Bar hanging from two springs with unequal masses

[Diracobama2181]

Suppose we have two different masses, m1m1 and m2m2 with each at the end of a massless rod of length ll, with each mass being attached to a separate spring of constant k such that both springs stem from the same point on the ceiling? What would be the Lagrangian of this system. I have tried several different approaches, but none seem to be quite right.
What I got thus far, r1=(x1,y1)(coordinate of the first mass) r2=(x2,y2) (coordinate of the second mass)
So T=(1/2)m1T=(1/2)m1((x˙1)^2+(y˙1)^2)+(1/2)m2((x˙2)^2+(y˙2)^2)
U=-mgy1-mgy2+(1/2)k((x1)^2+(y1)^2)+(1/2)k((x2)^2+(y2)^2).
From here, I get L=T-V. Is there any issues with my setup?

 

[BvU]

Hello DIrac, !

For this kind of problem we have an introductory physics homework forum with a decent template -- please use it.

And try $\LaTeX$ to make your expressions legible. If not, at least ue the sub/superscript buttons under ...

Questions to you:

You don't specify it's a 2D problem, but from your expressions it appears you do treat it as such, right ? Is it 2D ?

How many degrees of freedom are there (considering the constraint) ? So how many generalized coordinates ?

From here, I get L=T-V.

I don't see the logic. Do you mean With L = T-U I would get the following expression ... ?

What is you zero point for U ?

 

[Diracobama2181]

Yes, it is constrained to two dimensions. The problem itself seems to be a variation on a Goldstein problem. In that problem, a coordinate system can be taken with respect to the center of mass. The problem has three degrees of freedom, the center of mass of the bar in the $x$ and $y$ directions as well as an angular component about the center of mass. $L=T-U$ is the Lagrangian. I was thinking I could instead use coordinates about the center of mass, so T=$(1/2)(m_1+m_2)((\dot x)^2+(\dot y)^2)+(1/2)I(\dot \theta)^2$. However, now I seem to have trouble finding the extensions on the springs.

 

[BvU]

Yes, it seems to go messy.
Perhaps try a simpler case first (identical masses & springs)

[edit] are you looking for the full langrangian or one for small deviations from equilibrium ?

[edit]@haruspex : any suggestions ?

 

[Dr.D]

Yes, it is constrained to two dimensions. The problem itself seems to be a variation on a Goldstein problem. In that problem, a coordinate system can be taken with respect to the center of mass. The problem has three degrees of freedom, the center of mass of the bar in the xx and yy directions as well as an angular component about the center of mass. L=T−UL=T-U is the Lagrangian. I was thinking I could instead use coordinates about the center of mass, so T=(1/2)(m1+m2)((˙x)2+(˙y)2)+(1/2)I(˙θ)2(1/2)(m_1+m_2)((\dot x)^2+(\dot y)^2)+(1/2)I(\dot \theta)^2. However, now I seem to have trouble finding the extensions on the springs.

As is so often the case, you have leaped into the dynamics without paying adequate attention to the kinematics. Until you figure out how to express the positions of the masses in terms of your generalized coordinates, you will not make any progress. Ask yourself first, what are x, y, and theta? Secondly, how is the position of the first mass expressed in terms of these?

Over the years, I've seen countless students stumble on this very issue, the failure to address the kinematics. It is absolutely essential!

 

[Diracobama2181]

Rethought the problem and decided to use coordinates of the center of the rod. That way, the kinetic energy will be the translational kinetic energy plus the energy of rotation about the center.

Thus, setting $m=m_1+m_2$
$K=(1/2)m(\dot x)^2+(1/2)m(\dot y)^2+(1/2)I(\dot\theta)^2$
where $I=I_{cm}+md^2$
and $d=(l/2) \frac {m_1-m_2} {m_1+m_2}$
Then, assuming the system is initially in equilibrium,

$U=(1/2)k( ((x+(l/2)sin \theta )^2+(y+(l/2)sin \theta)^2))+((x-(l/2)cos \theta)^2+(y-(l/2)sin \theta)^2)$

From here, I can find the Lagrangian.

 

[Diracobama2181]

Rethought the problem and decided to use coordinates of the center of the rod. That way, the kinetic energy will be the translational kinetic energy plus the energy of rotation about the center.

Thus, setting $m=m_1+m_2$
$K=(1/2)m(\dot x)^2+(1/2)m(\dot y)^2+(1/2)I(\dot\theta)^2$
where $I=I_{cm}+md^2$
and $d=(l/2) \frac {m_1-m_2} {m_1+m_2}$
Then, assuming the system is initially in equilibrium,

$U=(1/2)k( ((x+(l/2)sin \theta )^2+(y+(l/2)sin \theta)^2))+((x-(l/2)cos \theta)^2+(y-(l/2)sin \theta)^2)$

From here, I can find the Lagrangian.

Correction, I only need to use the moment of inertia about the center of the bar. The parallel axis theorem is unnecessary.

 

[Dr.D]

Correction, I only need to use the moment of inertia about the center of the bar. The parallel axis theorem is unnecessary.

The correction is actually incorrect. The bar itself is massless, so you do need to use the parallel axis theorem to get the MMOI for the system about the system CM.