Tensor of inertia of a cylinder
I consider a solid cylinder with mass m, radius r, and height h. My objective is to determine its moment of inertia tensor, which I represent as:
\mathbf{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{bmatrix}
This tensor describes how the mass of the cylinder is distributed relative to the chosen coordinate system.
I use integration to calculate each component of the tensor. The general formula for a component I_{ij} is:
I_{ij} = - \int_V \rho (x_i x_j) \, \mathrm dV
where \rho is the density, V is the volume, and x_i, x_j are coordinates. I use cylindrical coordinates (\rho', \phi, z) because they are well-suited to the cylinder’s geometry.
Let’s look at a few key calculations. For I_{xx}, I have:
I_{xx} = \rho \int_{-h/2}^{h/2} \int_{0}^{2\pi} \int_{0}^{r} (y^2 + z^2) \, \rho' \, \mathrm d\rho' \, \mathrm d\phi \, \mathrm dz
Since y = \rho' \sin\phi, I can rewrite this integral. After some calculations I arrive at:
I_{xx} = \frac{1}{12}m (3r^2 + h^2)
Due to the cylinder’s symmetry, I_{yy} is the same as I_{xx}:
I_{yy} = I_{xx} = \frac{1}{12}m (3r^2 + h^2)
For I_{zz}, the calculation is a bit different:
I_{zz} = \rho \int_V (x^2 + y^2) \, \mathrm dV = \frac{1}{2} m r^2
Finally, because of symmetry, the off-diagonal terms I_{xy}, I_{xz}, and I_{yz} are all zero.
Putting it all together, I find the moment of inertia tensor for the cylinder is:
\mathbf{I} = \begin{bmatrix} \frac{1}{12}m (3r^2 + h^2) & 0 & 0 \\ 0 & \frac{1}{12}m (3r^2 + h^2) & 0 \\ 0 & 0 & \frac{1}{2}m r^2 \end{bmatrix}
This tensor provides a complete description of how the cylinder resists rotational acceleration about any axis.
For more insights into this topic, you can find the details here.