Moment of Inertia – Definition, Formula, Equation and Unit

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]

Radius of Gyration

It is defined as the distance from the axis of rotation at which, if the whole mass of the body were supposed to be concentrated, the moment of inertia would be the same as with the actual distribution of the mass of the body into small particles. It is denoted by $K$.

If $M$ is the mass of the body and $K$ is its radius of gyration about the axis $OZ$,

 

 

then the moment of inertia of the body about the axis $OZ$ is:

\[
I = M K^2
\]

From equation (1), we get:

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

If each of the $n$ particles constituting the rigid body is of mass $m$, then

$\displaystyle M{{K}^{2}}$ $\displaystyle =mr_{1}^{2}+mr_{2}^{2}+mr_{3}^{2}+\ldots +mr_{n}^{2}$ $\displaystyle =m\left[ {r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+\ldots +r_{n}^{2}} \right]$

Since $m \times n = M$, the mass of the body, we have:

$\displaystyle M{{K}^{2}}$ $\displaystyle =M\left( {\frac{{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+\ldots +r_{n}^{2}}}{n}} \right)$

or

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle K=\sqrt{{\frac{{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+\ldots +r_{n}^{2}}}{n}}}}}$

i.e.,

$\displaystyle K$ = root mean square distance of the constituting particles from the axis of rotation

Thus, the radius of gyration of a body about an axis of rotation may also be defined as the root mean square average of the particles from the axis of rotation and its square when multiplied with the mass of the body gives the moment of inertia of the body about that axis.

From the study of moment of inertia and the radius of gyration, the following conclusions can be drawn:

1. The moment of inertia of the body depends upon the mass of the body as well as the manner in which mass is distributed about the axis of rotation. It is because the moment of inertia of the body changes with the change in the position of the axis of rotation.
2. The radius of gyration of a body is not a constant quantity. Its value changes with the change of location of the axis of rotation.

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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 $AQ$,

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

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

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

---

Theorem of Parallel Axes

According to this theorem, moment of inertia of a rigid body about any axis AB is equal to moment of inertia of the body about another axis KL passing through center of mass C of the body in a direction parallel to AB, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes.

 

 

If h is the perpendicular distance between the axes AB and KL, then according to the theorem of parallel axes,

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle {{I}_{{AB}}}={{I}_{{KL}}}+M{{h}^{2}}}}$

Theorem of Perpendicular Axes

According to this theorem, the moment of inertia of a plane lamina (i.e. a two dimensional body of any shape or size) about any axis OZ perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually perpendicular axes OX and OY in the plane of the lamina, meeting at a point where the given axis OZ passes through the lamina.

In the figure below,

 

 

$\displaystyle {{I}_{x}}$= moment of inertia of the lamina about OX

$\displaystyle {{I}_{y}}$= moment of inertia of the lamina about OY

$\displaystyle {{I}_{z}}$= moment of inertia of the lamina about OZ

According to the theorem of perpendicular axes,

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle {{I}_{z}}={{I}_{x}}+{{I}_{y}}}}$

Calculate Moment of Inertia

Calculating Moment of Inertia of Straight Rod

Consider a uniform straight rod AB of length $l$, mass $M$ and center C. Then, mass per unit length of the rod $= \dfrac{M}{l}$

(i) Moment of inertia about an axis through its center.

Let the moment of inertia of the rod about the axis YY’ passing through its centre C and perpendicular to its length be $I$. Consider a small element of length $dx$ of the rod at a distance $x$ from the point C.

 

 

The mass of the elementary portion of the rod $= \dfrac{M}{l} dx$

The moment of inertia of the elementary portion of the rod about the axis YY’ is:

$\displaystyle =$ mass of the elementary portion × $\displaystyle {{(\text{distance from the axis Y{Y}’})}^{2}}$

\[
= \frac{M}{l} dx \times x^2 = \frac{M}{l} x^2 dx
\]

The moment of inertia ($I$) of the rod AB about the axis YY’ can be found by integrating the above for all such elementary portions between the points A and B, i.e. by integrating the above between the limits $x = -l/2$ to $x = +l/2$:

$\displaystyle I=\int_{{-l/2}}^{{+l/2}}{{\frac{M}{l}}}{{x}^{2}}dx$ $\displaystyle =\frac{M}{l}\int_{{-l/2}}^{{+l/2}}{{{{x}^{2}}}}dx$ $\displaystyle =\frac{M}{l}\left[ {\frac{{{{x}^{3}}}}{3}} \right]_{{-l/2}}^{{+l/2}}$

$ \displaystyle =\frac{M}{l}{{\left( {\frac{l}{2}} \right)}^{3}}-{{\left( {-\frac{l}{2}} \right)}^{3}}$ $\displaystyle =\frac{M}{l}\times \frac{{2{{l}^{3}}}}{8}$

$ \displaystyle =$ $\displaystyle \frac{{M{{l}^{2}}}}{{12}}$

or

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\frac{1}{{12}}M{{l}^{2}}}}$

(ii) Moment of inertia about an axis through one end.

Let $I’$ be the moment of inertia of the rod about an axis LM passing through the end A and perpendicular to its length. The axis LM is at a distance $\frac{l}{2}$ from the axis YY’ through the center of mass of the rod.

 

 

Therefore, according to the theorem of parallel axis, the moment of inertia of the rod about the axis LM is given by:

$ \displaystyle {I}’=I+M{{\left( {\frac{l}{2}} \right)}^{2}}$

$ \displaystyle {I}’=\frac{1}{{12}}M{{l}^{2}}+\frac{1}{4}M{{l}^{2}}$

or

$ \bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle {I}’=\frac{1}{3}M{{l}^{2}}}}$

Calculating Moment of Inertia of a Circular Ring

Consider a ring of mass $M$, radius $R$ and center $O$. Then, mass per unit length of the circumference of the ring $= \dfrac{M}{2 \pi R}$

(i) Moment of inertia about an axis through its center and perpendicular to its plane

Let the moment of inertia of the circular ring about the axis $ZZ’$ passing through its center $O$ and perpendicular to its plane. Consider an elementary portion of the ring.

 

 

Then, mass of the elementary portion $= \dfrac{M}{2\pi R} dx$

As the ring rotates about the axis $ZZ’$, the elementary portion is always at a distance $R$ from it. Therefore, the moment of inertia of the elementary portion about $ZZ’$:

$ \displaystyle =\left( {\frac{M}{{2\pi R}}dx} \right){{R}^{2}}=\frac{M}{{2\pi }}Rdx$

The moment of inertia $I$ of the ring about axis $ZZ’$ can be found by integrating the above for the whole circumference of the ring i.e. between limits $x = 0$ to $x = 2 \pi R$:

$\displaystyle I=\int_{0}^{{2\pi R}}{{\frac{M}{{2\pi }}}}Rdx$

$ \displaystyle \Rightarrow I=\frac{M}{{2\pi }}R\left[ x \right]_{0}^{{2\pi R}}$

$ \displaystyle \Rightarrow I=\frac{M}{{2\pi }}R[2\pi R-0]$

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

(ii) Moment of inertia about its diameter

Let $I_d$ be the moment of inertia of the circular ring about the diameter $XX’$. Let $YY’$ be another diameter of the ring perpendicular to the diameter $XX’$.

 

 

Since the ring is symmetrical about any of its diameters, the moment of inertia of the ring about the diameter $YY’$ is also equal to $I_d$. According to the theorem of perpendicular axes:

M.I. of the ring about ZZ’ = M.I. about XX’ + M.I. about YY’

i.e.

$\displaystyle I={{I}_{d}}+{{I}_{d}}=2{{I}_{d}}$

$\displaystyle \Rightarrow {{I}_{d}}=\frac{1}{2}I$

or

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle {{I}_{d}}=\frac{1}{2}M{{R}^{2}}}}$

---

Example: The moment of inertia of a circular ring about an axis passing through its centre and perpendicular to its plane is $200\, \text{g cm}^2$. If the radius of the ring is $5 \, \text{cm}$, find the mass of the ring.

Solution:

Here, $I = 200 \, \text{g cm}^2$, $R = 5 \, \text{cm}$

Now, moment of inertia of the ring:

\[
I = M R^2
\]

\[
M = \frac{I}{R^2} = \frac{200}{(5)^2} = 8 \, \text{g}
\]

---

Calculating Moment of Inertia of Circular Disc

Consider a circular disc of mass $M$, radius $R$, and center $O$. Then,

\[
\text{mass per unit area} = \frac{M}{\pi R^2}
\]

(i) Moment of inertia about an axis through its center and perpendicular to its plane.

Let the moment of inertia of the circular disc about the axis $ZZ’$ passing through its center $O$ and perpendicular to its plane be $I$.

 

 

The disc can be assumed to be made up of a large number of concentric rings. Consider one such ring of radius $x$ and an infinitesimally small thickness $dx$. Then,

area of ring = circumference × width $\displaystyle =2\pi xdx$

mass of ring $\displaystyle =\frac{M}{{\pi {{R}^{2}}}}\times 2\pi xdx$ $\displaystyle =\frac{{2M}}{{{{R}^{2}}}}xdx$

Since, the moment of inertia of this ring about the axis $ZZ’$ passing through its center and perpendicular to its plane is $\displaystyle I=M{{R}^{2}}$

M.I. of ring = mass of ring × radius²

$\displaystyle =\frac{{2M}}{{{{R}^{2}}}}{{x}^{2}}dx=\frac{{2M}}{{{{R}^{2}}}}{{x}^{3}}dx$

The moment of inertia $I$ of the whole disc is the sum of the moment of inertia of all such rings, which have radii between $x = 0$ and $x = R$ and is accordingly obtained by integrating the above between the limits $x = 0$ and $x = R$.

Therefore,

\[
I = \int_0^R \frac{2M}{R^2} x^3 \, dx
\]

or

$\displaystyle I=\frac{{2M}}{{{{R}^{2}}}}\times \left[ {\frac{{{{x}^{4}}}}{4}} \right]_{0}^{R}$

$ \displaystyle \Rightarrow I=\frac{{2M}}{{{{R}^{2}}}}\times \frac{{{{R}^{4}}}}{4}=\frac{{2M}}{{{{R}^{2}}}}\times \frac{{{{R}^{4}}}}{4}$

$ \bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle I=\frac{{M{{R}^{2}}}}{2}}}$

(ii) Moment of inertia about its diameter.

As deduced in section 2.05, it can be obtained that the moment of inertia of the circular disc about its diameter:

\[
I_d = \frac{1}{2} I = \frac{1}{2} \times \left( \frac{1}{2} M R^2 \right)
\]

or

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle {{I}_{d}}=\frac{1}{4}M{{R}^{2}}}}$

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Example: Calculate the moment of inertia about a transverse axis through the centre of a disc, whose radius is $20 \, \text{cm}$. Its density is $9 \, \text{g cm}^{-3}$ and its thickness is $7 \, \text{cm}$.

Solution:

Here, radius of disc, $R = 20 \, \text{cm}$; thickness of disc, $t = 7 \, \text{cm}$

Density of the material of ring, $\rho = 9 \, \text{g cm}^{-3}$

Therefore, the mass of disc:

$\displaystyle M=\pi {{R}^{2}}t\rho $ $\displaystyle =\frac{{22}}{7}\times {{(20)}^{2}}\times 7\times 9\text{g}$

Moment of inertia of the disc about transverse axis (axis perpendicular to its plane) and through its center is given by

$\displaystyle \begin{array}{l}I=\frac{1}{2}M{{R}^{2}}\\=\frac{1}{2}\times \frac{{22}}{7}\times {{(20)}^{2}}\times 7\times 9\times {{(20)}^{2}}\\=1.584\times {{10}^{7}}\text{ g c}{{\text{m}}^{2}}\end{array}$