Pendulum problem using Lagrangian approach

[Luminous Blob]

Hi,

I'm trying to do a problem that goes something like this:

There is a mass (m) attached to one end of a massless rod (length l). The other end of the rod is attached to a frictionless pivot. The rod is released from rest at an angle F0 < pi/2. At what angle F does the force in the rod change from compressive to tension?

Attached is a diagram of the situation, as well as what I have done so far. I apologize for not putting my working out in the body of my post, but I haven't used LaTex before, and I don't have the time at the moment to learn how to use it.

Another thing: The problem should be done using a Lagrangian approach (although it isn't entirely necessary).

Any help would be highly appreciated!

[Luminous Blob]

Oops, forgot to attach the file. And now that I've tried, I realise that I don't know how. The "Manage Attachments" button under "Additional Options" doesn't seem to do anything, and I can't see any other button/link for attachments. The FAQ says:

"To attach a file to a new post, simply click the [Browse] button at the bottom of the post composition page, and locate the file that you want to attach from your local hard drive."

Only problem is, I can't find any "Browse" button.

I know it's probably right in front of me, but I can't see it anywhere. Can anyone help me out here? My "Posting Rules" thing does say that I may post attachments.

Edit: Here is the attachment. I just realised I have my browser configured to block popups. I know, I am an idiot :)

[arildno]

I can't find the attachment, but I would assume that the angle is measured in a right-handed system with \frac{\pi}{2} the direction of the upwards vertical (antiparallell to the direction of the force of gravity)

Since the trajectory is circular, we may write:
\vec{r}(t)=l\vec{i}_{r},\vec{i}_{r}(t)=\cos\theta( t)\vec{i}+\sin\theta(t)\vec{j}
The acceleration may then be written as:
\vec{a}=-l{\omega}^{2}\vec{i}_{r}+l\dot{\omega}\vec{i}_{\th eta},\omega=\dot{\theta}

Newton's laws of motion may then be written as:
T-mg\sin\theta=-ml\omega^{2} (radial component)
-mg\cos\theta=ml\dot{\omega} (transverse component)

Multiplying the last equation by omega, and integrating, yields:
-g(\sin\theta(t)-\sin\theta_{0})=\frac{l\omega(t)^{2}}{2}\rightarro w\omega(t)^{2}=\frac{2g}{l}(\sin\theta_{0}-\sin\theta(t))

Clearly, T changes from compression to tension (at \theta=\hat{\theta})when it is zero:
2\sin\theta_{0}=3\sin\hat{\theta}\rightarrow
\hat{\theta}=\sin^{-1}\frac{2}{3}\sin\theta_{0}

[arildno]

Major error in first reply edited.

[Luminous Blob]

I can't find the attachment, but I would assume that the angle is measured in a right-handed system with \frac{\pi}{2} the direction of the upwards vertical (antiparallell to the direction of the force of gravity)

Since the trajectory is circular, we may write:
\vec{r}(t)=l\vec{i}_{r},\vec{i}_{r}(t)=\cos\theta( t)\vec{i}+\sin\theta(t)\vec{j}
The acceleration may then be written as:
\vec{a}=-l{\omega}^{2}\vec{i}_{r}+l\dot{\omega}\vec{i}_{\th eta},\omega=\dot{\theta}

Newton's laws of motion may then be written as:
T-mg\sin\theta=-ml\omega^{2} (radial component)
-mg\cos\theta=ml\dot{\omega} (transverse component)

Multiplying the last equation by omega, and integrating, yields:
-g(\sin\theta(t)-\sin\theta_{0})=\frac{l\omega(t)^{2}}{2}\rightarro w\omega(t)^{2}=\frac{2g}{l}(\sin\theta_{0}-\sin\theta(t))

Clearly, T changes from compression to tension (at \theta=\hat{\theta})when it is zero:
2\sin\theta_{0}=3\sin\hat{\theta}\rightarrow
\hat{\theta}=\sin^{-1}\frac{2}{3}\sin\theta_{0}

Thanks for that, and sorry I took so long to get back to this. I get everything you did except for the integration bit. Could you explain in more detail how that was done?

Thanks a lot.

[arildno]

All right:
1. Initial conditions are:
\theta(0)=\theta_{0},\dot{\theta}(0)=\omega(0)=0

That is, we start from rest.

2. Using transverse component:
Remember that \omega=\dot{\theta}
i.e omega is the temporal derivative of the angle.

By multiplying the transverse component with omega, we have:
-mg\dot{\theta}\cos\theta=ml\omega\dot{\omega}

Hence, we see that both sides are exactly a temporal derivative (by using the chain rule of differentiation); hence, we may integrate both sides from t=0, to an arbitrary time value.
Using the initial conditions yields the answer.

[Luminous Blob]

Aaahh, now I get it.

You've been a great help, thank you very much.