Find the moment of inertia of a uniform rign of mass M and radius R about a diameter.

Explanation

The total mass M is given by the integral of the surface density `sigma` over the area of the shell `M=intsigmadA`. Since the shell is homogeneous, the surface density `sigma` is constant.
Therefore,
`M=intsigmadA=sigmaA=4piR^(2)sigma`.
This implies that
`sum=M/(4pir^(2))`.
Calculations: In the fig. the hoop shown has a radius `r=Rsintheta` and therefore a circumference of `2piRsintheta`. The width of the hoop is `Rd theta`, so the area `dA=(2piRsintheta)(Rd theta)`, and the mass `dM=sigmadA=(M//4piR^(2))(2R^(2)sinthetad theta)=(M//2)sin thetad theta`.

We use dI to denote the moment of inertia of the hoop about the axis zz.. The perpendicular distance from the axis to each part of the hoop is `Rsintheta`. Therefore, `dI=dm(Rsintheta)^(2)anddM=(MR^(2)l^(2))sin^(3)thetad theta`. The moment of inertia of the entire shell is the sum of moment of inertia of the various hoops.
`dl=M/2sinthetad thetaR^(2)sin^(2)theta`.
Integrating over whole sphere, we get
`I=(MR^(2))/2int_(0)^(pi)sin^(3)thetad theta`.
We can write as
`sin^(3)theta=(-sin3theta+3sintheta)/4`.
Substituting in the above equation, we get
`I=(MR^(2))/8[int_(0)^(pi)sin3thetad theta+3int_(0)^(pi)sin thetad theta]`
= `(MR^(2))/8[-2/3+6]`
= `2/3MR^(2)`.
We have derived the result by adding moment of inertia of hoops.