Question

Derivation of moment of inertia of solid sphere.

Answer 1

An uniform solid sphere has a radius R and mass M. calculate its moment of inertia about any axis through its centre.

Note: If you are lost at any point, please visit the beginner’s lesson or comment below.

First, we set up the problem.

Slice up the solid sphere into infinitesimally thin solid cylinders

Sum from the left to the right

Recall the moment of inertia for a solid cylinder:

I=12MR2I=12MR2

Hence, for this problem,

dI=12r2dmdI=12r2dm

Now, we have to find dm,

dm=ρdVdm=ρdV

Finding dV,

dV=πr2dxdV=πr2dx

Substitute dV into dm,

dm=ρπr2dxdm=ρπr2dx

Substitute dm into dI,

dI=12ρπr4dxdI=12ρπr4dx

Now, we have to force x into the equation. Notice that x, r and R makes a triangle above. Hence, using Pythagoras’ theorem,

r2=R2–x2r2=R2–x2

Substituting,

dI=12ρπ(R2–x2)2dxdI=12ρπ(R2–x2)2dx

Hence,

I=12ρπ∫−RR(R2–x2)2dxI=12ρπ∫−RR(R2–x2)2dx

After expanding out and integrating, you’ll get

I=12ρπ1615R5I=12ρπ1615R5

Now, we have to find what is the density of the sphere:

ρ=MVρ=MV

ρ=M43πR3ρ=M43πR3

Substituting, we will have:

I=25MR2I=25MR2

And, we’re done!

Answer 2

$\textbf{\large{Answer -}}$

The moment of inertia of solid sphere is

$\boxed {I =\frac {2}{5}{MR}^{2}}$

Refer to the attachment for explanation.