Question 11.1: Apply Lagrange’s equations (11.15) to derive the equations o...

The three independent generalized coordinates and generalized  velocity components for the unconstrained motion of a particle in terms of cylindrical coordinates are defined by

\left(q_1, q_2, q_3\right)=(r, \phi, z), \quad\left(\dot{q}_1, \dot{q}_2, \dot{q}_3\right)=(\dot{r}, \dot{\phi}, \dot{z}).                       (11.19a)

It should be noted that the generalized coordinates and velocities are not the respective physical scalar components of either the actual  position vector   \mathbf{x}=r \mathbf{e}_r  +  z \mathbf{e}_z,  or the velocity vector   \mathbf{v}=\dot{r} \mathbf{e}_r+r \dot{\phi} \mathbf{e}_\phi+\dot{z} \mathbf{e}_z  in cylindrical coordinates. The kinetic energy function in cylindrical coordinates is given by

T=\frac{1}{2} m \mathbf{v} \cdot \mathbf{v}=\frac{1}{2} m\left(\dot{r}^2  +  r^2 \dot{\phi}^2  +  \dot{z}^2\right)                           (11.19b)

Hence , with (11.19a) and (11.15) in mind, we use (11.19b) to first  derive

\begin{array}{r}\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{r}}\right)-\frac{\partial T}{\partial r}=\frac{d}{d t}(m \dot{r})  –  m r \dot{\phi}^2=m\left(\ddot{r}  –  r \dot{\phi}^2\right), \\\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{\phi}}\right)  –  \frac{\partial T}{\partial \phi}=\frac{d}{d t}\left(m r^2 \dot{\phi}\right)=m \frac{d}{d t}\left(r^2 \dot{\phi}\right), \\ \frac{d}{d t}\left(\frac{\partial T}{\partial \dot{z}}\right)  –  \frac{\partial T}{\partial z}=\frac{d}{d t}(m \dot{z})=m \ddot{z} .\end{array}                 (11.19c)

To complete the formulation of the equations of motion from the Lagrange equations (11.15), we next consider the generalized  forces in (11.15). The virtual work (11.17) done by the total force   \mathbf{F}=F_r \mathbf{e}_r  +  F_\phi \mathbf{e}_\phi  +  F_z \mathbf{e}_z  in the virtual displacement   \delta \mathbf{x}=\delta r \mathbf{e}_r  +  r \delta \phi \mathbf{e}_\phi  +  \delta z \mathbf{e}_z  is given by

\delta \mathscr{W}=\mathbf{F} \cdot \delta \mathbf{x}                 (11.17)

\delta \mathscr{W}=F_r \delta r+r F_\phi \delta \phi+F_z \delta z (11.19d)

By (11.18), the virtual work done by the generalized forces (Q_1, Q_2, Q_3) =(Q_r, Q_\phi, Q_z) acting over the generalized virtual displacements (\delta q_1, \delta q_2, \delta q_3) =(\delta r, \delta \phi , \delta z) is

\delta \mathscr{W}=Q_r \delta r+ Q_\phi \delta \phi+Q_z \delta z (11.19e)

Since there are no constraints, the virtual work relations in (11.19d) and (11.1ge) must be equal for all arbitrary virtual displacements \delta r, \delta \phi , \delta z, and hence the generalized force components are given by

Q_r=F_r, \quad Q_\phi=r F_\phi, \quad Q_z=F_z.               (11.19f)

Notice that   \left[Q_\phi\right]=[F L]  has  dimensional units of torque . We thus see that the generalized forces need not have dimensional units of force , as do  Q_r  and  Q_z.

Use of the two sets of results (11.19c) and (11.9f) in the Lagrange equations (11.15) now yields the familiar equations of  unconstrained motion for a particle in terms of its cylindrical coordinates:

m\left(\ddot{r}-r \dot{\phi}^2\right)=F_r, \quad \frac{m}{r} \frac{d}{d t}\left(r^2 \dot{\phi}\right)=F_\phi, \quad m \ddot{z}=F_z.                 (11.19g)

These agree with equations (6.4) based on Newton’s second law.