How to Apply Lagrange Equation, Really Confused

[refrigerator]

Homework Statement

I am having trouble understanding how to apply Lagrange's equation. I will present a simplified version of one of my homework problems.

Imagine an inverted pendulum, consisting of a bar attached at a hinge at point A. At point A is a torsional spring with spring constant K. The bar has length L and a moment of inertia of I about point A. The free end of the bar experiences a load P that is always pointing downward. Find the equation of motion for this system.

Homework Equations

L= T-V

The Attempt at a Solution

To have something to check against, I first did the problem using angular momentum balance. My coordinate of choice is the angle the bar forms with respect to vertical, with positive θ being counterclockwise.

I$\ddot{θ}$-PLsin(θ)-Kθ=0

OK. Now:

T= 1/2 I$\dot{θ}$²
V= 1/2 K$\dot{θ}$²

L= T-V

Eventually I get:

I$\ddot{θ}$-Kθ=0

Clearly something is not right. I need somehow to take into account load P, but I don't know where in Lagrange's equation it would show up. I thought maybe instead of setting Lagrange's equation equal to zero I should set it equal to the forcing term? But then how do I take into account the direction of the force? The other thought was, maybe I somehow need to take the load into account through the potential term, but I really have no clue how.

I am pretty sure this system is conservative, though considering how much I suck at physics I could be completely wrong...

I would appreciate any help... I am so lost.

 

[genericusrnme]

What you want to do is model the force P as the gradient of a potential energy, think about what you would do if this force P were gravity (it is just gravity).

The system is conservative!

Unless you see the word 'slip' or 'friction' you'll be pretty safe in assuming that any problem is conservative.

 

[refrigerator]

Thanks for your response.

I suppose if I thought of P as "mg" (except at the tip, and not the center of mass of the bar), then potential energy would become:

V= 1/2 K$\dot{θ}$² + PLcos(θ)

which would indeed get me the correct answer.

I guess it's not clear to me why exactly you can do that, though. It's not like the force is storing potential energy. If it were an inverted pendulum with gravity, then clearly the bar would have gravitational potential energy in addition to that stored in the spring. Can I apply this same method to any conservative force? Doesn't this ever get you in trouble?

Forgive me, I'm new to this and for some reason I just don't feel comfortable with this yet. I've actually skipped a class that covers energy methods and am now thinking I should have done my course schedule differently.

What happens if the force is always sideways instead of being always vertical? Can you say that this force has some potential with respect to the ground?

I realize I haven't been able to articulate my question, so I apologize in advance. It is a result of my being confused and not knowing what I don't understand.

 

[GothFraex]

In this case P would be due to gravity. It's usually safe to assume that in problems involving pendulums you'll have to take gravity into account.

Maybe this page will clear some things up:

http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html

 

[genericusrnme]

You can do that because the force IS storing a potential energy.
If you forgot about everything else in the system except for the load P, and lifted everything up, you would get energy out of the system by letting it fall.

If you have a block, with a constant force being applied horizontally, say in the negative x direction, then that force can be derived from the gradient of a potential, namely U(x)=Fx
Why?
Because if you pull the block to in the positive x direction you have to do work on the system, you must put energy into the system, this is taken account of by potential energy.
It's the exact same thing in the case of your load P on your pendulum except the force is in the negative verticle direction and U(y) = - m g y.

Hope this helps :D

 

[refrigerator]

Thank you to everyone who responded; I think I understand now.

 

[genericusrnme]

I'm glad I was able to help!