Determination of moment of inertia

1. Mar 15, 2014

ajcoelho

1. The problem statement, all variables and given/known data

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 [Broken]

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!

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 6, 2017
2. Mar 15, 2014

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.

3. Mar 15, 2014

ajcoelho

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

4. Mar 16, 2014

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.

5. Mar 16, 2014

ajcoelho

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

So, what really makes torque isnt Tx??

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

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6. Mar 16, 2014

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 $\tau$

7. Mar 17, 2014

ajcoelho

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

8. Mar 17, 2014

BvU

Where is that $\cos \phi$??

9. Mar 17, 2014

ajcoelho

Ty = Mg/2 = T cosϕ

therefore: T = mg/2cosϕ

therefore: Tx = mg sin ϕ / 2 cosϕ

10. Mar 17, 2014

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.