# Angular velocity of a pendulum of sorts

1. Nov 23, 2006

### El Hombre Invisible

I feel really dumb asking this... it seems like a really basic problem, but it's due tomorrow, I've got heaps of other stuff to do and I'm not getting anywhere with it.

Here is the problem:

"A uniform rod of length l is freely pivoted at one end. It's initially held horizontal, then released from rest.

First, give angular momentum L as a function of angle."

Now, L is certainly a function of angular velocity: L = Iw. I for a rod is (Ml^2)/3, so L = (Ml^2)w/3. This isn't what I'd have expected for a question asking for L as a function of theta, but a) others I've spoken to claim to have done it this way, and b) I can't figure out a way of getting just theta in there.

"Next, give T (torque) as a function of theta, where theta = 0 when the rod is in its initial position."

Easy enough. Letting x be the distance of an element dm from the pivot point, dT = g cos(theta) x dx, so integrating from x = 0 to x = l gives T = Mgl cos(theta)/2. Also, T = Iw^2.

"Show that w^2 = (3g sin(theta)/l."

Eek! Using the two expressions for T:

T = (Mgl cos(theta))/2 = (Ml^2 w^2)/3,

which I'm presuming is cheating since L is not required here, gives:

w^2 = (3g cos(theta)/2l

which isn't good. I have a cos where I ought to have a sin which comes from my equation for T, but since at theta=0 Mg is perpendicular to the rod, the torque must be highest here so it must be proportional to cos(theta). I also have a 2 I don't want.

I also tried T = dL/dt = (3g cos(theta)/(2l), but I can't figure out a way of writing theta as a function of time to perform the required integral without it swallowing its own tail.

I do have a habit of occassionally going about a simple problem in completely the wrong way at first and I usually come to my senses quite quickly, but I've left the planet with this one. What's annoying me is that the problem should be more complicated than the methods I'm using, but I see no better ones to use.

Any help by tomorrow would be most appreciated.

Thanks,

El Hombre

EDIT: sincerest apologies. In my absence I seem to have forgotten how to use latex. I hope you can read this.

Last edited: Nov 23, 2006
2. Nov 24, 2006

### OlderDan

3. Nov 25, 2006

### El Hombre Invisible

Yeah, I cracked it yesterday. Thanks. Also, thanks for not ridiculing my reinvention of the laws of physics (T = Iw^2?? Wha?). I'd had zero sleep the night before I posted and I'd been working on the problem for so long that funny things were happening. After a good nights sleep I got it straight away.