Derivation of moment of inertia of solid sphere about diameter
Answers
Answer:
Explanation:
Derivation of moment of inertia of an uniform solid sphere
moment of inertia of sphere
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=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=MV
ρ=M43πR3
Substituting, we will have:
I=25MR2
And, we’re done!