Normal Modes - Pendulum on a Moving Block

[Plutoniummatt]

Homework Statement

A block of mass M can move along a smooth horizontal track. Hanging from the
block is a mass m on a light rod of length l that is free to move in a vertical plane
that includes the line of motion of the block. Find the frequency and displacement
patterns of the normal modes of oscillation of the system firstly by `spotting' the
normal modes of the system and

then secondly by writing the Lagrangian L = T - U
for the system and solving the Euler-Lagrange equations.

Homework Equations

[tex]\frac{d}{dt}\frac{\delta L}{\delta\dot{q}_i} = \frac{\delta L}{\delta\dot{q}}[/tex]

The Attempt at a Solution

Mode 1- Translation
Mode 2- Pendulum swings, at the same time the Block also oscillates from side to side?

Kinetic Energy:

[tex]\frac{1}{2} M\dot{x}^2 + \frac{1}{2}ml^2\dot{\theta}^2[/tex]

Potential Energy:

[tex]mg(l - l cos\theta)[/tex]

Then write down the Lagrangian as L = T-U

applying the Euler-Lagrange equations for variables [tex]x[/tex] and [tex]\theta[/tex]

I get [tex]M\dot{x} = 0[/tex]

and for the [tex]\theta[/tex] coordinate, I just get the trivial pendulum equation...with [tex]\omega = \sqrt{g/l}[/tex]


Is that it? because this doesn't seem to have yielded me the answer they're looking for I guess...

 

[Gokul43201]

Homework Statement

A block of mass M can move along a smooth horizontal track. Hanging from the
block is a mass m on a light rod of length l that is free to move in a vertical plane
that includes the line of motion of the block. Find the frequency and displacement
patterns of the normal modes of oscillation of the system firstly by `spotting' the
normal modes of the system and

then secondly by writing the Lagrangian L = T - U
for the system and solving the Euler-Lagrange equations.

Homework Equations

[tex]\frac{d}{dt}\frac{\delta L}{\delta\dot{q}_i} = \frac{\delta L}{\delta\dot{q}}[/tex]

The Attempt at a Solution

Mode 1- Translation
Mode 2- Pendulum swings, at the same time the Block also oscillates from side to side?

Kinetic Energy:

[tex]\frac{1}{2} M\dot{x}^2 + \frac{1}{2}ml^2\dot{\theta}^2[/tex]

Is that all? Aren't you missing a term here (for the hanging mass)?

 

[Plutoniummatt]

Is that all? Aren't you missing a term here (for the hanging mass)?

am I?

isnt that taken into account by the second term?

 

[Plutoniummatt]

Is that all? Aren't you missing a term here (for the hanging mass)?

aha! you mean by the translation of the hanging mass as well as the swinging right?

 

[vela]

You've omitted the kinetic energy of the pendulum due to translation, for one thing. One approach to obtaining the Lagrangian is to write down the position of each mass, differentiate it with respect to time to get the velocity of the mass, and use that result to express the kinetic energy. For instance, the horizontal position of the mass m is equal to, using your symbols, [itex]x+l \sin\theta[/itex]. The horizontal component of its velocity is then [tex]\dot{x}+(l \cos\theta) \dot{\theta}[/tex]. You can already see the kinetic energy of mass m will have contributions from the pendulum's motion, from the whole system translating, and from a cross term.

 

[Plutoniummatt]

You've omitted the kinetic energy of the pendulum due to translation, for one thing. One approach to obtaining the Lagrangian is to write down the position of each mass, differentiate it with respect to time to get the velocity of the mass, and use that result to express the kinetic energy. For instance, the horizontal position of the mass m is equal to, using your symbols, [itex]x+l \sin\theta[/itex]. The horizontal component of its velocity is then [tex]\dot{x}+(l \cos\theta) \dot{\theta}[/tex]. You can already see the kinetic energy of mass m will have contributions from the pendulum's motion, from the whole system translating, and from a cross term.

can I write the KE as:

[tex]\frac{1}{2} m\dot{x}^2 + \frac{1}{2} ml^2\dot{\theta}^2[/tex]

or do I have to write it as x and y components?

 

[vela]

When you include all the components, it'll reduce down to those two components plus, it appears, another cross term. That cross term is important because it corresponds to the interaction between system's translational motion and the oscillatory motion of the pendulum.

 

[Plutoniummatt]

When you include all the components, it'll reduce down to those two components plus, it appears, another cross term. That cross term is important because it corresponds to the interaction between system's translational motion and the oscillatory motion of the pendulum.

After everything, I got:

[tex]M \ddot{x} + m \ddot{x} + ml \ddot{\theta}cos \theta - ml\dot{\theta}^2 sin\theta = 0[/tex]

[tex]m \ddot{x} lcos\theta - m \dot{x} l \dot{\theta} sin \theta + ml^2 \ddot{\theta} - mglsin \theta = 0[/tex]


how the heck can I solve this?

 

[vela]

I haven't actually worked it out, so maybe there's a better way, but this is what I'd try: One of the equations should have been d/dt(something)=0, so the something is a constant. You can use that to eliminate [itex]\dot{x}[/itex] from the other equation, leaving only [itex]\theta[/itex]s. You also want to assume [itex]\theta[/itex] is small, so you can use the small-angle approximations for sine and cosine.