# Moments of Inertia & Radius of Gyration

• MHB
• harpazo
In summary, The moments of inertia about the x-axis, y-axis, and origin can be found by setting up the double integrals of r^2 * rho * dy * dx and finding the appropriate limits of integration. The radius of gyration about the x-axis and y-axis can then be determined by plugging in the expression for rho and finding the distance from the area chunk to the given axis of rotation. It is important to note that the choice of axis of rotation greatly affects the value of the moment of inertia.
harpazo
Find the moments of inertia about the x-axis, y-axis and the origin. Also, find the radius of gyration about the x-axis and y-axis.

y = 0, y = b, x = 0, x = a

Rho = ky

1. Is ky the density function?

2. Do I integrate over dxdy or dydx?

3. Are the limits of integration y = 0, y = b, x = 0, and x = a?

4. For Rho = ky, is k constant?

Harpazo said:
Find the moments of inertia about the x-axis, y-axis and the origin. Also, find the radius of gyration about the x-axis and y-axis.

y = 0, y = b, x = 0, x = a

Rho = ky

1. Is ky the density function?

Yep!

Harpazo said:
2. Do I integrate over dxdy or dydx?

I don't think it's going to matter, because the region of integration is rectangular.

Harpazo said:
3. Are the limits of integration y = 0, y = b, x = 0, and x = a?

Yep!

Harpazo said:
4. For Rho = ky, is k constant?

I would assume so.

$$I=\int r^2 \, dm,$$
and then understand that $\rho = \dfrac{dm}{dA}$, or $\rho \, dA = dm$. Can you continue?

Incidentally, I'm a fan of using $\lambda$ for linear mass density, $\sigma$ for area mass density, and $\rho$ for volume mass density. This problem is squarely in the area mass density, so I would normally write $\sigma \, dA = dm$.

Ackbach said:
Yep!
I don't think it's going to matter, because the region of integration is rectangular.
Yep!
I would assume so.

$$I=\int r^2 \, dm,$$
and then understand that $\rho = \dfrac{dm}{dA}$, or $\rho \, dA = dm$. Can you continue?

Incidentally, I'm a fan of using $\lambda$ for linear mass density, $\sigma$ for area mass density, and $\rho$ for volume mass density. This problem is squarely in the area mass density, so I would normally write $\sigma \, dA = dm$.

Sorry, where did r^2 come from?

Harpazo said:
Sorry, where did r^2 come from?

That's from the definition of the moment of inertia. If you consider a small chunk of mass $dm$, then $r$ is the perpendicular distance from that chunk of mass to the axis of rotation. So you can see that the moment of inertia is highly dependent on your choice of the axis of rotation.

Ackbach said:
That's from the definition of the moment of inertia. If you consider a small chunk of mass $dm$, then $r$ is the perpendicular distance from that chunk of mass to the axis of rotation. So you can see that the moment of inertia is highly dependent on your choice of the axis of rotation.

Can you set up the double integrals? I can then solve the problem on my own.

So,
$$I=\int r^2 \, dm = \iint r^2 \, \rho \, dA = \iint \, r^2 \rho \, dy \, dx.$$
You need to figure out the limits for $x$ and $y$, plug in the expression for $\rho$, and finally you need to figure out what the radius is from the area chunk $dx\,dy$ to the specified axis of rotation, plug that into $r^2$, and then you're on your way.

Forget it.

Harpazo said:
Forget it.

Excuse me? Are you unsure of some of the steps that I've outlined? If so, please let me know, and I'll help you through them.

Harpazo said:

The best thing here is to point out which parts of the previous replies are unclear, and to ask for clarification.

Harpazo said:
Try writing in a student-friendly matter.

That's not a fair statement to make, is it? If something isn't clear, then please and by all means...ask for clarification, but don't accuse the person trying to help of writing in an unfriendly manner, unless you can point to a statement made that is clearly unfriendly. I don't see anything like that in this thread.

For what it's worth, Adrian (Ackbach) has well over a decade of experience with online math sites providing help as well as working professionally as an educator for many years. He firmly believes that the student should do the heavy lifting, but is willing to explain further if something he posted is not clear to the student.

I didn't like the "Excuse me?" way in which he started his reply.

Harpazo said:
I didn't like the "Excuse me?" way in which he started his reply.

Well, let's try to be objective and examine the context in which it was said. He posted judicious help (i.e., guidance), to which you replied "Forget it." To me, this seemed to be dismissive, implying the help he provided was inadequate or lacking in some way. He provided a suggestion for continuing, and when you did not follow those suggestions, he perhaps found it confounding that you were not willing to follow through there.

I can assure you Adrian was not trying to be rude...he was likely just wondering why you would dismiss his suggestions rather than make an attempt to follow them. (Wondering)

Harpazo said:
Can you set up the double integrals? I can then solve the problem on my own.

Setting up the integrals is the whole point of these type of problems - it's more than $70\%$ of the work. Often cranking through the integrals themselves is routine. That's why I haven't posted any more of the solution. What about what I've written so far is unclear to you? I'd be happy to clarify anything that's not clear.

Considering that this question goes back to May 9th of this year, we should move on now. I am in the chapter dealing with cylindrical and spherical triple integrals. If I get stuck somewhere in the chapter, I will return to this forum with my questions. Agree? I do not want to argue about my question posted more than one month ago. I thank everyone here for their math help and guidance.

Harpazo said:
Considering that this question goes back to May 9th of this year, we should move on now. I am in the chapter dealing with cylindrical and spherical triple integrals. If I get stuck somewhere in the chapter, I will return to this forum with my questions. Agree?

Sure! No problem.

Harpazo said:
I do not want to argue about my question posted more than one month ago. I thank everyone here for their math help and guidance.

## What is the moment of inertia?

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is a property that depends on an object's mass and its distribution relative to its axis of rotation.

## How is the moment of inertia calculated?

The moment of inertia is calculated by multiplying the mass of each particle of an object by the square of its distance from the axis of rotation, and then summing up these values for all particles in the object.

## What is the radius of gyration?

The radius of gyration is a measure of the distribution of an object's mass around an axis of rotation. It is the distance from the axis at which the object's total mass could be concentrated to have the same moment of inertia as the object itself.

## How is the radius of gyration related to the moment of inertia?

The radius of gyration is directly proportional to the square root of the moment of inertia. This means that the larger the radius of gyration, the larger the moment of inertia, and vice versa.

## What are the practical applications of moments of inertia and radius of gyration?

Moments of inertia and radius of gyration are important in engineering and physics, as they help determine the stability, strength, and behavior of rotating objects. They are also used in designing structures like bridges, buildings, and vehicles to ensure their safe and efficient operation.

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