Moment of inertia of sphere.. ✍✍✍✍✍✍✍✍✍

The moment of inertia of a sphere expression is obtained in two ways.

• First, we take the solid sphere and slice it up into infinitesimally thin solid cylinders.
• Then we have to sum the moments of exceedingly small thin disks in a given axis from left to right.

We will look and understand the derivation below.

First, we take the moment of inertia of a disc that is thin. It is given as;

I = ½ MR²

In this case, we write it as;

dI (infinitesimally moment of inertia element) = ½ r²dm

Find the dm and dv using;

dm = p dV

p = moment of a thin disk of mass dm

dv = expressing mass dm in terms of density and volume

dV = π r² dx

Now we replace dV into dm. We get;

dm = p π r² dx

And finally, we replace dm with dI.

dI = ½ p π r⁴ dx

The next step involves adding x to the equation. If we look at the diagram we see that r, R and x forms a triangle. Now we will use Pythagoras theorem which gives us;

r² = R² – x²

Now if we substitute the values we get;

dI = ½ p π (R² – x²)² dx

This leads to:

I = ½ p π _-R∫^R (R² – x²)² dx

After integration and expanding we get;

I = ½ pπ 8/15 R⁵

Additionally, we have to find the density as well. For that we use;

p = m / V

p = m / m/v πR³

If we substitute all the values;

I = 8/15 [m / m/v πR³ ] R⁵

I = ⅖ MR²