Let us consider a point P in cylindrical co-ordinate system, having cylindrical co-ordinates \( (\rho,\ \phi,\ z) \). The cartesian co-ordinates of the point P is \( (x,\ y,\ z) \). Let’s draw a vertical line PQ on the X-Y plane, which is parallel to the z-axis, as shown in Fig.1.
OQ is represented by \( \rho \), which makes an angle \( \phi \) with the X-axis.
From Fig. 1 we can write,
\( x=\rho\ \cos\phi \)
\( y=\rho\ \sin\phi \)
and, \( z=z \)
If \( \vec{r} \) is the position vector with respect to the origin O, then we can write
\( \vec{r}=x\ \hat{i}+y\ \hat{j}+z\ \hat{k}\tag{1} \)
where, \( \hat{i} \), \( \hat{j} \) and \( \hat{k} \) is the unit vector along the X-axis, Y-axis and Z-axis respectively.
Now, equation (1) can be written as,
\( \vec{r}=\rho\ \cos\phi\ \hat{i}+\rho\ \sin\phi\ \hat{j}+z\ \hat{k}\tag{2} \)
The unit vectors \( \hat{\rho} \), \( \hat{\phi} \) and \( \hat{z} \) in the cylindrical pola co-ordinate system in terms of the unit vecctors in certesian co-ordinate can be written as,
\( \hat{\rho}=\cos\phi\ \hat{i}+ \sin\phi\ \hat{j}\tag{i} \)
\( \hat{\phi}=-\sin\phi\ \hat{i}+\cos\phi\ \hat{j}\tag{ii} \)
and, \( \hat{z}=\hat{k}\tag{iii} \)
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Again,
Differentiating \( \hat{\rho} \) with respect to time \( t \) we get,
\( \frac{d\hat{\rho}}{dt}=\frac{d}{dt}(\cos\phi\ \hat{i}+ \sin\phi\ \hat{j}) \)
\( or,\ \frac{d\hat{\rho}}{dt}=\frac{d}{d\phi}(\cos\phi\ \hat{i}+ \sin\phi\ \hat{j})\frac{d\phi}{dt} \)
\( or,\ \frac{d\hat{\rho}}{dt}=(-\sin\phi\ \hat{i}+\cos\phi\ \hat{j})\dot{\phi} \)
\( or,\ \displaystyle{\frac{d\hat{\rho}}{dt}=\dot{\phi}\ \hat{\phi}}\tag{iv} \) using equation (ii)
Differentiating \( \hat{\phi} \) with respect to time \( t \), we get
\( \frac{d\hat{\phi}}{dt}=\frac{d}{dt}(-\sin\phi\ \hat{i}+\cos\phi\ \hat{j}) \)
\( or,\ \frac{d\hat{\phi}}{dt}=\frac{d}{d\phi}(-\sin\phi\ \hat{i}+\cos\phi\ \hat{j})\frac{d\phi}{dt} \)
\( or,\ \frac{d\hat{\phi}}{dt}=-(\cos\phi\ \hat{i}+\sin\phi\ \hat{j})\dot{\phi} \)
\( or,\ \displaystyle{\frac{d\hat{\phi}}{dt}=-\dot{\phi}\hat{\rho}}\tag{v} \) using equation (i).
Differentiating \( \hat{z} \) with respect to time \( t \) we get,
\( \displaystyle{\frac{d\hat{z}}{dt}=0}\tag{vi} \)
Multiplying both side of equation (i) by \( \cos\phi \) and equation (ii) by \( \sin\phi \) and then subtracting, we get
\( \hat{i}=\cos\phi\ \hat{\rho}-\sin\phi\ \hat{\phi}\tag{vii} \)
Multiplying both side of equation (i) by \( \sin\phi \) and equation (ii) by \( \cos\phi \) and then adding, we get
\( \hat{j}=\sin\phi\ \hat{\rho}+\cos\phi\ \hat{\phi}\tag{viii} \)
Now putting value of \( \hat{i} \), \( \hat{j} \) and \( \hat{k} \) in the equation (2), by using equaions (vii), (viii) & (iii), we get
\( \vec{r}=\rho\ \cos\phi\ (\cos\phi\ \hat{\rho}-\sin\phi\ \hat{\phi})+\rho\ \sin\phi\ (\sin\phi\ \hat{\rho}+\cos\phi\ \hat{\phi})+z\ \hat{z} \)
\(or,\ \vec{r}=(\rho\ \sin^2\phi+\rho\ \cos^2\phi)\hat{\rho}+(\rho\ \cos\phi\ \sin\phi-\rho\ \cos\phi\ \sin\phi)\hat{\phi}+z\ \hat{z} \)
\( or,\ \vec{r}=\rho\ \hat{\rho}+z\ \hat{z}\tag{3} \)
Now eqation (3) can be written as,
\( \vec{r}=r_{\rho}\ \hat{\rho}+r_{\phi}\ \hat{\phi}+r_{z}\ \hat{z}\tag{4} \)
where, \( r_{\rho}=\rho \), \( r_{\phi}=0 \) and \( r_{z}=z \) are the components of the position vectors \( \vec{r} \) along the direction of the unit vectors \( \hat{\rho} \), \( \hat{\phi} \) and \( \hat{z} \) respectively.
Expression for velocity in cylindrical polar co-ordinate system:
Since velocity is the time derivative of the displacement vector,
So, velocity \( \displaystyle{\vec{v}=\frac{d\vec{r}}{dt}} \)
Now differentiating equation (4) with respect to time t, we get
\( \displaystyle{\frac{d\vec{r}}{dt}=\frac{dr_{\rho}}{dt}\ \hat{\rho}+r_{\rho}\frac{d\hat{\rho}}{dt}+\frac{dr_{\phi}}{dt}\hat{\phi}+r_{\phi}\frac{d\hat{\phi}}{dt}+\frac{dr_{z}}{dt}\hat{z}+r_{z}\frac{d\hat{z}}{dt}} \)
\( or,\ \displaystyle{\frac{d\vec{r}}{dt}=\frac{dr_{\rho}}{dt}\ \hat{\rho}+r_{\rho}\ \dot{\phi}\ \hat{\phi}+\frac{dr_{\phi}}{dt}\hat{\phi}+r_{\phi}(-\dot{\phi}\hat{\rho})+\frac{dr_{z}}{dt}\hat{z}} \)
[ using equations (iv), (v) & (vi) ]
\( or,\ \displaystyle{\frac{d\vec{r}}{dt}=\frac{d\rho}{dt}\ \hat{\rho}+\rho\ \dot{\phi}\ \hat{\phi}+\dot{z}\ \hat{z}} \)
\( or,\ \displaystyle{\frac{d\vec{r}}{dt}=\dot{\rho}\ \hat{\rho}+\rho\ \dot{\phi}\ \hat{\phi}+\dot{z}\ \hat{z}}\tag{5} \)
So equation (5) is the expression for the velocity of a particle in a cylindrical polar co-ordinate system.