#### Question

Prove the theorem of parallel axes about moment of inertia

#### Solution

Consider a rigid of mass ‘M’ rotating about an axis passing through a point ‘O’ and perpendicular to the plane of the figure.

Let ‘I_{o}’ be the moment of inertia of the body about an axis passing through point ‘O’.

Take another parallel axis of rotation passing through the centre of mass of the body.

Let ‘I_{c}’ be the moment of inertia of the body about point ‘C’.

Let the distance between the two parallel axes be OC = h.

OP = r and CP = r_{o}

Take a small element of body of mass ‘dm’ situated at a point P. Join OP and CP, then

`I_@=intOP^2dm=intr^2dm`

`I_C=intCP^2dm=intr_0^2dm`

From point P draw a perpendicular to OCproduced.

Let CD = x

From the figure,

OP^{2} = OD^{2 }+ PD^{2}

∴OP^{2}=(h + CD)^{2} + PD^{2}

^{ }=h^{2} + CD^{2} + 2hCD + PD^{2}

∴OP^{2}= CP^{2} + h^{2} + 2hCD ............(CD^{2} + PD^{2}=CP^{2})

∴`r^2=r_0^2+h^2+2hx`

Multiplying the above equation with 'dm' on both the sides and integrating, we get

`intr^2dm=intr_0^2dm+inth^2dm+int2hxdm`

∴`intr^2dm=intr_0^2+inth^2dm+2hintxdm`

`intxdm=0 `as 'C' is the centre of mass and algebriac sum of moments of all the particles about the center of mass always zero, for body in equilibrium.

`therefore intr^2dm=intr_0^2dm+h^2intdm+0.........(Equ.1)`

the particles about the centre of mass is always zero, for body in equilibrium.

But`intdm= "mass of the body"`

`intr^2dm=I_@and intr_0^2dm=I_c`

Substituing in equation 1, we get

I_{0}=I_{C} + Mh^{2}

This proves the theorm of parallel axes about moment of inertia.