Moment of Inertia and its Physical Significance

The moment of inertia of a rigid body about a given axis of rotation is the sum of the products of the masses of the various particles and squares of their perpendicular distances from the axis of rotation.

The value of the moment of inertia depends upon:

1. position of the axis of rotation.
2. orientation of the axis of rotation.
3. shape of the body.
4. size of the body.
5. distribution of mass of the body about the axis of rotation.

Moment of Inertia Formula

Suppose a rigid body consisting of n number of particles of masses $\displaystyle {{m}_{1}}$, $\displaystyle {{m}_{2}}$, $\displaystyle {{m}_{3}}$,…,$\displaystyle {{m}_{n}}$ situated at distances $\displaystyle {{r}_{1}}$, $\displaystyle {{r}_{2}}$, $\displaystyle {{r}_{3}}$,…,$\displaystyle {{r}_{n}}$, respectively from the axis of rotation OX as given in the figure below.

Then, by the definition, the moment of inertia of the rigid body about the axis OZ is given by

$\displaystyle I$ $\displaystyle ={{m}_{1}}r_{1}^{2}+{{m}_{2}}r_{2}^{2}+{{m}_{3}}r_{3}^{2}+\ldots +{{m}_{n}}r_{n}^{2}$

or

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\sum\limits_{{i=1}}^{{i=n}}{{{{m}_{i}}r_{i}^{2}}}}}$ $\displaystyle \ldots \left( 1 \right)$

where $\displaystyle {{{m}_{i}}}$ is the mass and $\displaystyle {{{r}_{i}}}$ is the distance of the ith particle of the rigid body from the axis of rotation OZ.

Since, the body rotates about the z-axis, therefore, the z-co-ordinate of the particles remains unchanged. If $\displaystyle \left( {{{x}_{i}},\text{ }{{y}_{i}}} \right)$ are the co-ordinates of the ith particle at any time and $\displaystyle {{{r}_{i}}}$ is the distance from the axis OZ, then

$\displaystyle r_{i}^{2}=x_{i}^{2}+y_{i}^{2}$

or

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\sum\limits_{{i=1}}^{{i=n}}{{{{m}_{i}}\left( {x_{i}^{2}+y_{i}^{2}} \right)}}}}$

If the body has a continuous uniform distribution of mass, the above equation (1) can be replaced by integration and can be rewritten as

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\int{{{{r}^{2}}dm}}}}$

Units of Moment of Inertia

The unit of moment of inertia in cgs system is g cm² and kg m²

Its dimensional formula is [M L² T⁰]

Example: Three mass points $m_1$, $m_2$, $m_3$ are located at the vertices of an equilateral triangle of length $a$ as shown in figure. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through mass point $m_1$?

Solution:

Let the mass points $m_1$, $m_2$, and $m_3$ be situated at vertices $A$, $B$, and $C$ of the equilateral triangle of each side $a$. $AQ$ is the altitude of the triangle passing through mass point $m_1$.

Then, the moment of inertia of the system about the altitude $AQ$,

$\displaystyle I$ $\displaystyle ={{m}_{1}}\times {{(\text{distance of }{{m}_{1}}\text{ from }AQ)}^{2}}$ $\displaystyle +{{m}_{2}}\times {{(\text{distance of }{{m}_{2}}\text{ from }AQ)}^{2}}$ $\displaystyle +{{m}_{3}}\times {{(\text{distance of }{{m}_{3}}\text{ from }AQ)}^{2}}$

$\displaystyle ={{m}_{1}}{{(0)}^{2}}+{{m}_{2}}{{\left( {\frac{a}{2}} \right)}^{2}}+{{m}_{3}}{{\left( {\frac{a}{2}} \right)}^{2}}$

$\displaystyle =\frac{1}{2}({{m}_{2}}+{{m}_{3}}){{a}^{2}}$