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 = \(\frac{1}{2}\)MR2

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.

Moment Of Inertia Of Solid Cylinder

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 – \(\pi\)r2

dA = \(\pi\)(r2 + 2rdr + (dr)2) – \(\pi\)r2

Notably, here the (dr)2 = 0.

dA = 2\(\pi\)r dr

2. Substitution of dA into dV we get;

dV = 2\(\pi\)r L dr

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

dm = 2\(\pi\)r L dr

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

dl = 2\(\pi\)r3 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 \(\int_{R1}^{R2}\)r3 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 = \(\frac{1}{2}\)MR2