Lagrangian Equation of motion for rod on pivot in gravitational field

[K29]

Homework Statement

I am trying to get an equation of motion for the following (seemingly simple) setup. You place on a rod on a pivot. The rod's centre of mass is precisely over the pivot. Think of balancing a ruler horizontally on your finger. Gravity, of course acts downward.

The Attempt at a Solution

The equation of motion is surely not obtained by working out the Lagrangian using the Kinetic energy as $$T=\frac{1}{2}I\omega^{2}$$. where $$\omega = \dot{\alpha}$$ Alpha is just the angle rotated from the x-axis which I aligned along the rod when its horizontally at rest equilibrium. Clearly there is no potential energy to speak of here since the centre of mass is fixed.
After solving Lagrange's equation of motion it can quickly be seen that:
$$\ddot{\alpha}=0$$ which just can't be the case, since if you think about balancing a ruler on your finger, if you push one end down slightly it oscillates.
Thus I suppose the right form of Lagranges equation to be used is
$$\frac{d}{dt}\frac{\partial T}{\partial \dot{\alpha}}-\frac{\partial T}{\partial\alpha}=Q_{\alpha}$$
where Q is the generalised component of force.
But I can't seem to figure out how to formulate a Force of constraint s.t. $$Q_{\alpha}$$ comes out. As you lower one side of the rod gravity acts on the centre of masses of both sides such that the net force on one side is greater than the other causing it to oscillate about the horizontal axis? Am I allowed to consider the two halves of the rod having their own centres of mass? (I have tried this and still get an answer that doesn't match reality)
This is probably not even the right way of doing things.

 

[JesseC]

Is this not identical to the problem of a simple pendulum? Just upside down.

Image the rod to be squashed to a point mass at the position of the centre of mass. It would be as if a pendulum was suspended in the air, the length of the rigid 'string' would be half the length of the rod. (assuming constant mass per unit length)

You don't need to worry about the initial conditions of the system (the fact it is upright) until you actually come to solve the equation of motion.

I would also argue that there HAS to be potential energy. If there was none, and you perturbed this system slightly (gave it a tiny kick) how would it gain kinetic energy without there being potential energy to lose?

 

[K29]

I don't think that's quite it.The setup is much simpler, (yet I'm really having trouble with it) I should have been a bit clearer. If the rod is of length L, and you let it lie on the horizontal plane, you lower it onto a pivot at L/2 so that it balances perfectly like this:
_______________________________________
|___________________O___________________| <----rod

At the centre of the rod is the pivot point. The rod is uniform.
I don't actually even understand why if I were to push down slightly on the left hand side it would oscillate (and if there is air resistance, return to the horizontal equilibrium position)

I have no problem at all getting the equation of motion if I shift the centre of mass to ANY point to the left or right of the pivot. All the textbook problems are like that, and it is easy to solve. But as soon as we have this horizontal equilibrium and give it a nudge... well I'm lost as to understanding why it oscillates about the horizontal axis(it does do this right?) let alone how to use Lagrange's equations correctly for this case to get the eqn of motion

 

[JesseC]

Ahhh i see what you mean now.

Have you tried treating it as two point masses at the X's?

_______________________________________
|________X__________O_________X_________| <----rod

The point masses being connected by a massless rod, fixed at their combined centre of mass?

You'd need two generalized coordinates to describe the position of the two masses.

Edit: Heard of rigid rotor? this problem seems somewhat similar to a rigid rotor but in a gravitational potential.

 

[K29]

Thanks for the advice. I've tried that. My equation of motion ends up looking like two rods individually hanging from the pivot.In other words they both end up hanging like single rod-pendulums.
There is some kind of constraint due to the rod's "other half" that keeps it "up"
Assuming we can "split" the rod into two "half-rods" this is what happens.
If I write the kinetic energy of the right "half-rod" in terms of the moment of inertia around the pivot
Lets say the "half-rod's" mass is $$m=M/2$$ and length is $$l=L/2$$ where M and L are the masses and lengths of the entire rod.
We get for the Kinetic energy:
$$T=\frac{1}{2}I\dot{\alpha}^{2}.$$
(The kinetic energy is pure rotation around the centre point(no translation))
In the above equaition$$I=\frac{1}{3}ml^{2}$$ for a rod rotating about it's end point, which in this case is the pivot point.

I have taken alpha to be an angle off of the horizontal axis, and rotates anti-clockwise(yes I am being a bit pedantic)
The gravitational potential energy would then be (If the potential is set at zero at the plane where if l hangs vertically down it just touches)
$$V=mg(l+\frac{l}{2}sin\alpha)$$
(Potential is defined by the height of the centre of mass of a rigid body. In this casse we are looking at the centre of mass of the right-half of the rod)

Using $$L=T-V$$
and $$\frac{d}{dt} \frac{\partial L}{\partial \dot{\alpha}} = \frac{\partial L}{\alpha}$$
We get
$$\frac{1}{3}ml^{2}\ddot{\alpha}=mg\frac{l}{2}cos\alpha$$

This is the correct equation of motion for a uniform rod that is swinging like a pendulum(If you look at other examples you will see a sin instead of a cos. This is because their generalised co-ordinate was alpha taken off of a vertical axis, not a horizontal one, ).
Remember how I positioned $$\alpha$$. So when it is horizontal $$cos(\alpha)=cos(0)=1$$
and $$\ddot{\alpha}$$ appears.
So the left "half-rod" will have a similar equation of motion, which we can write in terms of [tex]\alpha[\tex] too. It will just pick up a negative.
I need to introduce a force of constraint or something due to the "other end" of the rod on the other side of the pivot that keeps it up in its horizontal equilibruim or oscillating about the horizontal axis.
Problem is the only force I have available is gravity. And that is acting right through the centre of the pivot

 

[K29]

I suppose if I'm going to get help from a Homework Helper I should put down my attempt considering the entire rod.
The rod is of mass M and length L
$$I=\frac{1}{12}ML^{2}$$ for a rod rotating about its centre of mass.
The kinetic energy is
$$T=\frac{1}{2}I\dot{\alpha}^{2}$$ where alpha is the same generalised co-ordinate as in the previous post (anti-clockwise off the horizontal axis)
Potential energy is at then centre nof mass of the rigid body. This is at the pivot point and never changes, so w emay as well set it to zero there.
Thus $$L=T$$
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{\alpha}}=\frac{\partial L}{\partial \alpha}$$
$$\frac{1}{12}ML^{2}\ddot{\alpha}=0$$
$$\ddot{\alpha}=0$$
This implies that there is never an angular acceleration. This is impossible, because if you take the rod out of its horizontal equilibruim position it oscillates around it (I think)

 

[JesseC]

Did post an answer last night when I was a bit tired... messed it up and removed it now. I fully see your problem... currently thinking about it!

I would add... you seem certain you should expect an SHM solution? You're right that gravity is acting through the center of mass, therefore it cannot exert a torque or restoring force. If gravity can't do it and there are no other forces, then it shouldn't oscillate! Your initial assertion that gravity has no effect may indeed be correct!

The balanced ruler on your finger analogy fails because there is too much friction involved, both air resistance and friction with your finger, i think this accounts for the oscillations that occur.

You seem convinced that your answer of the form:

$$(constants) \ddot{\alpha} = (constants- or- zero)$$

is wrong. I don't see why it's wrong... why does the system (under perfect conditions) have to oscillate?

 

[K29]

Firstly I want to refine the problem by explicitly stating that the pivit goes directly through the centre of mass. Although the ruler is still a good approximation to this since the centre of mass will sit above the pivot for small angles [and even big angles](unless its really really small pivot point), and friction prevents it from slipping off when the centre of mass moves. But let's completely drop the idea of the ruler.

We just have a rod, supported on a frictionless pivot, in a uniform gravitational field acting downwards, with no air resistance. Here is why I think the system should oscillate:
Fowles and Cassiday p467: "Intuition tells us that the potential energy must be a minimum in all cases for stable equilibrium. That this is so can be argued from energy consiederations. If the system is conservative, the total energy T+V is constant, so for a small change near equilibrium $$\Delta T = -\Delta V$$. Thus, T decreases if V increases; that is, the motion tends to slow down and return to the equilibrium position, given a small displacement. The reverse is true if the potential enegry is a maximum"

So my argumant for oscillation of our system:

Now if we assume there is no oscillations for the rod setup then $$\ddot{a}=0$$ or some constant. This implies that if the rod is stationary then every angle that the rod is at is a stable equilibrium. Then it is at a minimum potential at every possible postion. Now let's give the rod a a nudge, giving it Kinetic energy. From our energy conservation, the potential energy must now decrease. But this is impossible because the potential was at a minimum. Contradiction. Thus the system must oscillate. qed.
(This argment of course if for a conservative system. In our argument the fact that a constant minimum potetial exists everywhere implies that the system must be conservative.)
!If I applied itthe argument to an object in free space, under no gravitational field, we know that the angular accelaration is also a constant. However here, no potential function is defined in the free space set-up. My above argument assumes that a potential exists. Does a potential function exist?

Also, its easy to see that if you balance an object centre of mass with its centre of mass over a pivot in a gravitational field in reality, if it is stable, it MUST sit horizontally. You just can't balance a uniform symmmetrical object in reality without doing it in a horizontal position. In any other position you feel a strange torque pushing it towards this equilibrium.

I have a feeling the potential is not actually a gravitational potential. If you add up the torques on the left and righthand sides of the rod due to the gravitational components acting normal to the rod, the net sum is zero. The potential seems to come out of nowhere. A And of course increases on either side of the horizontal equilibrium, whereas gravitational potential would do this about a vertical equilibrium. I've been thinking about this and in my head(havent't actually written this down yet) I can get the right form of the equation of motion if I construct such a potential. It would then depend only on displacement of the rod. But it could have other dependances. It matters because the potential won't vanish in the equation of motion because it depends on alpha. Since I'm pulling it out of you know where, I would have to set out on a great theoretical and experimantal conquest to prove what the potential function is.

Alternatively, no potential function exists. Then we can safely allow the acceleration to be constant and no oscillation occurs and every angle is a nstable equilibrium. This completely tightens up the theory, without any contradictions. However, this would not explain what happens in reality

 

[JesseC]

That is an impressive discussion, you've really gone to a lot of trouble to understand this problem which is commendable.

Clearly we are talking about an ideal system, a purely 1 dimensional rod of uniform mass per unit length. A pivot placed exactly at its centre of mass, the mid point. So I agree let's forget about real life analogues for the moment because they are not ideal enough.

You say:

Now let's give the rod a a nudge, giving it Kinetic energy. From our energy conservation, the potential energy must now decrease.

I strongly disagree with this. Imagine a box on a frictionless horizontal plane. If we give this box a small nudge, it gains kinetic energy yes but does it lose an potential energy? By Newtons 1st law the box will continue to move with the energy we've given it until something else acts on it.

How is this related? Well in the case of the box, the centre of mass (COM) of the box moves only tangentially to the action of the potential. Think of the gravitational potential as acting purely in the z direction, normal the the x-y plane on which the box is sliding. Because the COM of the box does not move in the z direction it cannot gain or lose potential energy.

You could even think of the Lagrangian of the box, it will have no potential energy term as long as it is constrained to move on the horizontal plane. I feel it is not even necessary to get into Lagrangian dynamics to analyse this system!

The same is true for our rod on a pivot. You yourself said, at any angle of the rod the net torque is zero, and I agree. The COM is constrained to lie at constant z, thus the potential energy of the system cannot change and you cannot get oscillatory motion.

I would argue that if you could build a good enough rod and pivot in reality, you'd see that it is stable at any angle and does not oscillate.

 

[K29]

Alright, I have everything worked out and understood, from the theory to the reality.
Firstly my arguments in the last post I made are actually completely nulled. The net torque on the rod is zero. Thus this is free-motion. I shouldn't be looking for some fictitious torque. As a result of this we have a constant angular velocity (or a rest) and all energy considerations are fine. As you said, it will be stable at any angle and will not oscillate.

As for reality. When you supposedly "pivot" a ruler on your finger, what you are actually doing with a little help from friction, is lifting up the centre of mass into the air. You are rolling the ruler around your finger, not pivoting it. Of course gravity pulls back down on the center of mass and you get the oscillations. :D

Thanks for taking an interest and for all the calculations and input! They did help guide my thinking and keep me sane.

 

[JesseC]

Glad I could be of use!