Electro-mechanics (Part 2)

This is an example of damped oscillation driven by a constant force. (A case of overdamping).

The current flowing through the circuit is given by the equation

$Ri+L\dfrac{di}{dt}=BDv\implies R\dfrac{di}{dt}+L\dfrac{d^2i}{dt^2}=BD\dfrac{dv}{dt}$ .

The equation of motion of the rod is given by

$F-iDB=m\dfrac{dv}{dt}$ .

Combining these two, we get

$\dfrac{d^2I}{dt^2}+2b\dfrac{dI}{dt}+\omega_0^2I=0$ , where

$I=i-\dfrac{F}{BD}, b=\dfrac{R}{2L}, \omega_0=\dfrac{BD}{\sqrt {mL}}$ .

Solving this equation we get

$i=\dfrac{F}{BD}+Ae^{-bt}\sin (\omega t+α)$ , where $A, α$ are constants of integration to be determined, and $\omega =\sqrt {\omega_0^2-b^2}$ .

Substituting values, we get

$i=1+Ae^{-\frac{t}{2}}\sin (\frac{\sqrt {399}}{2}t+α), v=i+\dfrac{di}{dt}=\dfrac{1}{10}\left (1+10Ae^{-\frac{t}{2}}\sin (\omega t+α+\tan^{-1} 2\omega)\right )$ .

Initial conditions yield

$A\sin α=-1, 10A\sin (α+\tan^{-1} 2\omega)=-1\implies A=\dfrac{20}{\sqrt {399}}, α=\tan^{-1} 2\omega$ .

Hence,

$v=\dfrac{1}{10}\left (1+\frac{200}{\sqrt {399}} e^{-\frac{t}{2}}\sin (\frac{\sqrt {399}}{2}t -\tan^{-1} \frac{\sqrt {399}}{199})\right )$ .

Velocity will be minimum when the sine function in it's expression will be $-1$ . Hence

$t=\dfrac{2}{\sqrt {399}}\left (\frac{3π}{2}+\tan^{-1} (\frac{\sqrt{399}}{199})\right )$

Then $v_{min}=\dfrac{1}{10}\left (1-\frac{200}{\sqrt {399}}e^{-\frac{t}{2}}\right ) \approx \boxed {-0.686886}$ .

Consider the given situation at a general instant of time $t$ . Let the current through the circuit be $I$ and the velocity of the rail as it moves to the right be $v$ . Applying Kirchoff's voltage law gives:

$-BDv + IR + L\dot{I}=0 \ \dots (1)$

Here $E = BDv$ is the magnitude of induced voltage due to electromagnetic induction.

Applying Newton's second law to the rail gives:

$F - IDB = m\dot{v} \ \dots (2)$

Re-arranging the above equations in a matrix form gives:

$\left[\begin{matrix}\dot{I} \\ \dot{v}\end{matrix}\right]=\left[\begin{matrix}-R/L & BD/L \\ -BD/m &0\end{matrix}\right]\left[\begin{matrix}I \\ v\end{matrix}\right] + \left[\begin{matrix}0 \\ 1/m\end{matrix}\right]F$

Let:

$x = \left[\begin{matrix}I \\ v\end{matrix}\right]$

Substituting all values:

$\dot{x} = \left[\begin{matrix}-1 & 10 \\ -10 &0\end{matrix}\right]x + \left[\begin{matrix}0 \\ 10\end{matrix}\right]$

Initial conditions:

$x(0) = \left[\begin{matrix}0 \\ 0\end{matrix}\right]$

For the sake of convenience, I solved this numerically beyond this point. A plot of velocity vs. time is as such:

One can see that the rod moves left-ward and rightward and this oscillation is damped until it settles to its steady-state value of 0.1.