Question

momentum of inertia for hollow and solid sphere

Answer 1

The moment of inertia of a hoop or thin hollow cylinder of negligible thickness about its central axis is a straightforward extension of the moment of inertia of a point mass since all of the mass is at the same distance R from the central axis
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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 cylindersSum from the left to the right

Recall the moment of inertia for a solid cylinder:

I=12MR2

Hence, for this problem,

dI=12r2dm

Now, we have to find dm,

dm=ρdV

Finding dV,

dV=πr2dx

Substitute dV into dm,

dm=ρπr2dx

Substitute dm into dI,

dI=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–x2

Substituting,

dI=12ρπ(R2–x2)2dx

Hence,

I=12ρπ∫−RR(R2–x2)2dx

After expanding out and integrating, you’ll get

I=12ρπ1615R5

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

ρ=MVρ=M43πR3

Substituting, we will have:

I=25MR2

And, we’re done!

Back To Mechanics (UY1)

Answer 2

for hollow sphere= 2(MR^2)/3
for solid sphere = 2(MR^2)/5