Classical Mechanics, constraint motion problem

[fluidistic]

Homework Statement

A particle of mass m moves under a uniform gravitational field along a rod which moves in a vertical plane with a constant angular velocity [itex]\vec \Omega[/itex]. Write down the motion equations of the particle and calculate the constraint force. Is the energy conserved? Discuss.

Homework Equations

Lagrangian and... not really sure, maybe some modified Lagrange equation?

The Attempt at a Solution

I have a very strong feeling the energy isn't conserved. I understand how the particle will move and without someone to move the rod the motion will eventually end so there's some dissipation of energy I believe. I'm not sure the Lagrangian is still T-V.
Anyway, assuming it's still T-V, I chose polar coordinates r, theta to describe the position of the particle. My reference frame is in the middle of the rod (I assume it's "fixed").
[itex]T=\frac{m}{2}r^2 \dot \theta ^2[/itex] and [itex]V=mgr \sin \theta[/itex].
Now is my bigger problem. I don't know AT ALL how to modify the Lagrange equation!
I mean, I know that [itex]\frac{\partial L}{\partial q_i}- \frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q_i} \right ) \neq 0[/itex], unlike in conservative systems.
I guess I have to use some Lagrange multipliers but I'm not even sure. I'm stuck here. How to calculate the constraint force? I know it's the force the rod exert on the particle so it would be the normal force, namely [itex]mg \cos \theta[/itex] but I'd have to solve for [itex]\theta (t)[/itex]. Hmm... wait it seems rather easy. Since [itex]\dot \theta = \Omega[/itex], I have that [itex]\theta ( t)= \Omega t + \theta _0[/itex].
I guess r(t) would be much harder to find, but do I need it?
Also, how do I prove that the energy is not conserved? Maybe deriving the Lagrangian with respect to time and if it's not worth 0 it means the Lagrangian is not constant and thus (I believe the implication is true), E isn't constant. Is this a good approach? Am I missing stuff?

 

[ideasrule]

I have a very strong feeling the energy isn't conserved.

Good intuition!

I understand how the particle will move and without someone to move the rod the motion will eventually end so there's some dissipation of energy I believe. I'm not sure the Lagrangian is still T-V.

Actually, the motion wouldn't end in a perfect frictionless world. However, there's no way the rod can maintain a constant rotational speed all on its own, because the particle keeps on applying a force to it and accelerating it in this direction or that.

I don't know AT ALL how to modify the Lagrange equation!
I mean, I know that [itex]\frac{\partial L}{\partial q_i}- \frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q_i} \right ) \neq 0[/itex], unlike in conservative systems.

Actually, you don't have to modify the Lagrange equation at all. A non-conservative case is where an explicit external force (like friction) is introduced into the system. Here, you don't have to deal with an external force; you just need to deal with an internal constraint.

I guess I have to use some Lagrange multipliers but I'm not even sure.

Yes, Lagrange multipliers is the right approach to take. You've already figured out what your constraint should be: [itex]\theta ( t)= \Omega t + \theta _0[/itex]. The amazing thing about Lagrange multipliers is that the value of the multiplier gives you the constraint force. Here, your constraint forces the particle to remain on the rod, so the multiplier gives you the force that the rod exerts on the particle to make it stay on the rod.

Also, how do I prove that the energy is not conserved? Maybe deriving the Lagrangian with respect to time and if it's not worth 0 it means the Lagrangian is not constant and thus (I believe the implication is true), E isn't constant. Is this a good approach? Am I missing stuff?

Yes, that's the right approach. Noether's theorem says that if the Lagrangian is invariant under time translation, energy is conserved. Here, our constraint explicitly depends on time, so the Lagrangian is obviously not invariant.

 

[fluidistic]

Thanks for your reply, that's very helpful/encouraging.
I've been looking into Goldstein's book (1st edition) about Lagrange multipliers. I don't think I understand well, but here's what I've done:
The constraint equation is [itex]\theta - \Omega t - \theta _0 =0[/itex].
[itex]a_{\theta}=1[/itex].
Thus [itex]m(2 \dot r r \dot \theta + r^2 \ddot \theta )+mg r \cos \theta - \lambda=0[/itex].
I notice that [itex]\ddot \theta =0[/itex].
I also have that [itex]r=\frac{g \sin \theta}{\Omega ^2}, \dot r = \frac{g \Omega \cos (\theta )\Omega ^2}{\Omega ^4}=\frac{g \cos \theta }{\Omega }[/itex].
This gives me the equation [itex]\frac{mg^2 \sin \theta \cos \theta}{\Omega ^2 } \left ( \frac{2}{\Omega } +1 \right ) - \lambda=0[/itex].
I'm not sure I'm doing things right. I know I'm not finished but I don't know what to do next.

 

[ideasrule]

Thus [itex]m(2 \dot r r \dot \theta + r^2 \ddot \theta )+mg r \cos \theta - \lambda=0[/itex]. I notice that [itex]\ddot \theta =0[/itex].

Yup, that's right. Note that [itex]\lambda = 2m \dot r r \dot \theta+mg r \cos \theta[/itex]

I also have that [itex]r=\frac{g \sin \theta}{\Omega ^2}, \dot r = \frac{g \Omega \cos (\theta )\Omega ^2}{\Omega ^4}=\frac{g \cos \theta }{\Omega }[/itex].

That's pretty much what I got, except I have an extra factor of 1/2 for r.

This gives me the equation [itex]\frac{mg^2 \sin \theta \cos \theta}{\Omega ^2 } \left ( \frac{2}{\Omega } +1 \right ) - \lambda=0[/itex].
I'm not sure I'm doing things right. I know I'm not finished but I don't know what to do next.

Yup, you're definitely doing things right, and I think you are finished. You know how theta behaves--it's just wt+theta_0. You've already found the equation of motion for r. The constraint force is just lambda. Note that lambda is the sum of gravity and the Coriolis force, which is exactly what you'd expect with Newtonian mechanics.

 

[fluidistic]

Yup, that's right. Note that [itex]\lambda = 2m \dot r r \dot \theta+mg r \cos \theta[/itex]


That's pretty much what I got, except I have an extra factor of 1/2 for r.



Yup, you're definitely doing things right, and I think you are finished. You know how theta behaves--it's just wt+theta_0. You've already found the equation of motion for r. The constraint force is just lambda. Note that lambda is the sum of gravity and the Coriolis force, which is exactly what you'd expect with Newtonian mechanics.

Thank you very much for everything! Sorry for having took so much time to reply, I was busy with another course.
My Lagrangian is [itex]L=\frac{mr^2 \dot \theta ^2 }{2}-mgr \sin \theta[/itex]. When I do [itex]\frac{\partial L }{\partial r }[/itex] and I isolate r, I reach [itex]r = \frac{g \sin \theta }{\Omega ^2}[/itex]. May I know how you got your extra 1/2 factor?

 

[ideasrule]

My Lagrangian is [itex]L=\frac{mr^2 \dot \theta ^2 }{2}-mgr \sin \theta[/itex]. When I do [itex]\frac{\partial L }{\partial r }[/itex] and I isolate r, I reach [itex]r = \frac{g \sin \theta }{\Omega ^2}[/itex]. May I know how you got your extra 1/2 factor?

The Lagrangian should be [itex]L=\frac{mr^2 \dot \theta ^2 }{2}+\frac{m\dot r ^2}{2}-mgr \sin \theta[/itex]. You have to account for motion along the radial direction.

 

[fluidistic]

The Lagrangian should be [itex]L=\frac{mr^2 \dot \theta ^2 }{2}+\frac{m\dot r ^2}{2}-mgr \sin \theta[/itex]. You have to account for motion along the radial direction.

Ah I see, I understand.
Well it's a bit too late for me now (past 2 am) so I'm going to try tomorrow. It doesn't seem as easy as I thought to get r though. Lagrange's equation gives [itex]\ddot r - r \dot \theta ^2 +g \sin \theta =0[/itex]. I'll work on it and post here tomorrow.

 

[ideasrule]

Ah I see, I understand.
Well it's a bit too late for me now (past 2 am) so I'm going to try tomorrow. It doesn't seem as easy as I thought to get r though. Lagrange's equation gives [itex]\ddot r - r \dot \theta ^2 +g \sin \theta =0[/itex]. I'll work on it and post here tomorrow.

Yeah, that's why I was surprised that you got the answer without showing any work, considering that the rest of the solution was very detailed.

You could go through the motions of solving this second-order linear inhomogenous differential equation, or you could take a random guess and hope that it works. It turns that this "random guess" approach is more useful, especially since I already gave away the answer.

 

[fluidistic]

Yeah, that's why I was surprised that you got the answer without showing any work, considering that the rest of the solution was very detailed.

You could go through the motions of solving this second-order linear inhomogenous differential equation, or you could take a random guess and hope that it works. It turns that this "random guess" approach is more useful, especially since I already gave away the answer.

Sorry for being so late to thank you, but thanks anyway.
I'm going to take the second approach for now. :) If I have some problems I'll post here.