Moment Of Inertia Of Sphere

Moment of inertia of sphere is normally expressed as;

Here, R and M are the radius and mass of the sphere respectively. Students have to keep in mind that we are talking about the moment of inertia of a solid sphere about its central axis above. Additionally, if we talk about the moment of inertia of the sphere about its axis on the surface it is expressed as;

Moment Of Inertia Of Sphere Derivation

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 = ρ dv

ρ = density of a thin disk of mass dm

dv = volume of the thin disk

dv = π r² dx

Now we replace dV into dm. We get;

dm = ρ π r² dx

And finally, we replace dm with dI.

dI = (½) ρ 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 the Pythagoras theorem which gives us;

r² = R² – x²

Now if we substitute the values we get;

dI = ½ ρπ (R² – x²)² dx

This leads to:

I = ½ ρ π _-R∫^R (R² – x²)² dx

After integration and expanding we get;

I =ρ π (8/15) R⁵

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

ρ = M / V

ρ  = [M / (4/3) πR³]

If we substitute all the values;

I = 8/15 [M / (4/3) πR³] πR⁵

I = ⅖ MR²

⇒ Check Other Object’s Moment of Inertia:

• Moment Of Inertia Of A Hollow Sphere
• Moment Of Inertia Of Hollow Cone
• Moment Of Inertia Of A Quarter Circle
• Moment Of Inertia Of A Cube
• Moment Of Inertia Of Flywheel

Parallel Axis theorem