Moment of inertia of sphere is normally expressed as;
|I = (⅖ )MR2|
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;
|I = (7/5) MR2|
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