#### Question

State the theorem of parallel axes about moment of inertia.

#### Solution

**Defination of moment of inertia :**

A measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element's distance from the axis.

**Theorem of parallel axes:-**

The moment of inertia of a body about any axis is equal to the sums of its moment of inertia about a parallel axis passing through its centre of mass and the product of its mass and the square of the perpendicular distance between the two parallel axes.

Mathematically , I_{o} = I_{c} + Mh^{2}

where I_{o} = M. I of the body about any axis passing through centre O.

I_{c} = M. I of the body about parallel axis passing through centre of mass.

h = distance between two parallel axes.

**Proof : **

i) Consider a rigid body of mass M rotating about an axis passing through a point O as shown in the following

figure.

Let C be the centre of mass of the body, situated at distance h from the axis of rotation.

ii) Consider a small element of mass dm of the body, situated at a point P.

iii) Join PO and PC and draw PD perpendicular to OC when produced.

iv) M.I of the element dm about the axis through O is (OP)^{2} dm

∴ M.I of the body about the axis thorugh O is given by

I_{o} = ∫ (OP)^{2} dm ......(1)

v) M.I of the element dm about the axis through c is CP^{2}

dm

∴ M.I of the body about the axis through C

I_{c} = ∫ (CP)^{2} dm .....(2)

vi) From the figure,

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

= (OC + CD)^{2} + PD^{2}

= OC^{2} + 2OC . CD + CD^{2} + PD^{2}

∵ CP^{2} = CD^{2} + PD^{2}

∴ OP^{2} = OC^{2} + 2 OC . CD + CP^{2} .....(3)

vii) From equation (1)

I_{o} = ∫ (OP)^{2} dm

From equation (3)

`"I"_o = int ("OC"^2 + 2 "OC" . "CD" + "CP"^2) "dm"`

`therefore "I"_o = int ("h"^2 + 2"hx" + "CP"^2) "dm"`

`= int "h"^2 "dm" + int 2 "h.x" "dm" + int "CP"^2 "dm"`

`= "h"^2 int "dm" + 2"h" int "x" "dm" + int "CP"^2 "dm"`

`"I"_o = "h"^2 int "dm" + 2"h" int "x" "dm"`

[From equation (2)]

`therefore "I"_"o" = "I"_"c" + "h"^2 int "dm" + "2h" int "x dm"` .......(4)

viii) Since ∫ dm = M and ∫ x dm = 0 and

algebraic sum of the moments of the masses of its individual particles about the centre of mass is zero

for body in equilibrium.

∴ Equation (4) becomes

I_{o} = I_{c} + Mh^{2}

Hence proved.