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 = 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