Equation of motion for a rigid rotating rod

[Cole.]

Hi all,

Homework Statement

My problem is a pretty basic one, in a exercise a rigid rod of mass M is rotating around a horizontal pivot point i one end. The rod has the length L. I now need to derive the equation of motion using the Lagrangian formalism.

My question is:
Can i view the rod as a pendulum with a bob mass equal to that of the rod?


In this case the equation would simply be:
$\theta '' + \sqrt{\dfrac{g}{L}} sin(\theta)=0$

Best Regards
Cole.

 

[Doc Al]

My question is:
Can i view the rod as a pendulum with a bob mass equal to that of the rod?

Not exactly. An extended mass such as a rod cannot be treated as a simple pendulum. Treat it as a physical (or compound) pendulum.

 

[Cole.]

Ach thanks your reference to compound pendulum got me on the right track.
It's all about the center of mass:
$I \theta '' = -M_{cm} g L_{cm} sin \theta<br /> \Leftrightarrow \theta '' = -\dfrac{3 g}{2 L} sin (\theta)$

 

[Doc Al]

Good. Now do it using the Lagrangian.

 

[Cole.]

Thanks for the quick reply, I am slightly confused as how to in corporate the moment of inertia for the rod. If I use the kinetic energy $T=\dfrac{1}{2}M r'^{2}$ and potential energy $U=-mg r \mathrm{cos} \theta$ there is no difference from the regular pendulum and i don't get the 3/2 factor!
Could I use $T=\dfrac{1}{2} I r'^{2}$ where I is the inertia? A hint is appreciated :)

Regards
Cole.

 

[Doc Al]

You must use the rotational KE = 1/2Iω², not 1/2Mv^2. (You'll express in terms of θ', of course.)

 

[Cole.]

Great i figured it out!
However I have run into a new problem. Suppose the rod has a bead of mass m attached at the pivot moving down is it rotates. I can't figure out what the kinetic energy of the system is.. Me guess is $T_{tot}=T_{rod}+T_{bead}=\dfrac{1}{2}\dot{\theta}^{2}\left( I_{rod}+I_{bead}\right) + \dfrac{1}{2}m\left( \dot{r}+r\dot{\theta}\right) ^2$ because the bead must contribute with kinetic energy from the radial speed. ($\bar{v}=v_{r}+v_{\theta}=\dot{r}+r\dot{\theta}$) However my equations of motion don't seem right.. Any hints? You've been more than helpful so far!

Regards
Cole.

 

[Doc Al]

The KE of the rod is the same. Careful how you write the KE of the bead: Don't count it twice and don't add perpendicular velocity components together before squaring them.

 

[Cole.]

Everything appears correct except for the velocity factor. When I expand it I get:
$\dfrac{1}{2}m\ddot{r}^{2}+m\dot{r}r\dot{\theta}+ \dfrac{1}{2} mr^{2}\dot{\theta}^{2}$
and when I use the part of the Legrender equation relevant for the energy I get:
$\dfrac{\mathrm{d}}{\mathrm{t}}\dfrac{\mathrm{d}L}{\mathrm{d}\dot{\theta}}=m\dot{r}r\ddot{\theta} +m r^{2}\ddot{\theta}$
I have deduced from another solution that the above result should simply be: $2mr \dot{r} \dot{\theta}$
Where am I going wrong?
Thanks a million for the help!

Regards
Cole.

 

[Doc Al]

Everything appears correct except for the velocity factor. When I expand it I get:
$\dfrac{1}{2}m\ddot{r}^{2}+m\dot{r}r\dot{\theta}+ \dfrac{1}{2} mr^{2}\dot{\theta}^{2}$

What did you expand to get this? If you are referring to the last term in the following, I'd say it was wrong:

Me guess is $T_{tot}=T_{rod}+T_{bead}=\dfrac{1}{2}\dot{\theta}^{2}\left( I_{rod}+I_{bead}\right) + \dfrac{1}{2}m\left( \dot{r}+r\dot{\theta}\right) ^2$

Reread my last post.

 

[Cole.]

Yes I am referring to the last term:
$\dfrac{1}{2}m( \dot{r}+r \dot{\theta} )^{2}=\dfrac{1}{2}m(\dot{r}^{2}+2r\dot{r} \dot{\theta}+r^{2}\theta ^{2})$
What i think you mean in the above is to say:
$T_{b}=\dfrac{1}{2}I_{b}\dot{\theta}+\dfrac{1}{2}m \dot{r} ^{2}+\dfrac{1}{2}mr^{2} \dot{\theta}^{2}$
but that doesn't either give me the expression I'm looking for..

 

[Doc Al]

What i think you mean in the above is to say:
$T_{b}=\dfrac{1}{2}I_{b}\dot{\theta}+\dfrac{1}{2}m \dot{r} ^{2}+\dfrac{1}{2}mr^{2} \dot{\theta}^{2}$

Yes, that's what I would put as the KE term. Except that I_b should be I_rod in the first term. And the $\dot{\theta}$ should be squared.

but that doesn't either give me the expression I'm looking for..

Why is that?

 

[Cole.]

Ahh but it does! Silly mistake from my part.
Thank you so much it has been a real journey :)

Regards
Cole.