How can I calculate the moment of inertia of any object?

[cwill53]

Apologies if I make anyone frustrated.

To start, I've only had up to Calculus II so far but I was curious how to use and evaluate integrals used for moment of inertia. I know that the moment of inertia is basically an object's resistance to rotation, and is the rotational analog of mass. I know that it can be expressed for both uniform and non-uniform objects by, assuming you can divide the object into point masses, the summation of point masses multiplied by each of their respective distances from the axis of rotation, or for a continuous mass by integration, which is the part I'm shaky on. The moment of inertia I can be defined as:

$$\sum_{i}^{}m_{i}r_{i}=\int \int \int r^2dm$$​

where I is the moment of inertia and r is the distance of each point from the axis of rotation. I don't really understand how I would use the above integral and WHY there are three integrals in the first place, as the following expression can also be used to calculate moment of inertia:

$$I=\int r^2ρdV$$​

where ρρ is density and V is volume. m is mass.

If m=dρdVm=dρdV and dmdm can be expressed as $ρdV$ why would I integrate three times in the above integral?

I can give you an example of what confused me. I was reading through my physics textbook, working out problems, and every single moment of inertia problem involving calculus I simply couldn't wrap my head around.

For example, I viewed the solution online to the following problem:

A sphere with radius R= 0.200 m has density ρρ that decreases with distance r from the center of the sphere according to ρ=3.00×103kg/m3−(9.00×103kg/m4)rρ=3.00×103kg/m3−(9.00×103kg/m4)rρ=3.00×103kg/m3−(9.00×103kg/m4)rρ=3.00×103kg/m3−(9.00×103kg/m4)r (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

The solver solved the problem this way:

​

$$\int \int \int \rho (r)dV=\int dm\rightarrow \int_{0}^{2\pi }\int_{0}^{\pi }\int_{0}^{R}(a-br)r^2sin\phi .d\theta .d\phi .dr$$
As you can probably tell I had no idea what was going on here. I still copied the solution down into my notebook for future reference.

Another thing I've seen attached to moment of inertia is the word tensor, and the moment of inertia tensor. As far as I know a tensor, according to a books that I have, is a linear operator that maps a vector to something else, a first order tensor mapping a vector to a scalar, a second order mapping a vector to a vector, and a third order tensor mapping a vector to whatever that is that comes after vectors.

Another book gives a different kind of introduction:

Which is admittedly confusing.

 

[kuruman]

You don't really need tensors if you have never seen them to understand this. In Cartesian coordinates, an element of mass is $dm=\rho dV=\rho dx~dy~dz$. Its contribution to the moment of inertia would be $dI=dm(x^2+y^2+z^2)=\rho(x^2+y^2+z^2)dx~dy~dz$. The total moment of inertia is the sum, actually integral, of all such contributions. Since there is mass in all three directions, you need to do three additions, one per direction. Similarly for other coordinates but Cartesian is the easiest to visualize because all three directions are equivalent. Of course the moment of inertia does not depend on the choice of coordinate axes that one uses to do the integrals in.

Edited in view of post #6:
For example to calculate the moment of inertia about the $z$ axis, you set $z=0$ in the expression for $r^2$ since you are interested in the perpendicular distance from the element to the axis and the integral for the moment of inertia becomes$$I_z=\int (x^2+y^2) \rho(x,y,z) dx~dy~dz$$or, more formally,$$I_z=\int dz\int dy \int (x^2+y^2) \rho(x,y,z) dx.$$This last expression can be viewed as adding moments of inertia about $z$ of slices of thickness $dz$, parallel to the $xy$ plane with each slice at position $z$ relative to the origin. The moment of inertia of a single slice is expressed by the inner two integrals over $x$ and $y$.

 

[cwill53]

Apologies for the formatting issues. Every time I try to edit the format on another correct latex equation becomes undone for some reason.

 

[cwill53]

You don't really need tensors if you have never seen them to understand this. In Cartesian coordinates, an element of mass is $dm=\rho dV=\rho dx~dy~dz$. Its contribution to the moment of inertia would be $dI=dm(x^2+y^2+z^2)=\rho(x^2+y^2+z^2)dx~dy~dz$. The total moment of inertia is the sum, actually integral, of all such contributions. Since there is mass in all three directions, you need to do three additions, one per direction. Similarly for other coordinates but Cartesian is the easiest to visualize because all three directions are equivalent. Of course the moment of inertia does not depend on the choice of coordinate axes that one uses to do the integrals in.

So x, y and z are sort the axes for the height, length, and width for the object?
I know this might sound trivial, like a Google question, but, in a triple or quadruple integral, which integral is being evaluated with respect to which differential, and as you do each integral, are you integrating the function three times consecutively or separately?

 

[kuruman]

The convention is to place the differential after the integral sign and do the inner integral(s) first, e.g.
$$I=\int dx\int dy \int f(x,y,z)dz.$$ You do the $z$ integral first pretending that $x$ and $y$ are constant to get a function of $x$ and $y$ and so on. You cannot go wrong if you do the integrals consecutively. If the limits of integration are independent of the integration variables, they can be done separately if you know what you are doing. The web has many videos of how to do multiple integrals.

 

[vanhees71]

You don't really need tensors if you have never seen them to understand this. In Cartesian coordinates, an element of mass is $dm=\rho dV=\rho dx~dy~dz$. Its contribution to the moment of inertia would be $dI=dm(x^2+y^2+z^2)=\rho(x^2+y^2+z^2)dx~dy~dz$. The total moment of inertia is the sum, actually integral, of all such contributions. Since there is mass in all three directions, you need to do three additions, one per direction. Similarly for other coordinates but Cartesian is the easiest to visualize because all three directions are equivalent. Of course the moment of inertia does not depend on the choice of coordinate axes that one uses to do the integrals in.

A moment of inertia is NOT defined in the way indicated above. To define it you need a rigid body and a fixed axis of rotation. Choosing a Cartesian body-fixed basis with its 3-direction in direction of the axis, the corresponding moment of inertia is defined by
$$I=\int_V \mathrm{d}^3 r \rho(\vec{x}) (\vec{x}_1^2+\vec{x}_2^2),$$
i.e., it's the mass-density weighted integral over the perpendicular distance squared of each point of the body from the axis and not the distance to the origin of the body-fixed frame squared!

 

[kuruman]

A moment of inertia is NOT defined in the way indicated above. To define it you need a rigid body and a fixed axis of rotation. Choosing a Cartesian body-fixed basis with its 3-direction in direction of the axis, the corresponding moment of inertia is defined by
$$I=\int_V \mathrm{d}^3 r \rho(\vec{x}) (\vec{x}_1^2+\vec{x}_2^2),$$
i.e., it's the mass-density weighted integral over the perpendicular distance squared of each point of the body from the axis and not the distance to the origin of the body-fixed frame squared!

You are, of course, correct. I was intent on explaining multiple integrals and cut one corner too many by showing only $dI$ and not showing how the integral for the moment of inertia about an axis is assembled. I edited the post to atone for my omission.

 

[cwill53]

A moment of inertia is NOT defined in the way indicated above. To define it you need a rigid body and a fixed axis of rotation. Choosing a Cartesian body-fixed basis with its 3-direction in direction of the axis, the corresponding moment of inertia is defined by
$$I=\int_V \mathrm{d}^3 r \rho(\vec{x}) (\vec{x}_1^2+\vec{x}_2^2),$$
i.e., it's the mass-density weighted integral over the perpendicular distance squared of each point of the body from the axis and not the distance to the origin of the body-fixed frame squared!

I'm sorry, but I do not follow. Could you break it down further?

 

[kuruman]

I'm sorry, but I do not follow. Could you break it down further?

Look at my edited post #2. It might shed some light.

 

[cwill53]

Look at my edited post #2. It might shed some light.

I think my issue lies in the fact that I I haven't studied multivariate calculus yet. I've only dealt with single variable so far.

 

[kuruman]

I think my issue lies in the fact that I I haven't studied multivariate calculus yet. I've only dealt with single variable so far.

Once you study it, you will realize that it is a generalization of a single variable. So be patient and take a course in multvariate calculus at your earliest opportunity. It will come together in the end.