \hat{\bf r} + ( r \, \alpha + 2 \, \dot r \, \omega ) \; \hat{\boldsymbol{\theta}} + \ddot z \, \hat{\bf z} \end{eqnarray}$

The $$- r \, \omega^2 \, \hat{\bf r}$$ term is the centripetal acceleration. Since $$\omega = v_\theta / r$$, the term can also be written as $$- (v^2_\theta / r) \, \hat{\bf r}$$.

The $$2 \dot r \omega \, \hat{\boldsymbol{\theta}}$$ term is the Coriolis acceleration. It can also be written as $$2 \, v_r \, \omega \, \hat{\boldsymbol{\theta}}$$ or even as $$(2 \, v_r \, v_\theta / r ) \hat{\boldsymbol{\theta}}$$, which stresses the product of $$v_r$$ and $$v_\theta$$ in the term.

### Centripetal Accelerations in the Tire

The centripetal acceleration of a tire traveling at 70 mph is remarkably high. 70 mph is 31.3 m/s, and this is $$v_\theta$$. For a tire with a 0.3 m radius, the centripetal acceleration is

${v^2_\theta \over r} \quad = \quad {\text{31.3 m/s}^2 \over \text{0.3 m}} \quad = \quad \text{3,270 m}^2\text{/s} \quad = \quad \text{333 g's}$

### Cylindrical Acceleration Example

This example uses the function, $$r=[17-\cos(4\theta)]/16$$, with $$\theta = t$$, and calculates acceleration components.

The function looks like

The derivatives of $$r$$ are

$\begin{eqnarray} \dot r & = & {\partial \, r \over \partial \, t} = \left( {\partial \, r \over \partial \, \theta} \right) \left( {\partial \, \theta \over \partial \, t} \right) \\ \\ \\ & = & {1 \over 4} \sin(4 \theta) * \omega \qquad \text{where } \quad \omega = 1 \end{eqnarray}$
and the 2nd derivative is

$\begin{eqnarray} \ddot r & = & \cos(4 \theta) \end{eqnarray}$
So the acceleration vector is

$\begin{eqnarray} {\bf a} & = & ( \ddot r - r \, \omega^2 ) \, \hat{\bf r} + ( r \, \alpha + 2 \, \dot r \, \omega ) \; \hat{\boldsymbol{\theta}} + \ddot z \, \hat{\bf z} \\ \\ \\ & = & \left[ \cos(4 \theta) - {17 - \cos(4 \theta) \over 16 } \right] \hat{\bf r} + {1 \over 2} \sin(4 \theta) \hat{\boldsymbol{\theta}} \\ \\ \\ & = & { 17 \over 16} \left[ \cos(4 \theta) - 1 \right] \, \hat{\bf r} + {1 \over 2} \sin(4 \theta) \hat{\boldsymbol{\theta}} \end{eqnarray}$

### 2nd Cylindrical Acceleration Example

A bar is rotating at a rate, $$\omega$$. A collar starts at $$R_o$$ and is being flung off with zero friction. So the radial acceleration is zero. Therefore

$a_r = \ddot r - r \, \omega^2 = 0$
This is a 2nd order differential equation, whose solution is

$r = A e^{\omega \, t} + B \, e^{-\omega \, t}$
Assume the initial conditions are $$r(0) = R_o$$ and $$\dot r(0) = 0$$. This leads to

$R_o = A + B$ $\;\;\;0 = A - B$
And

$A = B = {R_o \over 2}$
So the solution is

$r = {1 \over 2} R_o e^{\omega \, t} + {1 \over 2} R_o e^{-\omega \, t}$
which can also be written as

$r = R_o \cosh(\omega \, t)$
Remember, this gives zero net radial acceleration for the case where $$\omega = constant$$.

Recall that the circumferential acceleration is

$a_\theta = r \, \alpha + 2 \dot r \omega$
$$\alpha$$ is zero because $$\omega$$ is a constant. $$\dot r$$ is

$\dot r = R_o \, \omega \sinh(\omega \, t)$
and this all combines to give

$a_\theta = 2 R_o \, \omega^2 \sinh(\omega \, t)$
which is a large circumferential acceleration due entirely to the Coriolis effect, even though $$\omega$$ is constant.

## Relationships in Cylindrical Coordinates

This section reviews vector calculus identities in cylindrical coordinates. (The subject is covered in Appendix II of Malvern's textbook.) This is intended to be a quick reference page. It presents equations for several concepts that have not been covered yet, but will be on later pages.

$\nabla = {\partial \over \partial \, r} \hat{{\bf r}} + {1 \over r} {\partial \over \partial \, \theta} \hat{\boldsymbol{\theta}} + {\partial \over \partial z} \hat{{\bf z}}$
where $$\hat{{\bf r}}$$, $$\hat{{\bf \theta}}$$, and $$\hat{{\bf z}}$$ are the three unit vectors.

The gradient of a scalar function, $$f$$, is

$\nabla f = {\partial f \over \partial \, r} \hat{{\bf r}} + {1 \over r} {\partial f \over \partial \, \theta} \hat{\boldsymbol{\theta}} + {\partial f \over \partial z} \hat{{\bf z}}$
The Laplacian of a scalar function is

$\nabla^2 \! f = {1 \over r} {\partial \over \partial \, r} \left( r {\partial f \over \partial \, r} \right) + {1 \over r^2} {\partial^2 \! f \over \partial \, \theta^2} + {\partial^2 \! f \over \partial z^2}$
The divergence of a vector is

$\nabla \cdot {\bf v} = {1 \over r} {\partial \over \partial \, r} \left( r \, v_{r} \right) + {1 \over r} {\partial \, v_{\theta} \over \partial \, \theta} + {\partial \, v_{z} \over \partial z}$
The curl of a vector is

$\nabla \times {\bf v} = \left( {1 \over r} {\partial \, v_{z} \over \partial \, \theta} - {\partial \, v_{\theta} \over \partial z} \right) \hat{{\bf r}} + \left( {\partial \, v_{r} \over \partial \, z} - {\partial \, v_{z} \over \partial \, r} \right) \hat{\boldsymbol{\theta}} + \left( {1 \over r} {\partial \over \partial \, r} (r \, v_\theta) - {1 \over r} {\partial \, v_{r} \over \partial \, \theta} \right) \hat{{\bf z}}$
The divergence of a tensor - in this case the stress tensor, $$\boldsymbol{\sigma}$$ - is given by

$\begin{eqnarray} \nabla \cdot \boldsymbol{\sigma} & = & \left[ {1 \over r} {\partial \over \partial \, r} \left( r \, \sigma_{\!rr} \right) + {1 \over r} {\partial \, \sigma_{\!r\theta} \over \partial \, \theta} + {\partial \, \sigma_{\!rz} \over \partial z} - {\sigma_{\theta \theta} \over r} \right] \hat{{\bf r}} \\ \\ \\ & + & \left[ {1 \over r} {\partial \over \partial \, r} \left( r \, \sigma_{\!r\theta} \right) + {1 \over r} {\partial \, \sigma_{\!\theta\theta} \over \partial \, \theta} + {\partial \, \sigma_{\!\theta z} \over \partial z} + {\sigma_{r \theta} \over r} \right] \hat{\boldsymbol{\theta}} \\ \\ \\ & + & \left[ {1 \over r} {\partial \over \partial \, r} \left( r \, \sigma_{\!rz} \right) + {1 \over r} {\partial \, \sigma_{\!\theta z} \over \partial \, \theta} + {\partial \, \sigma_{\!zz} \over \partial z} \right] \hat{{\bf z}} \end{eqnarray}$
The gradient of a vector produces a 2nd rank tensor.

$\nabla {\bf v} = \left[ \matrix { {\partial \, v_r \over \partial \, r} & {1 \over r} {\partial \, v_r \over \partial \, \theta} - {v_{\theta} \over r} & {\partial \, v_r \over \partial z} \\ \\ \\ {\partial \, v_{\theta} \over \partial \, r} & {1 \over r} {\partial \, v_{\theta} \over \partial \, \theta} + {v_r \over r} & {\partial \, v_{\theta} \over \partial z} \\ \\ \\ {\partial \, v_{z} \over \partial \, r} & {1 \over r} {\partial \, v_{z} \over \partial \, \theta} & {\partial \, v_{z} \over \partial z} } \right]$
If the vector happens to be the velocity vector, $${\bf v}$$, then the tensor is called the velocity gradient, and represented by $${\bf L} = \nabla {\bf v}$$. The symmetric part of $${\bf L}$$ is the rate of deformation tensor, $${\bf D}$$, and the antisymmetric part is the spin tensor, $${\bf W}$$.

${\bf D} \quad = \quad (\nabla {\bf v})_{sym} \quad = \quad \left[ \matrix { {\partial \, v_r \over \partial \, r} & {1 \over 2} \left( {1 \over r} {\partial \, v_r \over \partial \, \theta} + {\partial \, v_{\theta} \over \partial \, r} - {v_{\theta} \over r} \right) & {1 \over 2} \left( {\partial \, v_r \over \partial z} + {\partial \, v_{z} \over \partial \, r} \right) \\ \\ & {1 \over r} {\partial \, v_{\theta} \over \partial \, \theta} + {v_r \over r} & {1 \over 2} \left( {\partial \, v_{\theta} \over \partial z} + {1 \over r} {\partial \, v_{z} \over \partial \, \theta} \right) \\ \\ \text{sym} & & {\partial \, v_{z} \over \partial z} } \right]$

${\bf W} \quad = \quad (\nabla {\bf v})_{anti} \quad = \quad \left[ \matrix { 0 & {1 \over 2} \left( {1 \over r} {\partial \, v_r \over \partial \, \theta} - {\partial \, v_{\theta} \over \partial \, r} - {v_{\theta} \over r} \right) & {1 \over 2} \left( {\partial \, v_r \over \partial z} - {\partial \, v_{z} \over \partial \, r} \right) \\ \\ & 0 & {1 \over 2} \left( {\partial \, v_{\theta} \over \partial z} - {1 \over r} {\partial \, v_{z} \over \partial \, \theta} \right) \\ \\ \text{anti} & & 0 } \right]$
The deformation gradient tensor is the gradient of the displacement vector, $${\bf u}$$, with respect to the reference coordinate system, $$(R, \theta, Z)$$.

${\bf F} \quad = \quad {\bf I} + \nabla {\bf u} \quad = \quad \left[ \matrix { 1 + {\partial \, u_r \over \partial \, R} & {1 \over R} {\partial \, u_r \over \partial \, \theta} - {u_\theta \over R} & {\partial \, u_r \over \partial Z} \\ \\ \\ {\partial \, u_\theta \over \partial \, R} & 1 + {1 \over R} {\partial \, u_\theta \over \partial \, \theta} + {u_r \over R} & {\partial \, u_\theta \over \partial Z} \\ \\ \\ {\partial \, u_z \over \partial \, R} & {1 \over R} {\partial \, u_z \over \partial \, \theta} & 1 + {\partial \, u_z \over \partial Z} } \right]$

The Green strain tensor, $${\bf E}$$, is related to the deformation gradient, $${\bf F}$$, by $${\bf E} = ( {\bf F}^T \cdot {\bf F} - {\bf I} ) / 2$$. This applies in cylindrical, rectangular, and any other coordinate system. However, the terms in $${\bf E}$$ become very involved in cylindrical coordinates, so they are not written here.

The equations of equilibrium are

$\begin{eqnarray} & & {1 \over r} {\partial \over \partial \, r} \left( r \sigma_{rr} \right) + {1 \over r} {\partial \, \sigma_{r\theta} \over \partial \, \theta} + {\partial \, \sigma_{rz} \over \partial z} - { \sigma_{\theta\theta} \over r} + \rho f_r = \rho \, a_r \\ \\ \\ & & {1 \over r} {\partial \over \partial \, r} \left( r \sigma_{r \theta} \right) + {1 \over r} {\partial \, \sigma_{\theta\theta} \over \partial \, \theta} + {\partial \, \sigma_{\theta z} \over \partial z} + {\sigma_{r \theta} \over r} + \rho f_\theta = \rho \, a_\theta \\ \\ \\ & & {1 \over r} {\partial \over \partial \, r} \left( r \sigma_{rz} \right) + {1 \over r} {\partial \, \sigma_{\theta z} \over \partial \, \theta} + {\partial \, \sigma_{zz} \over \partial z} + \rho f_z = \rho \, a_z \end{eqnarray}$
Note that the terms involving $$\boldsymbol{\sigma}$$ constitute the divergence of the stress tensor, so all three equations can be abbreviated, $$\nabla \cdot \boldsymbol{\sigma} + \rho \, {\bf f} = \rho \, {\bf a}$$.

The three components of the acceleration vector, $${\bf a}$$, are

$\begin{eqnarray} a_r & = & {\partial \, v_r \over \partial \, t} + v_r {\partial \, v_r \over \partial \, r} + {v_{\theta} \over r} {\partial \, v_r \over \partial \, \theta} + v_{z} {\partial \, v_r \over \partial z} - {v^2_{\theta} \over r} \\ \\ \\ a_{\theta} & = & {\partial \, v_{\theta} \over \partial \, t} + v_r {\partial \, v_{\theta} \over \partial \, r} + {v_{\theta} \over r} {\partial \, v_{\theta} \over \partial \, \theta} + v_{z} {\partial \, v_{\theta} \over \partial z} + {v_r v_{\theta} \over r} \\ \\ \\ a_{z} & = & {\partial \, v_{z} \over \partial \, t} + v_r {\partial \, v_{z} \over \partial \, r} + {v_{\theta} \over r} {\partial \, v_{z} \over \partial \, \theta} + v_{z} {\partial \, v_{z} \over \partial z} \\ \end{eqnarray}$
The $$v^2_{\theta} / r$$ term in the $$a_r$$ component is the centripetal acceleration that produces centripetal forces (not centrifugal).

The $$v_r {\partial \, v_{\theta} \over \partial \, r}$$ and the $$v_r v_{\theta} / r$$ terms in the $$a_\theta$$ component together make up the Coriolis acceleration.

### Thank You

Thank you for visiting this webpage. Feel free to email me if you have questions.

Bob McGinty
bmcginty@gmail.com

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