# Finding equilibrium of mechanical system

A system is described as follows:

Consider a system consisting of two rotating bars of length $l$ and with uniform mass density and each with total mass $m$. The bars are attached to a common axis at one end around which the can rotate. The distance between the bars on the axis of rotation is $a$.

My teacher has asked this question:

Consider the case $V_{0} < 0$. What is the equilibrium configuration of the system? Determine the frequency of oscillation around this equilibrium for small $r$.

I have the Lagrangian for this system, which is:

$$L=\frac{1}{2}I_{tot}\dot{\theta}_{R}^{2}+\frac{1}{2}I_{rel}\dot{\theta}_{r}^{2}-V_{0}\cos(\theta_{r})$$

So my main problem is, I'm not quite sure when it is in equilibrium? Usually it's when the derivative is equal to zero or something, but I can't just do that to the entire Lagrangian, can I?

• What exactly is your system. The equilibrium condition can presumably be deduced by physical reasoning. – Danu Aug 13 '14 at 14:07
• Supposing that the third term is the potential, the problem should be straightforward, but without context the question doesn't seem answerable. – jinawee Aug 13 '14 at 14:13
• Look up the definition of the Euler-Lagrange equations. Then see what your generalized coordinates and velocities are. Insert your L into the E-L equations. Then think about what equilibrium means (the sum of all forces is ...). – Aziraphale Aug 13 '14 at 14:22
• Ah, sorry... Added the system in the main post now. – Denver Dang Aug 13 '14 at 14:23
• @Danu, The lagrangian tells everything we needs to know about the system. So, we do not need the system, if we have the lagrangian. – Physicist137 Aug 13 '14 at 15:25

I'll assume you want to know the equilibrium points. The Lagrangian tells you everything you need to know about the system.

Because variation of generalized momentum is: $$\frac{dp_k}{dt} = Q_k + \frac{\partial T}{\partial q_k} = -\frac{\partial V}{\partial q_k}+ \frac{\partial T}{\partial q_k} = \frac{\partial L}{\partial q_k}$$

Then: $$\frac{dp_k}{dt} = 0 \quad\Longrightarrow\quad \frac{\partial L}{\partial q_k}=0$$

In your question, I think it makes more sense to talk about equilibrium in terms of potential. You are using generalized coordinates $\theta_r, \theta_R$, and the lagrangian tells us the potential involves a $\cos\theta_r$, because the other terms are cleary kinetic related. The points of equilibrium is when the gradient of the potential equals to zero, which means, when the force equals zero. Thinking in terms of variation of generalized momentum, it must be zero. We then have:

$$\frac{\partial L}{\partial q_k} = \frac{\partial L}{\partial\theta_r} = \frac{\partial}{\partial\theta_r} \cos(\theta_r) = -\sin\theta_r$$.

The equlibrium points $\theta_r$ occurs, when $\sin\theta_r = 0$. Simple equation to solve.

• Perhaps something should be said about the stability of these equilibria (notably the one at zero..) – Danu Aug 13 '14 at 17:56
• What I was trying to say in my comment, what if you are given a lagrangian is not of the type $L=T-V$. It seems that in that your approach would not be valid. So you should have to reason why your lagrangian is of that form. – jinawee Aug 13 '14 at 18:00
• @jinawee, This is true. But the same could be proved directly from Euler-Lagrange equations, making no imposition for $L = T - V$. For what I saw so far, the lagrangian was "designed" to fully describe the system, and all data can be taken from a lagrangian. I can be wrong... If you think I am wrong, and you have a proof, an example, anything, please tell me. That way I can learn. =D – Physicist137 Aug 13 '14 at 20:23
• If the lagrangian above is of the type $L=T-V$, an equivalent lagrangian would be $L=1/3T^2+2TV-V^2=1/12(I_{tot}\dot{\theta}_{R}^{2}+\frac{1}{2}I_{rel}\dot{\theta}_{r}^{2})^2+\left(I_{tot}\dot{\theta}_{R}^{2}+I_{rel}\dot{\theta}_{r}^{2}\right)V_{0}\cos\theta_{r}-V_{0}^2\cos^2(\theta_{r})$, then you would have to solve $-\left(I_{tot}\dot{\theta}_{R}^{2}+I_{rel}\dot{\theta}_{r}^{2}\right)V_{0}\sin(\theta_{r})+2V_{0}^2\cos(\theta_{r})\sin(\theta_{r})=0$. I guess in this case there could be a way to justify $L=T-V$, but in general, the formula you gave doesn't seem right. – jinawee Aug 15 '14 at 8:55