# Equation of motion for a rigid rotating rod

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
Mentor
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.

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

Doc Al
Mentor
Good. Now do it using the Lagrangian.

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

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
Mentor
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.

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

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
Mentor
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 Ib should be Irod 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?

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

Regards
Cole.