Small oscillations of a simple pendulum placed on a moving block

[MatinSAR]

Homework Statement

A simple pendulum with mass $m$ and length $L$ is placed on a block with mass $M$. The block can move freely on a frictionless horizontal surface. Find the equation of motion and the frequency of small oscillations.

Relevant Equations

Hamilton's principle and Lagrangian equation.

Hello. This is the figure of the problem:

First, we should determine the Lagrangian of the system. I have already completed this part without any issues. To respect everyone’s time, I won’t go into the details of how I accomplished it.

$$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$ To find equation of motion I should directly put the above lagrangian in lagrange's equation? Or is it possible to first simplify it for small oscillations?

Edit 1: I forgot to mention that for small oscillations we can ignore the higher powers of theta and its derivatives.

 

[erobz]

Homework Statement: A simple pendulum with mass $m$ and length $L$ is placed on a block with mass $M$. The block can move freely on a frictionless horizontal surface. Find the equation of motion and the frequency of small oscillations.
Relevant Equations: Hamilton's principle and Lagrangian equation.

Hello. This is the figure of the problem:
View attachment 343055
First, we should determine the Lagrangian of the system. I have already completed this part without any issues. To respect everyone’s time, I won’t go into the details of how I accomplished it. View attachment 343056
$$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$ To find equation of motion I should directly put the above lagrangian in lagrange's equation? Or is it possible to first simplify it for small oscillations?

I think you would be able to substitute now for $\cos \theta$

 

[MatinSAR]

I think you would be able to substitute now for $\cos \theta$

I wanted to simplify it for small oscillations then substitute it in lagrange's equation. But I was not sure. So you don't see a problem in doing this?
I will try it and share results as soon as I can ... Thank you.

 

[erobz]

I wanted to simplify it for small oscillations then substitute it in lagrange's equation. But I was not sure. So you don't see a problem in doing this?
I will try it and share results as soon as I can ... Thank you.

I could be wrong though ( you're already getting out of my wheelhouse of comfort), maybe try it both ways and see if you come to agreement (sometimes doing it twice is the best way to convince yourself)?

 

[Orodruin]

Yes, you can expand the Lagrangian for small $\theta$ and then apply the EL equation, but if you want linear EoMs, you need to expand to second order in $\theta$.

 

[MatinSAR]

I've forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives.

I could be wrong though ( you're already getting out of my wheelhouse of comfort)

No problem. Thanks for your help.

maybe try it both ways and see if you come to agreement?

I will.

 

[MatinSAR]

Yes, you can expand the Lagrangian for small $\theta$ and then apply the EL equation, but if you want linear EoMs, you need to expand to second order in $\theta$.

Good news ...
I forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives according to the question.
So ... Can I ignore $\dot \theta$ and $\theta^2$ at first place?

 

[PeroK]

Good news ...
I forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives according to the question.
So ... Can I ignore $\dot \theta ^2$ and $\theta^2$ at first place?

Definitely not!!

 

[MatinSAR]

Definitely not!!

Then I should do a lot of math to find its equations of motion, sadly. Thanks, I'll try and post the results soon.

 

[kuruman]

Personally, I prefer to first obtain the equations of motion and then consider approximations only because I don't trust myself not to throw out the baby with the bath water.

 

[MatinSAR]

Lagrangian of the system is: $$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$
For coordiante $x$ we have:
$$\frac {\partial L} {\partial x}-\frac {d}{dt}\frac {\partial L} {\partial \dot x}=0 $$ $$\frac {\partial L} {\partial \dot x}=A$$ A is a constant. $$(M+m)\dot x+mL\dot \theta cos \theta=A$$$$\dot x = \dfrac {A-ml\dot \theta \cos \theta}{M+m}$$$$x=\dfrac {1}{M+m}(At-ml\sin \theta)$$
I'm not sure with last part. Did I solve for $x$ correctly? Many thanks.

Personally, I prefer to first obtain the equations of motion and then consider approximations only because I don't trust myself not to throw out the baby with the bath water.

You’re right. I just intended to solve the question faster and get to the rest of the tasks.

 

[Orodruin]

Good news ...
I forgot to mention that for small oscillations we can also ignore the higher powers of theta and its derivatives according to the question.
So ... Can I ignore $\dot \theta$ and $\theta^2$ at first place?

Huh? Definitely not!

As I said you need to keep terms up to second order in the variable in the Lagrangian to obtain the linearised equations of motion. $\theta^2$ is second order so you cannot ignore it.

One thing to note about your Lagrangian is that it can be easily separated into a translational part and an oscillatory part. This makes the actual problem one-dimensional instead and quite easy to handle.

 

[kuruman]

I'm not sure with last part. Did I solve for $x$ correctly?

It doesn't look right. $\theta$ and $\dot{\theta}$ are both functions of time and you don't know what they look like, in fact you are looking for $\theta(t)$.

 

[MatinSAR]

One thing to note about your Lagrangian is that it can be easily separated into a translational part and an oscillatory part. This makes the actual problem one-dimensional instead and quite easy to handle.

I think this way is hard for me. I'm not familiar with oscillation that much. So I prefer to find equation of motion using Lagrange equations then simplify it the way that was mentioned in question.

It doesn't look right. $\theta$ and $\dot{\theta}$ are both functions of time and you don't know what they look like, in fact you are looking for $\theta(t)$.

Cant I wrote that as $d \theta / dt$? Then I can simplify integral to $cos \theta d\theta$.

I've seen the edit just now. Can I use the other equation that I can find from lagrangian for $\theta$ coordinate?

 

[kuruman]

That you can do to get $\int \cos(\theta(t))d\theta.$ Then what? You have an implicit function of time under the integral sign that you don't know what it looks like.

I thnk @Orodruin's suggestion is worth looking into. Note that the center of mass stays fixed in the horizontla direction. I would try defining new generalized coordinates $$X=\frac{Mx+mL\sin\theta}{M+m}~;~~\xi=\frac{Mx-mL\sin\theta}{M+m}$$and see what happens. Note that $X$ describes the translatinal motion of the CM, while $\xi$ the departure from the equilibrium position. It's more work, but it should decouple the equations of motion.

 

[Orodruin]

That you can do to get $\int \cos(\theta(t))d\theta.$ Then what? You have an implicit function of time under the integral sign that you don't know what it looks like.

I thnk @Orodruin's suggestion is worth looking into. Note that the center of mass stays fixed in the horizontla direction. I would try defining new generalized coordinates $$X=\frac{Mx+mL\sin\theta}{M+m}~;~~\xi=\frac{Mx-mL\sin\theta}{M+m}$$and see what happens. Note that $X$ describes the translatinal motion of the CM, while $\xi$ the departure from the equilibrium position. It's more work, but it should decouple the equations of motion.

Effectively, I would just do it … effectively …

Use the constant of motion from translational invariance to eliminate $\dot x$ from the constant of motion from time translation invariance (ie, energy). Then you are basically done.

 

[MatinSAR]

That you can do to get $\int \cos(\theta(t))d\theta.$ Then what? You have an implicit function of time under the integral sign that you don't know what it looks like.

I thnk @Orodruin's suggestion is worth looking into. Note that the center of mass stays fixed in the horizontla direction. I would try defining new generalized coordinates $$X=\frac{Mx+mL\sin\theta}{M+m}~;~~\xi=\frac{Mx-mL\sin\theta}{M+m}$$and see what happens. Note that $X$ describes the translatinal motion of the CM, while $\xi$ the departure from the equilibrium position. It's more work, but it should decouple the equations of motion.

So I cannot use $x$ and $\theta$ to solve, right? How did you find those generalized coordinates? By guessing?
Is there any other way?

 

[kuruman]

Not really guessing. Just having seen this sort of problem before and some common sense. In this case, you note that the CM of the system does not accelerate horizontally because there are no horizontal external forces acting on the parts making it up.

So if one of the generalized coordinates is the position of the CM, its equation of motion should be $\ddot X =0$ and easy to solve. Furthermore, simple harmonic motion is expected to some approximation. It follows that the other generalized coordinate should be the departure from the equilibrium position in the horizontal direction. You should expect its equation of motion to be $\ddot{ \xi}+\omega^2~\xi=0$.

Similar considerations apply to the cousin of this problem, namely the block of mass $m$ sliding down on a frictionless wedge of mass $M$ placed on a frictionless horizontal surface.

 

[Orodruin]

Is there any other way?

Sure:

Use the constant of motion from translational invariance to eliminate x˙ from the constant of motion from time translation invariance (ie, energy). Then you are basically done.

 

[MatinSAR]

Use the constant of motion from translational invariance to eliminate $\dot x$ from the constant of motion from time translation invariance (ie, energy). Then you are basically done.

Do you mean that I can use conservation of momentum $\vec P$ in horizontal direction?(Or using conservation of $T+U$?)
I'm going to try this. Hope it works ...

Not really guessing. Just having seen this sort of problem before and some common sense. In this case, you note that the CM of the system does not accelerate horizontally because there are no horizontal external forces acting on the parts making it up.

So if one of the generalized coordinates is the position of the CM, its equation of motion should be $\ddot X =0$ and easy to solve. Furthermore, simple harmonic motion is expected to some approximation. It follows that the other generalized coordinate should be the departure from the equilibrium position in the horizontal direction. You should expect its equation of motion to be $\ddot{ \xi}+\omega^2~\xi=0$.

Similar considerations apply to the cousin of this problem, namely the block of mass $m$ sliding down on a frictionless wedge of mass $M$ placed on a frictionless horizontal surface.

Thanks again. However, this method is complicated for me. Given that there’s a chance my teacher might ask me to explain what I’ve written, I’d prefer to use a method that’s more understandable to me.

 

[MatinSAR]

@Orodruin @kuruman
This is my attempt at solving the problem. To save time, please skip the calculations. I just need help with the last part because I’m unsure if this method works.
Lagrangian of the system is given by: $$L=\dfrac {M+m}{2}\dot x^2+ml\dot x \dot \theta \cos \theta + \dfrac 1 2 m l^2 \dot \theta^2 +mgl\cos \theta $$ For coordiante $x$ we have: $$\frac {\partial L} {\partial x}-\frac {d}{dt}\frac {\partial L} {\partial \dot x}=0 $$ $$\frac {\partial L} {\partial \dot x}=A$$ A is a constant. $$(M+m)\dot x+ml\dot \theta cos \theta=A$$
For coordiante $\theta$ we have :$$\dfrac {\partial L}{\partial \theta}=-ml \dot x \dot \theta \sin \theta - mgl \sin \theta$$ $$\dfrac {\partial L}{\partial \dot \theta}=ml\dot x \cos \theta+ml^2 \dot \theta$$ $$\frac {d}{dt} \dfrac {\partial L}{\partial \dot \theta}=ml\ddot x \cos \theta -ml\dot x \dot \theta \sin \theta +ml^2 \ddot \theta$$ Using lagrange equation I get: $$-mgl \sin \theta -ml\ddot x \cos \theta - ml^2 \ddot \theta=0$$

I have two equations now: $$(M+m)\dot x+ml\dot \theta cos \theta=A$$$$-g \sin \theta -\ddot x \cos \theta - l \ddot \theta=0$$ Can I find $\theta (t)$ from the second equation and substitute it into the first equation, then solve for $x(t)$?

Edit 1: I've just fixed a big problem in my solution. $\ddot x$ is not necessarily $0$.

 

[erobz]

I have two equations now: $$(M+m)\dot x+ml\dot \theta cos \theta=A$$$$\ddot \theta + \dfrac {g}{l} \sin \theta =0$$ Can I find $\theta (t)$ from the second equation and substitute it into the first equation, then solve for $x(t)$?

What makes you think $\ddot x = 0$?

When I apply conservation of momentum in the $x$ direction I'm not finding that $\ddot x = 0$. Maybe I'm messing that up. What do you get when you apply conservation of momentum?

 

[MatinSAR]

What makes you think $\ddot x = 0$?

When I apply conservation of momentum in the $x$ direction I'm not finding that $\ddot x = 0$. Maybe I'm messing that up. What do you get when you apply C.o.M.?

Sorry. I've just edited that.

 

[erobz]

Sorry. I've just edited that.

I'm getting this from conservation of momentum:

$$M \dot x - m( \dot x - l \dot \theta \cos \theta ) = 0 $$

$$ \implies \dot x = -\frac{m}{M-m} l \dot \theta \cos \theta $$

That doesn't seem to be equivalent to your expression?

 

[MatinSAR]

I'm getting this from conservation of momentum:

$$M \dot x - m( \dot x - l \dot \theta \cos \theta ) = 0 $$

$$ \implies \dot x = -\frac{m}{M-m} l \dot \theta \cos \theta $$

That doesn't seem to be equivalent to your expression?

I've used lagrange equation. You've used conservation. Should our equations look the same? Or they should be different to help us find $x(t)$?

 

[erobz]

I've used lagrange equation. You've used conservation. Should our equations look the same? Or they should be different to help us find $x(t)$?

I think $\dot x$ is $\dot x$; if they don't agree, we have a problem.

 

[MatinSAR]

I think $\dot x$ is $\dot x$; if they don't agree, we have a problem.

So there should be a problem with the equation I derived from Lagrange equation. I will try to do it again.

 

[erobz]

In the inertial frame the little mass is swinging to the left faster than the large block is moving to the right. Does my approach with the signs make sense to you?

 

[MatinSAR]

In the inertial frame the little mass is swinging to the left faster than the large block is moving to the right. Does my approach with the signs make sense to you?

I think that $m \dot x$ should be positive. Because velocity of the mass $m$ is given by its velocity in block M frame + velocity of block M in inertial refrence frame.

 

[erobz]

Ok, so you are saying with $\dot \theta < 0$ we should have this equation:

$$ M \dot x + m( \dot x + m l \dot \theta \cos \theta ) = 0 $$

Which gives:

$$ \dot x = -\frac{ml}{M+m} \dot \theta \cos \theta $$

I can buy that.