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The moment of inertia is defined as

$$I = \int r^2 dm$$

but I am not sure how to proceed with solving the above integral. All examples I have seen seem to be done with different strategies. They usually start: "well consider this part of the mass $dm$" and then suddenly just plug in $2 \pi r $ and the answer miraculously pops out.

I can not find some (good) guidelines on how to solve the integral. What function is actually integrated, because both mass $m$ and distance $r$ can be variable in one another:

$$I = \int f(m(r),r(m)) r^2 dr$$

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    $\begingroup$ It's a common misconception that $dm$ indicates what variable is integrated over ( = what variable changes in running through the integral). This is not the case. Instead, dm is a differential (differentials are maps that linearly assign values to tangent vectors of the underlying manifold). In this application, dm will tell you what infinitesimal value you can add to the integral when you "add" an infinitesimal tangent vector $e_r$ to the space you're integrating over. $\endgroup$ Jan 26 at 13:38
  • $\begingroup$ @Quantumwhisp while this may be technically true, here $dm$ is just the small amount of mass located at $\vec r$. If you want to think of $dm$ as a tangent vector, then in what space does it live in? You can make $dm$ into an abstract differential geometric quantity if you want, but it’s not terribly enlightening to do this. $\endgroup$ May 1 at 12:05
  • $\begingroup$ @ZeroTheHero I never said dm is a tangent vector, quite the opposite. I said it's a differential (-form, to be precise). It takes 3 tangent vectors (that specify an infinitesimal little cube) and assign to them the infinitesimal mass that you also mentioned. IMHO this is terribly enlightening, especially if you want to talk about transformation laws, stokes theorem and the connection to derivatives. $\endgroup$ May 2 at 7:28
  • $\begingroup$ @Quantumwhisp yes I misread your comment (not sure why I misread it but clearly I did). My broader point is all this tangent vector stuff is in fact NOT super enlightening unless you’re thinking of this mathematically. This form language is correct (and it’s quite convenient for the reasons you allude to, especially for transformations) but in physics very few people think of a volume as a form unless you need to understand the transformation of volumes. $\endgroup$ May 2 at 11:52
  • $\begingroup$ I suppose another way to put it is that whereas it's natural to think of $d\vec S$ or $d\vec \ell$ geometrical quantities, it's not so clear how natural it is to think of $dm=\rho dV$ in this way; maybe one should, but in most cases it's not that useful. $\endgroup$ May 2 at 13:58

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TL;DR This question is actually more related to mathematics than physics. Here I give just some basic guidelines which can help you in most textbook problems of finding moment of inertia. Please note that finding moment of inertia is not difficult, but it takes practice! In the most complicated cases we use specialized software to numerically compute moment of inertia.

Below you can find three examples on how to calculate moment of inertia for a cube and a cylinder. Make sure to understand these examples before you move on to more complex problems (objects). For anything more detailed than this, consult some book on calculus on how to deal with double and triple integrals.


Moment of inertia in general case

The idea is simple - imagine an object is composed of many many infinitesimally small particles of mass $m_i$ at distance $r_i$ from the axis of rotation. To find moment of inertia you just sum distance squared times mass of each particle

$$I = \sum_{i} r_i^2 m_i$$

The above equation works well in discrete case, but in continuous case the sum becomes integral

$$\boxed{I = \iiint_\mathcal{V} r^2 dm} \tag 1$$

where $\mathcal{V}$ is the volume that object occupies in 3D. Now the only thing you have to do is to represent the infinitesimally small mass $dm$ as a function of distance $r$. For objects whose density $\rho$ is constant (homogeneous objects), the mass can be expressed as

$$dm = \rho \cdot dV$$

where $dV$ is infinitesimally small volume. The integral in Eq. (1) now becomes

$$\boxed{I = \rho \iiint_\mathcal{V} r^2 dV} \tag 2$$

Moreover, if object's cross section does not change along the axis of rotation, the above equation can be further simplified:

$$dm = \rho \cdot dV = \rho \cdot (h \cdot dA) = \sigma \cdot dA$$

where $h$ is object length along axis of rotation, and $\sigma$ is mass-area density of the object. The above simplification converts triple integral (volume) to double integral (area), which is much easier to depict. The integral in Eq. (2) now becomes

$$\boxed{I = \sigma \iint_\mathcal{A} r^2 dA} \tag 3$$

where $\mathcal{A}$ is the area that object occupies in 2D. The double integrals are solved either in Cartesian coordinates or polar coordinates, whichever is more suitable for a specific problem. Except for these two, the triple integrals can also be solved in spherical coordinates, but I will focus here only on double integrals.


Cartesian and polar coordinates

In all cases discussed here we assume that the axis of rotation goes through the origin. You are allowed to place the origin wherever it is most convenient to solve a particular problem (integral)!

In Cartesian coordinates, the infinitesimally small area and distance from origin are defined as

$$dA = dx \cdot dy \qquad \text{and} \qquad r^2 = x^2 + y^2$$

and the Eq. (3) becomes

$$\boxed{I = \sigma \iint_\mathcal{A} (x^2 + y^2) dx dy} \tag 4$$

In polar coordinates, the infinitesimally small area is defined as

$$dA = (r+dr)^2 \frac{d\theta}{2} - r^2 \frac{d\theta}{2} = (2rdr + (dr)^2) \frac{d\theta}{2} \approx r dr d\theta$$

where $(dr)^2$ is neglected being much (much!) smaller than $2rdr$. With this Eq. (3) becomes

$$\boxed{I = \sigma \iint_\mathcal{A} r^2 \cdot r dr d\theta} \tag 5$$

It must be noted that Eq. (4) and Eq. (5) are only valid if:

  • object is homogeneous, i.e. its density $\rho$ is constant, and
  • object's cross section does not change along the axis of rotation!

In most textbook problems the above two requirements are satisfied. If not, you have to use Eq. (2) for homogeneous objects or Eq. (1) for non-homogeneous objects.


Example 1: Cube with all sides equal to $L$

Assume that the axis of rotation goes through center of mass and is perpendicular to two faces. Moment of inertia is

$$I = \sigma \int_{-L/2}^{L/2} \int_{-L/2}^{L/2} (x^2 + y^2) dx dy = \frac{1}{6} M L^2$$

where $M = \sigma L^2$ is the total mass of the cube. Integration is done in Cartesian coordinates, and the axis of rotation is placed at the origin.


Example 2: Cylinder with radius $R$

Assume that the axis of rotation goes through center of mass and is perpendicular to cylinder's base. Moment of inertia is:

$$I = \sigma \int_{0}^{R} \int_{0}^{2\pi} r^3 dr d\theta = \frac{1}{2} M R^2$$

where $M = \sigma R^2 \pi$ is the total mass of the cylinder. Integration is done in polar coordinates, and the axis of rotation is placed at the origin.


Parallel-axis theorem

The parallel axis theorem, also known as Huygens-Steiner theorem, relates moment of inertia of an object about axis through center of mass to moment of inertia about arbitrary parallel axis

$$\boxed{I = I_\text{cm} + M d^2} \tag 6$$

where $M$ is total mass of the object and $d$ is distance between the two parallel axis.


Example 3: Cube rotating about edge

In Example 1 we have found moment of inertia for cube that rotates about axis through center of mass. When axis goes along cube edge, moment of inertia is

$$I = \sigma \int_{0}^{L} \int_{0}^{L} (x^2 + y^2) dx dy = \frac{2}{3} M L^2$$

We could have calculated this using the parallel-axis theorem. The axis of rotation is at the distance $d = \frac{\sqrt{2}}{2} L$ from axis that goes through center of mass, hence

$$I = \frac{1}{6} M L^2 + M \bigl( \frac{\sqrt{2}}{2} L \bigr)^2 = \frac{2}{3} M L^2$$

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    $\begingroup$ Excellent answer $\endgroup$ Jan 28 at 12:43
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    $\begingroup$ @AccidentalTaylorExpansion Thanks, much appreciated! $\endgroup$ Jan 28 at 16:00
  • $\begingroup$ The cube should evaluate to $I = \frac{1}{12} M L^2$. Check your math please. $\endgroup$
    – JAlex
    Jan 28 at 20:39
  • $\begingroup$ @JAlex I double checked, and also checked in some online tables - they all seem to agree that moment of inertia of a cube is $I = \frac{1}{6} M L^2$.. Wikipedia - List of moments of inertia $\endgroup$ Jan 28 at 21:27
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    $\begingroup$ Yes, I see because all sides are the same $\tfrac{M}{12} (L^2 + L^2 ) = \tfrac{M}{6} L^2$. You are correct. $\endgroup$
    – JAlex
    Jan 28 at 21:46
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The mass properties of an object are found using a volume integral.

The general procedure is a mathematical summation of the volume of the solid. There are several steps needed to be taken to get the final result and they are outlined below: (caution, calculus ahead)

  1. Volume - The solid is parametrized in a way the volume can be summed by $$V = \int {\rm d}V$$

    • There are various common ${\rm d}V$ expressions depending on the geometry, such as ${\rm d}V = {\rm d}x\, {\rm d}y\, {\rm d z}$ or ${\rm d}V = r\,{\rm d}r\,{\rm d}\theta\,{\rm d}z$.
    • For the most general case if the interior points of the solid can be expressed in terms of a vector with three parameters $\vec{\rm pos}(u,v,w)$ then the volume differential is $${\rm d}V = \left| \frac{\partial \vec{\rm pos}}{\partial u} \cdot (\frac{\partial \vec{\rm pos}}{\partial v} \times \frac{\partial \vec{\rm pos}}{\partial w}) \right| {\rm d}u\,{\rm d}v\,{\rm d}w$$ Here $\cdot$ is the vector inner product, and $\times$ the vector cross product.
    • Volume Element in Wikipedia has more details on how ${\rm d}V$ is defined for solids.
  2. Mass - The volume integral is summed over the density function $$ m = \int \rho\,{\rm d}V$$ If density is assumed to be constant, then the above is usually flipped to set $\rho = m/V$ in the equations below.

  3. Center of Mass - The volume integral is summed over the position function $$ \vec{\rm com} = \frac{1}{m} \int (\vec{\rm pos})\,\rho {\rm d}V$$

  4. Mass moment of inertia (tensor) about origin - The MMOI of the body about the origin is evaluated with the volume integral summing the MMOI of each particle in the solid $$ \mathrm{I}_O = \int ( \vec{\rm pos} \cdot \vec{\rm pos} - \vec{\rm pos} \otimes \vec{\rm pos}) \, \rho\,{\rm d}V$$ Here $\cdot$ is the vector inner product, and $\otimes$ the vector outer product.

    an alternate form of the above is $$ \mathrm{I}_O = \int \begin{vmatrix} z^2+y^2 & -x y & -x z \\ -x y & x^2+z^2 & - y z \\ -x z & -y z & x^2+y^2 \end{vmatrix} \, \rho\,{\rm d}V$$

    for the case with uniform density, usually, MMOI is expressed in terms of a known mass, which is done by taking the density out of the integral $$ \mathrm{I}_O = \frac{m}{V} \int ( \vec{\rm pos} \cdot \vec{\rm pos} - \vec{\rm pos} \otimes \vec{\rm pos}) \,{\rm d}V$$

  5. Mass moment of inertia (tensor) about the center of mass - Finally the parallel axis theorem must be applied to transform the MMOI about the origin $\mathrm{I}_O$ to the MMOI about the center of mass $\mathrm{I}_C$ $$\mathrm{I}_C = \mathrm{I}_O - m ( \vec{\rm com} \cdot \vec{\rm com} - \vec{\rm com} \otimes \vec{\rm com})$$

The above describes MMOI in the most general approach possible, in order to be able to apply it to various scenarios.

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Ok, so I understand that you recently started learning about MOI. I will try to explain what is happening there.

Basically we define MOI of a point mass m or a discrete particle

at a distance of r from the axis as

mr^2

Now I called it a point mass because MOI is distant dependent. If the same mass m is distributed over a certain area, then the inertia would be different.

Now for 2D objects like disc, mass is distibuted areally. Therefore we find mass per unit area first. Now we take up all "pointed" elements over any variable distance x. These set of pointed elements gives us a ring(in case of disc).Now to find the mass of elemental ring, we need it's area.

Here comes an interesting step. We convert the ring into a rectangle by breaking at its joint. It's length is 2pir and breadth is dr. Hence area is 2pir*dr. Multiplying with areal density gives mass and hence integral can be calculated.

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Keep in mind objects have different moments of inertia depending on how they are spinning. Flip a coin, that's one moment of inertial. Spin it about an axis normal to the center of one of its faces, that's another. Offset that axis to the side a bit, and flip, that's yet another. An object may also have variable density. Also remember $r$ in question is a measure of the distance from the infinitesimal mass element to the axis of rotation and not distance from origin.

Example:

Spinning:

$\int_{-h/2}^{h/2}\int_0^{2\pi}\int_0^R\rho r^2\ rdr d \theta dz=2\pi\rho r^4h/4$. $\rho = M/\pi R^2h$, the density, so multiplying it out, $MOI=(1/2)MR^2$.

By Fubini's Theorem, you can rearrange the order of integration and do part of the triple integral to get:

$\int_0^R (2\pi r)h\rho (r^2dr)$

It's the same integral, simplified so you only have to worry about the $r$ integration. That's where circumference comes in because it is multiplied by an area term to give you volume.

Flipping:

$\int_{-h/2}^{h/2}\int_0^{2\pi}\int_0^R \rho [(r\sin \theta)^2+z^2] rdr d\theta dz$

$=\int_0^{2\pi}\int_0^R \rho (hr^2\sin^2\theta+h^3/12) rdrd\theta$

$=\int_0^R \rho (\pi hr^3 + \pi rh^3/6) dr$

$=\rho (\pi h R^4/4+\pi R^2h^3/12)$

$=M(R^2/4+h^2/12)$

This is one way things can get confusing. Whereas in the spinning case, $r$ was not only a necessary part of the volume term, it was also equal to the distance from the axis of rotation. In this flipping case, $r$ is still the distance from the origin in the plane, but the $r^2$ term for weighting the mass element is $(r\sin\theta)^2+z^2$ because that's the distance squared about the new axis of rotation.

Flipping about Off center axis: $(x-a)^2+y^2 = R^2, a<r, |z|<h/2$, flipping axis through the origin.

$MOI=\int_{-h/2}^{h/2}\int_{a-R}^{a+R}\int_{-\sqrt{R^2-(x-a)^2}}^{\sqrt{R^2-(x-a)^2}} \rho [x^2+z^2]dydxdz$

$=\int_{a-R}^{a+R}\int_{-\sqrt{R^2-(x-a)^2}}^{\sqrt{R^2-(x-a)^2}} \rho [hx^2+h^3/12]dydx$ Let $u=x-a$. Then $du=dx$.

$\int_{-R}^R\int_{-\sqrt{R^2-u^2}}^{\sqrt{R^2-u^2}}\rho h[u^2+2au+a^2+h^2/12]dydu$

$=\int_{-R}^R 2\rho h[u^2+2au+a^2+h^2/12]\sqrt{R^2-u^2}du$

$= 2\rho h[\pi R^4/8+ (a^2+h^2/12)\pi R^2]$

$=M( \pi R^2/4 +a^2 + h^2/12)$

Notice we retain the same moment of inertia from the regular flipping case plus an $Ma^2$ term, the mass times the off center distance squared. This is an example of The Parallel Axis Theorem

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How to obtain the inertia tensor for rotation symmetric figure.

First put the axes system on the symmetric lines thus the inertia tensor $~\mathbf I~$ is diagonal

$$\mathbf I=\begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z\\ \end{bmatrix}$$

with \begin{align*} &I_r = \rho \iiint_S r^2 dV=\frac{M}{V}\iiint r^2 \,dx\,dy\,dz \quad\Rightarrow\\ &I_x=\frac{M}{V}\iiint (y^2+z^2) \,dx\,dy\,dz \\ &I_y=\frac{M}{V}\iiint (x^2+z^2) \,dx\,dy\,dz \\ &I_z=\frac{M}{V}\iiint (x^2+y^2) \,dx\,dy\,dz \\ &V=\iiint \,dx\,dy\,dz \end{align*}

starting with the position vector $~\vec R~$ to the point on the surface
\begin{align*} &\vec{R}=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\mapsto \begin{bmatrix} x(q_1,q_2,q_3) \\ y(q_1,q_2,q_3) \\ z(q_1,q_2,q_3) \\ \end{bmatrix} \end{align*} where $~q_i~$ are the generalized coordinates
thus \begin{align*} &dx\,dy\,dz\mapsto \sqrt{det(J^T\,J)}\,dq_1\,dq_2\,dq_3\\ &J=\begin{bmatrix} \frac{\partial x}{\partial q_1} & \frac{\partial x}{\partial q_2} & \frac{\partial x}{\partial q_3} \\ \frac{\partial y}{\partial q_1} & \frac{\partial y}{\partial q_2} & \frac{\partial y}{\partial q_3} \\ \frac{\partial z}{\partial q_1} & \frac{\partial z}{\partial q_2} & \frac{\partial z}{\partial q_3} \\ \end{bmatrix} \end{align*}

Center of mass coordinate \begin{align*} &x_\text{cm}=\frac{1}{V}\iiint x\,dx\,dy\,dz=\frac{1}{V}\iiint x\,\sqrt{det(J^T\,J)}\,dq_1\,dq_2\,dq_3\\ &y_\text{cm}=\frac{1}{V}\iiint y\,dx\,dy\,dz=\frac{1}{V}\iiint x\,\sqrt{det(J^T\,J)}\,dq_1\,dq_2\,dq_3\\ &z_\text{cm}=\frac{1}{V}\iiint z\,dx\,dy\,dz=\frac{1}{V}\iiint x\,\sqrt{det(J^T\,J)}\,dq_1\,dq_2\,dq_3\\ \end{align*}
Inertia tensor at the CM coordinate \begin{align*} &\mathbf{I}_{\text{CM}}=\begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \\ \end{bmatrix} +M\, \left[ \begin {array}{ccc} 0&-z_{{{\it cm}}}&y_{{{\it cm}}} \\ z_{{{\it cm}}}&0&-x_{{{\it cm}}} \\ -y_{{{\it cm}}}&x_{{{\it cm}}}&0\end {array} \right] \, \left[ \begin {array}{ccc} 0&-z_{{{\it cm}}}&y_{{{\it cm}}} \\ z_{{{\it cm}}}&0&-x_{{{\it cm}}} \\ -y_{{{\it cm}}}&x_{{{\it cm}}}&0\end {array} \right] \end{align*}

Example cone

enter image description here

\begin{align*} &\vec{R}=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \left[ \begin {array}{c} {\frac { \left( h-u \right) r\cos \left( \varphi \right) }{h}}\\ {\frac { \left( h-u \right) r\sin \left( \varphi \right) }{h}}\\ u \end {array} \right]\quad, \vec{q}=\begin{bmatrix} r \\ \varphi \\ u \\ \end{bmatrix}\\ &J= \left[ \begin {array}{ccc} {\frac { \left( h-u \right) \cos \left( \varphi \right) }{h}}&-{\frac {r\cos \left( \varphi \right) }{h}}&-{ \frac { \left( h-u \right) r\sin \left( \varphi \right) }{h}} \\ {\frac { \left( h-u \right) \sin \left( \varphi \right) }{h}}&-{\frac {r\sin \left( \varphi \right) }{h}}&{\frac { \left( h-u \right) r\cos \left( \varphi \right) }{h}} \\ 0&1&0\end {array} \right] \\ &\text{determinate}~d={\frac { \left( h-u \right) ^{2}r}{{h}^{2}}} \\ &x^2+y^2={\frac { \left( h-u \right) ^{2}{r}^{2}}{{h}^{2}}}\\ \end{align*}
\begin{align*} &V=\int_{0}^{h}\int_{0}^{2\pi}\int_{0}^{\rho} {\frac { \left( h-u \right) ^{2}r}{{h}^{2}}}\,dr\,d\varphi\,du =\frac 13\,{\rho}^{2}h\pi \\ &I_z=\frac{M}{V}\iiint (x^2+y^2)\,{\frac { \left( h-u \right) ^{2}r}{{h}^{2}}}= \frac{M}{V}\int_{0}^{h}\int_{0}^{2\pi}\int_{0}^{\rho} {\frac { \left( h-u \right) ^{4}{r}^{3}}{{h}^{4}}}\,dr\,d\varphi\,du =\frac{3}{10}\,M\rho^2\\ &\text{analog}\\ &I_x=\frac{1}{20}\,M \left( 3\,{\rho}^{2}+2\,{h}^{2} \right) \\ &I_y=I_x\\ &\text{the CM coordinates are}\quad x_\text{cm}=y_\text{cm}=0~,z_\text{cm}=\frac{h}{4}\quad \Rightarrow\\ &\mathbf{I}_\text{CM}=\begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \\ \end{bmatrix}+ \left[ \begin {array}{ccc} -\frac{1}{16}\,M{h}^{2}&0&0\\ 0& -\frac{1}{16}\,M{h}^{2}&0\\ 0&0&0\end {array} \right] \end{align*}

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  • $\begingroup$ Don't you need to take the square root of the determinanat, and not the other way around $$ dx\,dy\,dz\mapsto \sqrt{ {\rm det}( J^T\,J)}\,dq_1\,dq_2\,dq_3 $$ $\endgroup$ Feb 1 at 13:54
  • $\begingroup$ yes this is a tipo $\endgroup$
    – Eli
    Feb 1 at 14:17

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