# Determination of moment of inertia

• ajcoelho
In summary, the conversation is about finding the period of a pendulum and discussing the equations and reasoning used in a provided source. The error in equation 5 is addressed and the correct torque equation is confirmed. There is also a discussion about the approximation of cos phi to 1.
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!

## The Attempt at a Solution

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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.

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

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.

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...

#### Attachments

• Sem título.jpg
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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##

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

Where is that ##\cos \phi##??

Ty = Mg/2 = T cosϕ

therefore: T = mg/2cosϕ

therefore: Tx = mg sin ϕ / 2 cosϕ

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.

1 person

## What is moment of inertia?

Moment of inertia, also known as rotational inertia, is a measure of an object's resistance to changes in its rotational motion. It is affected by the mass and distribution of mass around the axis of rotation.

## How is moment of inertia calculated?

The moment of inertia of a point mass is calculated by multiplying the mass by the square of its distance from the axis of rotation. For a continuous object, it can be calculated by integrating the mass distribution over its entire volume.

## What is the relationship between moment of inertia and angular acceleration?

According to Newton's Second Law of Motion, the angular acceleration of an object is directly proportional to the net torque acting on it and inversely proportional to its moment of inertia. This means that objects with larger moments of inertia require more torque to achieve the same angular acceleration as objects with smaller moments of inertia.

## How does the shape of an object affect its moment of inertia?

The distribution of mass around the axis of rotation is a major factor in determining an object's moment of inertia. Objects with more mass concentrated towards their axis of rotation have smaller moments of inertia, while objects with mass distributed further from the axis of rotation have larger moments of inertia.

## What are some real-world applications of moment of inertia?

Moment of inertia is used in many fields, such as engineering, physics, and sports. It is crucial in designing structures and machines that need to resist rotational forces, such as bridges and aircraft. In sports, understanding moment of inertia can help athletes optimize their movements and equipment for better performance.

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