Finding moment of inertia from Lagrange equation

I'm getting the following information:

Consider a system consisting of two rotating bars of length $\ell$ 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 $a$.

So I'm guessing it's just to bars revolving around the z-axis with the separation of $a$.

Now, I'm told that the Lagrangian for the system is:

$$\mathcal{L} = \frac{1}{2}I_{1}\dot\theta_{1}^{2}+\frac{1}{2}I_{2}\dot\theta_{2}^{2}-V(\theta_{1},\theta_{2}),$$

where $I_{1}=I_{2}$ is the moment of inertia, and $V$ an unspecified potential.

Now I'm asked to calculate $I_{1}=I_{2}$.

My idea was to just use the Euler-Lagrange equation, which I did, and ended up with $I_{1} = \frac{F_{\theta_{1}}\ell}{\alpha}=\frac{torque}{\alpha}$, where $\alpha$ is just the rotational acceleration. But if I just look up the moment of inertia for a rod with end points at the axis of rotation (As here), I find $I = \frac{m\ell^{2}}{3}$.

So I'm thinking I'm a bit off here. So any idea how to do this instead, or am I just missing something ?

• I'm having trouble picturing the bars and the axis. Could you clarify a little? I'd like to take a crack at this. – user35033 Feb 14 '14 at 6:29
• The moment of inertia is just a property of the rod. Like you said, it's $ml^2/3$ about one end, or $ml^2/12$ about the center. – user35033 Feb 14 '14 at 6:31
• I think I only have to calculate it at the end point actually. But, that is not "calculating" it, more of just looking it up. And it said "calculate" the moment of inertia, so... – Denver Dang Feb 14 '14 at 10:45