(a) Prove the theorem of perpendicular axes.
(Hint : Square of the distance of a point `(x, y)` in the x-y plane from an axis through the origin perpendicular to the plane is `x^(2)+y^(2)`).
(b) Prove the theorem of parallel axes.
(Hint : If the centre of mass is chosen to be the origin `Sigmam_(1)(r_(i)=0)`.

Explanation

(a) Proof of theorem of perpendicular axes :
The moment of inertia of a planar body about an axis (z) perpendicular to its plane `(x-y)` is equal to the sum of its moment of inertia about two perpendicular axes (x and y) concurrent with perpendicular axes (x and y) concurrent with perpendicular axis and lying in the plane of the body.
`therefore I_(z)=I_(x)+I_(y)`
If planar body in yz plane then
`I_(x)=I_(y)+I_(z)`
If planar body in xz plane then
`I_(y)=I_(x)+I_(y)`
This theorem is applied to planar body.
Suppose planar body consists of many particles each of mass of m. Suppose the coordinate of P having mass m is `(x,y)` and distance r from origin.

Momentum of inertia of a particle about an axis OZ.
`I=mr^(2)`
If the moment of inertia of whole planar body about z-axis is `I_(2)` then
`I_(z)=summr^(2)`
`=summ(x^(2)+y^(2))`
`therefore I_(z)=summx^(2)+summy^(2)....(1)`
Now momentum of inertia of whole planar body about an axis x and y is `I_(x) and I_(y)`
`I_(x)=summx^(2)andI_(y)=summy^(2)`
`therefore I_(z)=I_(x)+I_(y)` from eqn. (1) it is proved.
(b) Proof of theorem of parallel axes :
The theorem of parallel axes states that the moment of inertia `I_(C )` of the body about a parallel axis passing through its centre of mass and the product of its mass M and the square of the distance d between the two parallel axes
`therefore I=I_(C )+Md^(2)`
where `I_(C )` = Moment of inertia about an axis passing through the centre of mass
I = Moment of inertia about an axis passing through centre of mass parallel to any axis.
M = Total mass of body `underset(i=l)overset(n)summ_(i)`
d = perpendicular distance between two parallel axes

LL. axis passing through the centre of mass Total mass of system of n particles
`M=underset(i=l)overset(n)summ_(i)`
Suppose any particle of body P is, LL. is `r_(i)` distance away axis
The perpendicular distance of P from axis
`ZZ.=OC+CP`
`=d+r_(i)`
Moment of inertia about an axis passing through centre of mass
`I_(C)=underset(i=l)overset(n)summ_(i)r_(i)^(2)....(1)`
If `I_(i)` is the moment of inertia of a point P about an axis ZZ.
`I_(i)=m_(i)(OP)^(2)=m_(i)(r_(i)+d)^(2)`
If moment of inertia of whole body about an axis ZZ. is I, then
`I=underset(i=1)overset(n)sumI_(i)`
`=underset(i=1)overset(n)summ_(i)(r_(i)+d)^(2)`
`=underset(i=l)overset(n)summ_(i)(r_(i)^(2)+2r_(i)d+d^(2))`
`=underset(i=1)overset(n)summ_(i)r_(i)^(2)+underset(i=1)overset(n)sum2m_(i)r_(i)d+underset(i=1)overset(n)summ_(i)d^(2)`
`=I_(C)+0+Md^(2) [because underset(i=1)overset(n)summ_(i)r_(i)=0,underset(i=1)overset(n)summ_(i)=M]`
`I=I_(C)+Md^(2)` proved.