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How do I calculate the moment of inertia of a uniform solid cube about an axis passing through its center of mass?

I also wanted to know if the moment of inertia of a body is independent of its shape. Also, recently I read somewhere that the moment of inertia of a uniform solid cube is minimum about an axis passing through its COM because the mass is more concentrated at its center. Does the statement make any sense?

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    $\begingroup$ There are many axes through the center of a cube. Which one are you using? $\endgroup$ – DJohnM Mar 1 '14 at 17:36
  • $\begingroup$ web.phys.ntu.edu.tw/semi/ceos/general.files/… Read this and ask where you got confusion in the derivation. Also there can by found many resources to calculate Moment of inertia's of different shapes just google or refer to a text book. $\endgroup$ – user31782 Mar 1 '14 at 18:12
  • $\begingroup$ @ItachiUchiha Can you please mention the source? $\endgroup$ – Prasanna Sep 30 '18 at 10:51
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The moment of inertia can be defined as the volume integral of the density times the position vector (centered at the origin of the axis you choose): $$ I_{obj}=\int dV\,\rho\left(\mathbf{r}\right)\mathbf{r}^2 $$ which should work always.

As for your other questions, if we had a thin, solid cylinder and rotated it about its end point:

enter image description here

The moment of inertia would be $$ I_{cyl,end}=\frac13mL^2 $$ If we took the same rod and rotated it about it's center,

enter image description here

Then the moment of inertia ends up being $$ I_{cyl,mid}=\frac1{12}mL^2 $$ If we instead had a sphere rotating about its center,

enter image description here

the moment of inertia is $$ I_{sph}=\frac25mr^2 $$

So clearly the shape and the axis affect the moment of inertia.

If we rotated an object about an axis that does not line up with its center of mass, then we need to use the parallel-axis theorem. This tells us that the total moment of inertia is then $$ I_{tot}=I_{cm} + mr^2 $$ where the $r$ denotes the distance from the object's center of mass to the axis of rotation and $I_{cm}$ is the normal moment of inertia for the object (e.g., the above few ones).

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First you need to define the orientation of the cube relative to the axis you want to measure. Typically a 3×3 rotation matrix $E$ does the job transforming local coordinates along the principal axes to the world coordinates. The use the transformation $E I_{body} E^\intercal$

Example:

A single rotation $\theta$ about the world $z$ axis is

Example Cube

$$E = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

If the mass moment of inertia matrix about the principal axes $(x,y,z)$ is

$$I_{body} = \begin{vmatrix} I_{xx} & & \\ & I_{yy} & \\ & & I_{zz} \end{vmatrix} $$

then the mass moment of inertia about the world coordinates is

$$ I_{world} = E\,I_{body}\, E^\intercal $$

where $E^\intercal$ is the inverse rotation found by the transpose operator. The result is $$ I_{world} = \begin{vmatrix} I_{yy}+(I_{xx}-I_{yy})\cos^2\theta & (I_{xx}-I_{yy})\sin\theta\cos\theta & 0\\ (I_{xx}-I_{yy})\sin\theta\cos\theta & I_{xx}+(I_{yy}-I_{xx})\cos^2\theta & 0\\ 0 & 0& I_{zz} \end{vmatrix} $$

This represent the mass moment of inertia about the three world coordinates. To get the MMOI about a specific axis $\hat{e}$ you calculate $$I_{ee} = \hat{e}^\intercal I_{world} \hat{e} $$

So to get the MMOI about the world $X$ axis with $\hat{e}=(1,0,0)$ you find that $$I_{XX} = \begin{pmatrix}1 & 0 & 0 \end{pmatrix} I_{world} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = I_{yy}+(I_{xx}-I_{yy})\cos^2\theta $$

Alernatively You can find the local coordinates of the world $X$ axis as $\hat{x} = E^\intercal \hat{e}$ and then compute $$\boxed{ I_{XX} = \hat{x}^\intercal I_{body} \hat{x} }$$

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Two theorems worth knowing about when it comes to calculations like this:

1) Parallel axis theorem. For any object of mass $m$, the moment of inertia $I_A$ about an axis A that is parallel to but displaced by a distance $x$ from an axis $C$ through the center of mass, is given by $$I_A = mx^2 + I_C$$

From this, it follows immediately that the moment of inertia about an axis through the center of mass is a minimum (for that direction of rotation) since $mx^2$ will always be >0 for any nonzero value of $x$ (the offset).

2) Perpendicular axis theorem. For a lamina (thin sheet), the moment of inertia about an axis perpendicular to the sheet is equal to the sum of moments of inertia about two perpendicular axes in the sheet.

This theorem is handy to calculate the moment of inertia of a square plate of mass $m$ and side $s$. It is easy to calculate the moment of inertia about an axis in the plane of the plate and parallel to the sides of the square - about that axis the mass distribution of the object is no different than the distribution of a rod, for which we have the result

$$I_x = \frac{1}{12} m s^2$$

By symmetry, $I_y = I_x$. From the perpendicular axis theorem, $I_z = I_x + I_y$ so

$$I_z = \frac{1}{6} m s^2$$

Considering a cube as being made up of a stack of lamina, it follows that the moment of inertia of a cube about an axis through the center of mass (and through the face of the cube) is the same as the above.

You can see a pretty comprehensive listing of other moments of inertia for different shapes at this link

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protected by AccidentalFourierTransform Aug 11 '18 at 14:00

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