Chapter 1, Problem 1.47P

(a)

To determine

The sketch of three cylindrical coordinates, expressions for the coordinates in terms of the Cartesian coordinates, the definition of $\rho$ in words and the reason why it is unfortunate to use $r$ instead of $\rho$.

Answer to Problem 1.47P

The sketch of three cylindrical coordinates is

The expressions for the cylindrical coordinates in terms of the Cartesian coordinates are $\left(\rho ,\varphi ,z\right)=\left(\sqrt{x^2+y^2},\arctan \left(\frac{y}{x}\right)+\eta ,z\right)$ and the reason why it is unfortunate to use $r$ instead of $\rho$ is that $\overrightarrow{r}$ is the position vector of the particle at point P.

Explanation of Solution

The three cylindrical polar coordinates of the point P is shown in figure 1.

$\rho$ is the distance of P from the $z$ axis of the cylinder, $\varphi$ is the angle made by the $\rho$ vector with the $x$ axis and $z$ is the height of the point P from the $xy$ plane.

Write the expression for $\rho$ in terms of the Cartesian coordinates.

  $\rho =\sqrt{x^2+y^2}$

Write the expression for $\varphi$ in terms of the Cartesian coordinates.

  $\begin{array}{l}\varphi =\text{arctan}\left(\frac{y}{x}\right)+\eta \\ \text{where, }\eta =\{\begin{array}{c}\text{undefined when }x=0y=0\\ 0\text{ when }x\ge 0y\ge 0\\ \pi \text{ when }x<0\\ 2\pi \text{ when }x\ge 0\text{and }y<0\end{array}\end{array}$

Write the expression for $z$ in terms of the Cartesian coordinates

  $z=z$

It is unfortunate to use $r$ instead of $\rho$ since $\overrightarrow{r}$ is the position vector of the particle at point P.

Conclusion:

Therefore, the sketch of three cylindrical coordinates is shown in figure 1. The expressions for the cylindrical coordinates in terms of the Cartesian coordinates are $\left(\rho ,\varphi ,z\right)=\left(\sqrt{x^2+y^2},\arctan \left(\frac{y}{x}\right)+\eta ,z\right)$ and the reason why it is unfortunate to use $r$ instead of $\rho$ is that $\overrightarrow{r}$ is the position vector of the particle at point P.

(b)

To determine

The description of the three unit vectors $\widehat{\rho },\widehat{\varphi },\widehat{z}$ and the expansion of the position vector $\overrightarrow{r}$ in terms of the unit vectors.

Answer to Problem 1.47P

The unit vector $\widehat{\rho }$ points in the direction of increasing of the coordinate $\rho$ so that it will be directly away from the $z$ axis. The unit vector $\widehat{\varphi }$ is tangent to the horizontal circle through P centered on the $z$ axis. The unit vector $\widehat{z}$ will be parallel to the $z$ axis. The position vector $\overrightarrow{r}$ in terms of the unit vectors is $\overrightarrow{r}=\rho \widehat{\rho }+z\widehat{z}$ .

Explanation of Solution

The unit vector $\widehat{\rho }$ points in the direction of increasing of the coordinate $\rho$ so that it will be directly away from the $z$ axis. The unit vector $\widehat{\varphi }$ is tangent to the horizontal circle through P centered on the $z$ axis. The unit vector $\widehat{z}$ will be parallel to the $z$ axis.

Write the expansion of the position vector $\overrightarrow{r}$ in terms of the unit vectors.

  $\overrightarrow{r}=\rho \widehat{\rho }+\varphi \widehat{\varphi }+z\widehat{z}$

Here, $\overrightarrow{r}$ is the position vector.

The unit vector $\widehat{\rho }$ already contains the $\varphi$ direction information so that $\widehat{\varphi }$ component can be ignored from the expression for $\overrightarrow{r}$.

Rewrite the expression for $\overrightarrow{r}$ neglecting $\widehat{\varphi }$ component.

  $\overrightarrow{r}=\rho \widehat{\rho }+z\widehat{z}$        (I)

Conclusion:

Therefore, the unit vector $\widehat{\rho }$ points in the direction of increasing of the coordinate $\rho$ so that it will be directly away from the $z$ axis. The unit vector $\widehat{\varphi }$ is tangent to the horizontal circle through P centered on the $z$ axis. The unit vector $\widehat{z}$ will be parallel to the $z$ axis. The position vector $\overrightarrow{r}$ in terms of the unit vectors is $\overrightarrow{r}=\rho \widehat{\rho }+z\widehat{z}$.

(c)

To determine

The cylindrical components of the acceleration $\overrightarrow{a}=\ddot{\overrightarrow{r}}$ of the particle.

Answer to Problem 1.47P

The cylindrical components of the acceleration $\overrightarrow{a}=\ddot{\overrightarrow{r}}$ of the particle are $a_{\rho }=\ddot{\rho }-\rho {\dot{\varphi }}^2,a_{\varphi }=\rho \ddot{\varphi }+2\dot{\rho }\dot{\varphi },a_z=\ddot{z}$.

Explanation of Solution

Differentiate equation (I) with respect to time.

  $\dot{\overrightarrow{r}}=\dot{\rho }\widehat{\rho }+\rho \dot{\widehat{\rho }}+\dot{z}\widehat{z}+z\dot{\widehat{z}}$        (II)

Write the expression for $\dot{\widehat{\rho }}$ .

  $\dot{\widehat{\rho }}=\dot{\varphi }\widehat{\varphi }$        (III)

Write the expression for $\dot{\widehat{\varphi }}$ .

  $\dot{\widehat{\varphi }}=-\dot{\varphi }\widehat{\rho }$        (IV)

Write the expression for $\dot{\widehat{z}}$ .

  $\dot{\widehat{z}}=0$        (V)

Put equations (III) and (V) in equation (II).

  $\begin{array}{c}\dot{\overrightarrow{r}}=\dot{\rho }\widehat{\rho }+\rho \dot{\varphi }\widehat{\varphi }+\dot{z}\widehat{z}+0\\ =\dot{\rho }\widehat{\rho }+\rho \dot{\varphi }\widehat{\varphi }+\dot{z}\widehat{z}\end{array}$

Differentiate the above equation with respect to time.

  $\ddot{\overrightarrow{r}}=\ddot{\rho }\widehat{\rho }+\dot{\rho }\dot{\widehat{\rho }}+\dot{\rho }\dot{\varphi }\widehat{\varphi }+\rho \ddot{\varphi }\widehat{\varphi }+\rho \dot{\varphi }\dot{\widehat{\varphi }}+\ddot{z}\widehat{z}+z\dot{\widehat{z}}$

Put equations (III), (IV) and (V) in the above equation.

  $\begin{array}{c}\ddot{\overrightarrow{r}}=\ddot{\rho }\widehat{\rho }+\dot{\rho }\dot{\varphi }\widehat{\varphi }+\dot{\rho }\dot{\varphi }\widehat{\varphi }+\rho \ddot{\varphi }\widehat{\varphi }+\rho \dot{\varphi }\left(-\dot{\varphi }\widehat{\rho }\right)+\ddot{z}\widehat{z}+0\\ =\ddot{\rho }\widehat{\rho }+\dot{\rho }\dot{\varphi }\widehat{\varphi }+\dot{\rho }\dot{\varphi }\widehat{\varphi }+\rho \ddot{\varphi }\widehat{\varphi }-\rho {\dot{\varphi }}^2\widehat{\rho }+\ddot{z}\widehat{z}\\ =\left(\ddot{\rho }-\rho {\dot{\varphi }}^2\right)\widehat{\rho }+\left(\rho \ddot{\varphi }+2\dot{\rho }\dot{\varphi }\right)\widehat{\varphi }+\ddot{z}\widehat{z}\end{array}$        (VI)

Write the expression for the acceleration of the particle.

  $\overrightarrow{a}=\ddot{\overrightarrow{r}}$        (VII)

Here, $\overrightarrow{a}$ is the acceleration of the particle.

Write the expression for $\overrightarrow{a}$ in cylindrical polar components.

  $\overrightarrow{a}=a_{\rho }\widehat{\rho }+a_{\varphi }\widehat{\varphi }+a_z\widehat{z}$        (VIII)

Here, $a_{\rho },a_{\varphi },a_z$ are the cylindrical components of acceleration of the particle.

Put equations (VI) and (VIII) in equation (VII).

  $\begin{array}{c}{a}_{\rho }\widehat{\rho }+a_{\varphi }\widehat{\varphi }+a_z\widehat{z}=\left(\ddot{\rho }-\rho {\dot{\varphi }}^2\right)\widehat{\rho }+\left(\rho \ddot{\varphi }+2\dot{\rho }\dot{\varphi }\right)\widehat{\varphi }+\ddot{z}\widehat{z}\\ \Rightarrow a_{\rho }=\ddot{\rho }-\rho {\dot{\varphi }}^2\\ a_{\varphi }=\rho \ddot{\varphi }+2\dot{\rho }\dot{\varphi }\\ a_z=\ddot{z}\end{array}$

Conclusion:

Therefore, the cylindrical components of the acceleration $\overrightarrow{a}=\ddot{\overrightarrow{r}}$ of the particle are $a_{\rho }=\ddot{\rho }-\rho {\dot{\varphi }}^2,a_{\varphi }=\rho \ddot{\varphi }+2\dot{\rho }\dot{\varphi },a_z=\ddot{z}$.