State and prove parallel axis theorem.

Explanation

Parallel axis theorem : The moment of inertia of a rigid body about an axis passing through a point is the sum of moment of inertia about parallel axis passing through centre of mass `(I_G)` and mass of the body multiplied by square of distance `(MR^2)` between the axes i.e.,
`I = I_G + MR^2`
Proof : Consider a rigid body of mass M with 'G' as its centre of mass. `I_G` the moment of inertia about an axis passing through centre of mass. I = The moment of inertia about an axis passing through the point 'O ' in that plane.
Let perpendicular distance between the axes is OG = R (say)
Consider point P in the given plane. Join OP and GP. Extend the line OG and drop a normal from 'P' on to it as shown in figure.

The moment of inertia about the axis passing through centre of mass G.
`(I_G) = Sigma m GP^2 " " ....(1)`
M.O.I of the body about an axis passing through O (I) `= Sigma mOP^2 " " ...(2)`
From triangle OPD
`OP^2 = OD^2 + DP^2`
`rArr OD = OG +GD`
`therefore OD^2 = (OG +GD)^2 = OG^2 + GD^2 + 2OG, GD " "...(3)`
From equations (2) and (3)
` I = Sigmam OP^2 = Sigma m [(OG^2 + GD^2 + 2OG . GD)+ DP^2]`
`therefore I = Sigma m {GD^2 + DP^2 + OG^2 + 2OG . GN}`
But ` GD^2 + DP^2 = GP^2`
`therefore I = Sigma m { GP^2 + OG^2 + 2OG . GP}`
`therefore I = Sigma m GP^2 + Sigma m OG^2 + 2OG Sigma mGD " " ....(4)`
But the terms `Sigma mGP^2 = I_G`
`Sigma mOG^2 = MR^2 (becuase Sigmam = " M and OG = R)`
The term `2OG Sigma mGD` = 0 , Because it represent sum of moment of masses about centre of mass . Hence its value is zero .
` therefore I = I_G +MR^2`
Hence parallel axis theorem is proved .