Lagrangian of a sliding ladder

[physiks]

Homework Statement

A ladder of length 2l and mass m leans against a smooth wall and rests on a smooth floor. The ladder initially makes an angle θ₀ to the vertical. It slides downwards maintaining contact with both the wall and the floor. Calcula the the Lagrangian and the conjugate momentum, and find the equation of motion. (The moment of inertia of a rod of length a and mass M about an axis through its centre perpendicular to the rod is Ma²/12.)

The Attempt at a Solution

I have L=(2/3)ml²(dθ/dt)²-mglcosθ with θ the time varying angle to the vertical. I'm confident this is correct as a similar result is obtained here for a slightly different definition of θ http://mathhelpforum.com/calculus/129710-lagrangian-function-ladder-smooth-wall.html.

Therefore why is the question asking for a conjugate momentum?

 

[BvU]

Once you have the Lagrangian, Euler-Lagrange gives you the equation of motion. Conjugate momentum is the momentum associated with the generalized coordinates, in this case only $\theta$. It occurs in the EOM.

 

[physiks]

Once you have the Lagrangian, Euler-Lagrange gives you the equation of motion. Conjugate momentum is the momentum associated with the generalized coordinates, in this case only $\theta$. It occurs in the EOM.

Ah, so the conjugate momentum for theta is the partial derivative of L wrt theta dot. I got confused thinking a conjugate momentum had to be conserved, but I realize now it doesn't.

So this means the angular momentum of the ladder is varying due to the external torques?

 

[BiGyElLoWhAt]

Is that the correct Lagrangian? I feel like there should be a translational portion to the KE terms. Yea, the ladder is rotating, but it's center of mass is moving down, and (potentially neglegibly, depending on your prof) out away from the wall it's leaning on. I say potentially neglegibly because in my intermediate mechanics class we had a system of 2 rods fixed at one end and a spring connecting the other 2 ends with some masses in between the springs, and we were told to neglect the translational motion of the blocks as the springs expanded and contracted (as the springs expand, the blocks move closer to the fixed pivot, and vice versa)... But of course we were told this after the test... lol.

 

[CAF123]

Is that the correct Lagrangian? I feel like there should be a translational portion to the KE terms.

There is a translational component. If $\theta$ is the angle from the vertical, then the C.O.M can be described by $\ell \cos \theta \hat{y} + \ell \sin \theta \hat{x}$. From this you can extract the x and y components and take their time derivative to give the translational kinetic energy.

 

[BiGyElLoWhAt]

There is a translational component. If $\theta$ is the angle from the vertical, then the C.O.M can be described by $\ell \cos \theta \hat{y} + \ell \sin \theta \hat{x}$. From this you can extract the x and y components and take their time derivative to give the translational kinetic energy.

Ok that's what I thought, just wanted to make sure before OP left here with the wrong L.

Only 1 question, is the $\ell \cos (\theta) \hat{y}$ a typo? seems as though the $\hat{x}$ & $\hat{y}$ should be switched. Also I am almost certain it should be $\frac{\ell}{2}$ vs. $\ell$ (assuming that's the description of it's position, of course)

 

[CAF123]

Hi BiGyElLoWhAt,

Only 1 question, is the $\ell \cos (\theta) \hat{y}$ a typo? seems as though the $\hat{x}$ & $\hat{y}$ should be switched.

No, I don't think so, provided I interpreted the question correctly. See sketch.

Also I am almost certain it should be $\frac{\ell}{2}$ vs. $\ell$ (assuming that's the description of it's position, of course)

It is given that the length is 2l, so from the C.O.M to either end, we have to consider a length l.

 

[BiGyElLoWhAt]

If $\theta$ is the angle from the vertical,

OOOOO

A ladder of length 2l and mass m

AAAAA

And that reading thing strikes again...

 

[BvU]

Big: Physik has the correct kinetic energy of translation and rotation in the Lagrangian. That's how he/she got the 2/3 factor.

no typo, no error. $\theta$ is wrt vertical, length of ladder is 2 $\ell$. It is completely as described in post #1.

Ah, so the conjugate momentum for theta is the partial derivative of L wrt theta dot. I got confused thinking a conjugate momentum had to be conserved, but I realize now it doesn't.

So this means the angular momentum of the ladder is varying due to the external torques?

Right: there has to be a symmetry to get a conserved quantity. See Noether's[/PLAIN] theorem -- extremely important in physics.

And what happens to L is what happens to $\dot \theta$: Its time derivative is definitely not zero, as you can see from the EOM.

[edit]caf beat me to it