Understanding the set up of a lagrangian problem

[CAF123]

Homework Statement

A rod of length ##L## and mass ##M## is constrained to move in a vertical plane.
The upper end of the rod slides freely along a horizontal wire. Let ##x## be the
distance of the upper end of the rod from a fixed point, and let ##\theta## be the angle
between the rod and the downward vertical.
Show that the Lagrangian is $$L = \frac{1}{2}M \left(\dot{x}^2 + \frac{1}{3}L^2 \dot{\theta}^2 + L\dot{x}\dot{\theta} \cos \theta \right) + \frac{1}{2}MgL \cos \theta $$

Homework Equations

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L = T - V

The Attempt at a Solution

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I really can't make much sense out of the problem statement. I am picturing a rod fixed at the origin say, rotating in a vertical plane, thereby producing a cone shaped surface. I take 'upper part of rod' to mean like a bead attached to the rod, but not sure what it means to say it slides freely along a horizontal wire?

And the first term of the lagrangian is the kinetic energy of this 'bead' relative to the fixed point, but it gives it a mass M which I have understood to be the mass of the whole rod.

Many thanks for any clarity here!

 

[Orodruin]

Try to express the movement (both rotational and translational) of the rod in terms of the coordinates.

The first term is relatively easy to get correctly (just consider how the rod is moving if theta is constant and x changing), as is the term including only ##\dot\theta## (if x is fixed, the rod is just rotating around the endpoint). The really tough contribution is the cross term.

 

[CAF123]

Hi Orodruin, thanks for reply

The first term is relatively easy to get correctly (just consider how the rod is moving if theta is constant and x changing)

##x## is the coordinate of the upper part of the rod relative to the fixed point. At time ##t## its position is therefore ##x(t)## and its velocity ## \dot{x}(t)##. The kinetic energy at this instant is then ##\frac{1}{2} m \dot{x}(t)^2##. But I don't see why the m there is M, the mass of the rod. My difficulty I think is I am misinterpreting the problem statement.

as is the term including only ##\dot\theta## (if x is fixed, the rod is just rotating around the endpoint)

Yes, this term is clear. It is the rotational energy of the rod about the fixed point.

The really tough contribution is the cross term.

I tried reexpressing the position of the upper part in terms of cylindrical coordinates so that ##\mathbf{r} = x \cos (\theta - \pi/2) \hat{r} + x \sin (\theta - pi/2) \hat z## and then maybe computing the time derivative and squaring it.

 

[Orodruin]

There is no small m, only the mass M of the rod, which will have rotational and translational energy.

If you keep theta fixed and change x, the entire rod is being translated.

 

[shreyaarya]

Consider that when the rod moves along the horizontal wire(neglecting the rotational degree of freedom,for the moment) then the whole rod is being translated.Of course,the expected mass is definitely M.
Alternatively,to set up the Lagrangian-you could write the center of mass coordinates of the rod,take the time derivative and square the terms and also add the piece corresponding to the rotation about the center of mass.
You could do this using cartesian coordinates itself.

 

[CAF123]

Thanks to both. As I thought, I was clearly misinterpreting the problem and took the 'upper end of the rod to slide along a horizontal wire' to mean a bead or something moving along the wire. I have the correct answer now, thanks!

 

[CAF123]

I was wondering if I could get some help with the second part of the problem?

The rod is given an initial angular velocity ##\omega## when hanging vertically at rest. Show that the maximum angle, ##\theta_M##, through which it can rise, satisfies $$ \sin^2 \left(\frac{\theta_M}{2}\right) = \frac{\omega^2 L}{24g}$$

Attempt: The total energy of the system is $$E = T + V = \frac{1}{2}M \left( \dot{x}^2 + \frac{1}{3}L^2 \dot{\theta}^2 + L \dot{x} \dot{\theta} \cos \theta \right) - \frac{1}{2}Mg L \cos \theta .$$ Since the lagrangian is not a function of x explicitly, momentum in x direction is conserved, ##M \dot{x} = \text{const}## so ##\dot{x}## is a constant. The initial conditions given in ##\left\{\theta, \dot{\theta}\right\}## and given that ##\dot{x}## is constant should mean that the above equation can be solved for ##\theta## and hence maximised to get ##\theta_M##.

The initial conditions are ##\dot{\theta}(t=0) = \omega, \theta(t=0) = 0## so the energy is $$E = \frac{1}{2} M \left( \left(\frac{A}{M}\right)^2 + \frac{1}{3} L^2 \omega^2 + \frac{L}{M} A \omega \right) - \frac{MgL}{2},$$ where ##A## is the const above. Equate this to the above expression for ##E## and rearrange gives a second order non linear equation for ##\dot{\theta}##. The method seems sound, but the result at the end seems analytically not tractable.

Thanks!

 

[Orodruin]

Since the lagrangian is not a function of x explicitly, momentum in x direction is conserved, ##M \dot{x} = \text{const}## so ##\dot{x}## is a constant.

This is not true. While the Lagrangian does not depend on x explicitly, this only means that the canonical momentum
$$
p = \frac{\partial \mathcal L}{\partial \dot x} = M\dot x + \frac M2 L \dot\theta \cos\theta
$$
is constant. See if you can take it from there.

 

[CAF123]

This is not true. While the Lagrangian does not depend on x explicitly, this only means that the canonical momentum
$$
p = \frac{\partial \mathcal L}{\partial \dot x} = M\dot x + \frac M2 L \dot\theta \cos\theta
$$
is constant. See if you can take it from there.

Yes, I got it. At ##\theta_{\text{max}}##, ##\dot{\theta} = 0##, so it is easily solvable. So no need to deal with solving a second order non linear equation, thanks!