# The moment of inertia for a solid cylinder about the axis of symmetry is what?

Sep 3, 2017

It's $I = \frac{M {R}^{2}}{2}$

Where $M$ is the total mass and $R$ is the radius of the cylinder.

#### Explanation:

The moment of inertia may be defined as,

$I = \sum {m}_{i} {r}_{i}^{2}$ and if the system is continuous, then

$I = \int {r}^{2} \mathrm{dm}$

If $\rho$ is the mass density then, $\mathrm{dm} = \rho \mathrm{dV}$ where $\mathrm{dV}$ is an elementary volume.

Therefore,

$I = \int \rho {r}^{2} \mathrm{dV}$

Here we make the assumption that the mass density is constant

Therefore,

$I = \rho \int {r}^{2} \mathrm{dV}$

In order to evaluate this integral, we exploit the cylindrical symmetry of the system and employ cylindrical polar coordinates.

Thus, moment of inertia along axis of cylinder which is assumed to be the $z$ axis is,

${I}_{z} = \rho \int \int \int {r}^{2} \mathrm{dV}$

Limits of $s$ is from $0$ to $R$ where $R$ is the radius. Limits of $\theta$ are from $0$ to $2 \pi$ and limits of $z$ are from $0$ to $h$ where $h$ is the height of the cylinder.

$\implies {I}_{z} = \rho {\int}_{0}^{2 \pi} d \theta {\int}_{0}^{R} {s}^{3} \mathrm{ds} {\int}_{0}^{h} \mathrm{dz}$

Where in this case, ${r}^{2} = {s}^{2}$ and $\mathrm{dV} = s \mathrm{ds} d \theta \mathrm{dz}$ in cylindrical coordinates.

$\implies {I}_{z} = \frac{\rho 2 \pi {R}^{4}}{4} h$

But, $\rho = \frac{M}{\pi {R}^{2} h}$ where $M$ is total mass.

Therefore, ${I}_{z} = \frac{M {R}^{2}}{2}$