Determination of moment of inertia

[ajcoelho]

Homework Statement

I've been trying to find out what is the period os this kind of pendulum decribed here: http://www.eng.uah.edu/~wallace/mae364/doc/Labs/mominert.pdf

The thing is, I've came to the same result shown in equation (11) but my reasoning it's different. I would even say that there is an error on this website on equation 5: in this equaition where is the sin of theta?? It can be omitted, only if the angle between the force vector and position vetor is 90º, but it's not the case clearly!

Am I thinking well? In my reasoning I had also to the aproximattion of cos theta ≈ cos of phi ≈ 1 and in this site things are really simplified just because of that lack of sin of theta...

If you're interested i can also post my reasoning but I'd like to get some help on this equation 5 first...

Thanks a loot!

Homework Equations

The Attempt at a Solution

 

[BvU]

At best there is a cosΘ (which is very close to 1), but the suspension points are still 2a apart on the equilibrium line, so the F point a little more outward than drawn in the figure.

 

[ajcoelho]

the thing is: what goes to torque equation is Tx right? Beacause Ty and Mg/2 cancell one another

 

[BvU]

The torque equation would be numbered (5) ? That is for the z component of the torque vector $\tau = \vec r \times \vec F$ (I suppose you will learn that vector version much later on, for now: it is for the torque about a vertical axis going through C). The expression is correct, the drawing is not. F should point more outwards.

I can't find much wrong with (6) for the magnitude of F.

For the torque around the x-axis (front view, fig.1, from C towards you) you indeed have
(mg/2 a/2) + (mg/2)(-a/2)=0.

 

[ajcoelho]

But suppose you're viewing from end view. Then, the forces are above...

So, what really makes torque isn't Tx??

Then we make Ty=Mg/2, and so on...

 

[BvU]

From the pdf I gather the x-axis is (in the front view) pointing towards the viewer. Take C as the origin. If the y-axis is pointing to the right, then the z axis is pointing up. $I\ddot \theta$ is then the z component of the torque $\tau$

 

[ajcoelho]

I get it now... But i still have that cos (phi) that i have to approximate to 1 right?

 

[BvU]

Where is that $\cos \phi$??

 

[ajcoelho]

Ty = Mg/2 = T cosϕ

therefore: T = mg/2cosϕ

therefore: Tx = mg sin ϕ / 2 cosϕ

 

[BvU]

Ah, my mistake. Your T is the tension in the wires, not a torque as I was stupily reading. You are absolutely right. And yes, assume $\phi$ is so small that the cosine is 1.