Derive An Expression For Acceleration Of A Particle In Cylindrical Polar Co-Ordinate System.

Share:

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.

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, \( \hat{z}=\hat{k}\tag{iii} \)

[ To know the derivation (CLICK HERE) ]

Again we know that,

\( \displaystyle{\frac{d\hat{\rho}}{dt}=\dot{\phi}\ \hat{\phi}}\tag{iv} \)

\( \displaystyle{\frac{d\hat{\phi}}{dt}=-\dot{\phi}\hat{\rho}}\tag{v} \)

and, \( \displaystyle{\frac{d\hat{z}}{dt}=0}\tag{vi} \)

The expression for the velocity of a particle in a cylindrical polar co-ordinate system is

\( \displaystyle{\frac{d\vec{r}}{dt}=\dot{\rho}\ \hat{\rho}+\rho\ \dot{\phi}\ \hat{\phi}+\dot{z}\ \hat{z}} \)

[ To know the derivation of the expression for velocity of a particle in cylindrical polar co-ordinate system ( CLICK HERE ) ]

Expression for acceleration in cylindrical co-ordinate system:

Since the acceleration is the time derivative of the velocity,

So, acceleration \( \displaystyle{\vec{a}=\frac{d\vec{v}}{dt}} \)

\( \displaystyle{\vec{a}=\frac{d}{dt}(\dot{\rho}\ \hat{\rho}+\rho\ \dot{\phi}\ \hat{\phi}+\dot{z}\ \hat{z})} \)

\( or,\ \displaystyle{\vec{a}=\frac{d\dot{\rho}}{dt}\hat{\rho}+\dot{\rho}\frac{d\hat{\rho}}{dt}+\dot{\rho}\dot{\phi}\hat{\phi}+\rho\frac{d\dot{\phi}}{dt}\hat{\phi}+\dot{\rho}\dot{\phi}\frac{d\hat{\phi}}{dt}+\frac{d\dot{z}}{dt}\hat{z}+\dot{z}\frac{d\hat{z}}{dt}} \)

\( or,\ \vec{a}=\ddot{\rho}\hat{\rho}+\dot{\rho}\dot{\phi}\hat{\phi}+\dot{\rho}\dot{\phi}\hat{\phi}+\rho\ddot{\phi}\hat{\phi}-\rho\dot{\phi}\dot{\phi}\hat{\rho}+\ddot{z}\hat{z} \)
[using equations (iv), (v) & (vi) ]

\( or,\ \displaystyle{\vec{a}=(\ddot{\rho}-\rho{\dot{\phi}}^2)\hat{\rho}+(\rho\ddot{\phi}+2\dot{\rho}\dot{\phi})\hat{\phi}+\ddot{z}\hat{z}} \)

This is the expression for acceleration at any instant of time in the cylindrical polar co-ordinate system.

Share:


Subscribe to the Physics Notebook Newsletter and get the latest insights and updates delivered straight to your inbox.