 # Moment Of Inertia Of A Solid Cylinder

Moment of inertia of a solid cylinder about its centre can be found using the following equation or formula;

 $I = \frac{1}{2}MR^{2}$

Here, M = total mass and R = radius of the cylinder.

## Derivation Of Moment Of Inertia Of Solid Cylinder

We will take a solid cylinder with mass M, radius R and length L. We will calculate its moment of inertia about the central axis. Here we have to consider a few things:

• The solid cylinder has to be cut or split into infinitesimally thin rings.
• Each ring consists of the thickness of dr with length L.
• We have to sum up the moments of infinitesimally these thin cylindrical shells.

We will follow the given steps.

1. We will use the general equation of moment of inertia:

dI = r2 dm

Now we move on to finding the dm. It is normally given as;

dm = ρ dV

In this case, the mass element can be expressed in terms of an infinitesimal radial thickness dr by;

dm = 2r L dr

In order to obtain dm we have to calculate dv first. It is given as;

dV = dA L

Meanwhile, dA is the area of the big ring (radius: r + dr) minus the smaller ring (radius: r). Hence;

$dA = \pi (r + dr)^{2} – \pi r^{2}$ $dA = \pi (r + 2rdr + (dr)^{2}) – \pi r^{2}$

Notably, here the (dr)2 = 0.

$dA = 2\pi r dr$

2. Substitution of dA into dV we get;

$dV = 2\pi L dr$

Now, we substitute dV into dm and we will have;

$dm = 2\pi L dr$

The dm expression is further substituted into the dI equation and we get;

$dl = 2\pi r^{3} L dr$

3. Alternatively, we have to find the expression for density as well. We use the equation;

$p = \frac{M}{V}$

Now,

$p = \frac{M}{\pi R^{2}L}$

4. The final step involves using integration to find the moment of inertia of the solid cylinder. The integration basically takes the form of a polynomial integral form.

$I = 2P \pi r L\int_{R_{1}}^{R_{2}} r^{3} dr$ $I = 2P \pi\, L\, \frac{r^{4}}{4}$ $I = 2\pi [\frac{m}{\pi R^{2}L}] L \frac{R^{4}}{4}$

Therefore, $I = \frac{1}{2}MR^{2}$