The moment of inertia of a particle of mass m about an axis A-A is defined as \begin{align} I=mr^2, \nonumber \end{align} where r is the perpendicular distance of the particle from the axis A-A.

The moment of inertia of a system of particles is given by \begin{align} I=\sum_i m_i r_i^2. \nonumber \end{align}

The moment of inertia of a body having continuous mass distribution is given by \begin{align} I=\int_\text{body} r^2\,\mathrm{d}m, \nonumber \end{align} where $r$ is perpendicular distance of the mass element $\mathrm{d}m$ from the axis. The integration is carried out over the entire body.

Let $I_\mathrm{cm}$ be moment of inertia of a body of mass m about an axis passing through its centre of mass C. The moment of inertia of this body about a parallel axis at a perpendicular distance d from C is given by \begin{align} I_{\parallel}=I_\mathrm{cm}+md^2. \nonumber \end{align} The moment of inertia is the minimum for an axis passing through the centre of mass i.e., $ I_\mathrm{cm} \leq I_\parallel$.

Let a plane lamina lies in the x-y plane. The moment of inertia of the lamina about an axis perpendicular to its plane is given by \begin{align} I_z=I_x+I_y, \nonumber \end{align} where $I_x$ and $I_y$ are its moment of inertia about the x and y axes.

The radius of gyration ($k$) of a body of mass $m$ about an axis A-A is defined as \begin{align} k=\sqrt{I/m}, \nonumber \end{align} where $I$ is moment of inertia of the body about the axis A-A.

A symmetric lamina of mass $M$ consists of a square shape with a semicircular section over each of the edge of the square as shown in figure. The side of the square is $2a$. The moment of inertia of the lamina about an axis through its centre of mass and perpendicular to the plane is $1.6 M a^2$. The moment of inertia of the lamina about the tangent AB in the plane of the lamina is _____.

**Solution: **
Let $\mathrm{ZZ^\prime}$ be an axis perpendicular to the lamina and passing through its centre of mass O. Given $I_\mathrm{ZZ^\prime}=1.6Ma^2$.

**Problem (IIT JEE 1992):**
The moment of inertia of a thin square plate ABCD, of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is (where $I_1$, $I_2$, $I_3$ and $I_4$ are respectively moment of inertia about axes 1, 2, 3 and 4 which are in the plane of the plate.)

- $I_1+I_2$
- $I_3+I_4$
- $I_1+I_3$
- $I_1+I_2+I_3+I_4$

**Solution:**
Let I be the moment of inertia of the square plate about an axis passing through its centre O and perpendicular to its plane. The perpendicular axis theorem gives
\begin{align}
\label{iob:eqn:1}
&I_1+I_2=I, \\
\label{iob:eqn:2}
& I_3+I_4=I.
\end{align}
The mass is symmetrically distributed about axis 1 and axis 2 and also about axis 3 and axis 4. Thus,
\begin{align}
\label{iob:eqn:3}
&I_1=I_2,\\
\label{iob:eqn:4}
&I_3=I_4.
\end{align}
From above equations,
\begin{align}
I_1+I_3=\frac{1}{2}I+\frac{1}{2}I=I.
\end{align}

**Problem (IIT JEE 1998):**
Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to

- $I$
- $I\sin^2\theta$
- $I\cos^2\theta$
- $I\cos^2\left(\theta/2\right)$

**Solution:**
Let $\mathrm{A^\prime B^\prime}\perp \mathrm{AB}$ and $\mathrm{C^\prime D^\prime}\perp \mathrm{CD}$.

By perpendicular axis theorem, moment of inertia about an axis passing through centre of the plate and perpendicular to its plane is \begin{align} I_\mathrm{ZZ}&=I_\mathrm{AB}+I_\mathrm{A^\prime B^\prime} \\ &=2I_\mathrm{AB}=2I. \end{align} Also, \begin{align} I_\mathrm{ZZ}&=I_\mathrm{CD}+I_\mathrm{C^\prime D^\prime} \\ &=2I_\mathrm{CD}. \end{align} Thus, $I_\mathrm{CD}=I$.

**Problem (IIT JEE 2000):**
A thin wire of length L and uniform linear mass density $\rho$ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis $\mathrm{XX^\prime}$ is

- $\frac{\rho L^3}{8\pi^2}$
- $\frac{\rho L^3}{16\pi^2}$
- $\frac{5\rho L^3}{16\pi^2}$
- $\frac{3\rho L^3}{8\pi^2}$

**Solution:**
The mass of the wire is $m=\rho L$ and radius of the loop is $R={L}/{(2\pi)}$. The moment of inertia about an axis passing through the centre O and perpendicular to plane of the loop is
\begin{align}
I_z=mR^2.
\end{align}
Using symmetry and perpendicular axis theorem, $I_x+I_y=I_z$, we get
\begin{align}
I_x&=I_y=\frac{1}{2}I_z\\
&=\frac{1}{2}mR^2,\nonumber
\end{align}
where $I_x$ is the moment of inertia about an axis passing through O and lying in the plane of the loop. The centre of mass O of the loop lies at a perpendicular distance $d=R$ from the axis $\mathrm{XX^\prime}$. The parallel axis theorem gives
\begin{align}
I_\mathrm{XX^\prime}&=I_x+md^2\\
&=\frac{1}{2}mR^2+mR^2\\
&=\frac{3\rho L^3}{8\pi^2}.\nonumber
\end{align}

**Problem (IIT JEE 2001):**
One quarter section is cut from a uniform circular disc of radius R. This section has a mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

- $\frac{1}{2}MR^2$
- $\frac{1}{4}MR^2$
- $\frac{1}{8}MR^2$
- $\sqrt{2}MR^2$

**Solution:**
The moment of inertia of a disc of mass m and radius r about an axis passing through its centre and normal to its plane is
\begin{align}
I_\text{d}=\frac{1}{2}mr^2.
\end{align}
The mass of a quarter is m/4. Let $I_q$ be the moment of inertia of the quarter. By symmetry, $I_\text{d}=4I_q$ which gives
\begin{align}
I_q=\frac{1}{8}mr^2.
\end{align}
Substitute $m=4M$ and $r=R$ to get $I_q=\frac{1}{2}MR^2$.

You may be surprised to know that moment of inertia of any sector of a disc about an axis perpendicular to its plane and passing through its tip is $\frac{1}{2}MR^2$, where $M$ and $R$ are mass and radius of the sector. We encourage you to find moment of inertia of an infinitesimally thin triangle of mass $M$ and length $R$ about an axis perpendicular to its plane and passing through its tip. \emph{Hint:} $I=\frac{1}{2}MR^2$.

**Problem (IIT JEE 2006):**
A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to I, then the value of $r$ is equal to

- $\frac{2}{\sqrt{15}}R$
- $\frac{2}{\sqrt{5}}R$
- $\frac{3}{\sqrt{15}}R$
- $\frac{\sqrt{2}}{\sqrt{15}}R$

**Solution:**
The moment of inertia of a solid sphere of mass m and radius R about an axis passing through its centre is
\begin{align}
%\label{fob:eqn:1}
I_\text{sphere}=\frac{2}{5}mR^2. \nonumber
\end{align}
The mass m remains same when sphere is melted into a disc. The moment of inertia of a disc of mass m and radius r about its symmetry axis is
\begin{align}
%\label{fob:eqn:2}
I_\text{cm}=\frac{1}{2}mr^2. \nonumber
\end{align}
Using parallel axis theorem, moment of inertia of the disc about the desired axis is
\begin{align}
%\label{fob:eqn:3}
I_\text{disc}&=md^2+I_{cm} \\
&=mr^2+\frac{1}{2}mr^2 \\
&=\frac{3}{2}mr^2. \nonumber
\end{align}
Given $I_\text{sphere}=I_\text{disc}$. Substitute the values to get $r=2R/\sqrt{15}$.

**Problem (IIT JEE 2011):**
Four solid spheres each of diameter $\sqrt{5}$ cm and mass 0.5 kg are placed with their centers at the corner of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N\times 10^{-4}$ kg-m^{2}, then $N$ is_____

**Solution:**
The moment of inertia of each sphere about an axis passing through its centre is $\frac{2}{5}mr^2$.

**Problem ((IIT JEE 2015)):**
The densities of two solid spheres A and B of the radii R vary with radial distance r as $\rho_A(r)=k\left({r}/{R}\right)$ and $\rho_B(r)=k\left({r}/{R}\right)^5$, respectively, where $k$ is a constant. The moments of inertia of the individual spheres about axes passing through their centres are $I_A$ and $I_B$, respectively. If $I_B/I_A=n/10$, the value of $n$ is

**Solution:**
Consider a spherical shell of radius $r$ and small thickness $\mathrm{d}r$.

For planer bodies, the sum of the moments of inertia about two axes, perpendicular to each other but in the plane of the body, equals the moment of inertia of the body about the axis through the same point perpendicular to the plane.

The moment of inertia of an object can be found by suspending it from a support and allowing it to oscillate about the suspension. The time period happens to be proportional to the square root of the moment of inertia, \(T=k\sqrt{I}\).

You need a wooden plate with two bolts fixed on the sides and one bolt fixed at the center, a wire fixed with a nut, clamp-stand and stop watch.

- Suspend the plate by fixing the bolt in a nut on a side. Let the plate be suspended . Now twist the plate about the vertical axis and measure time period using stop watch. Make sufficient number of readings to be sure of the value. This gives moment of inertia \(I_x\) up to the proportionality constant.
- Open the nut and screw it in the bolt on other side. Get the time period of twist oscillations. From this find the moment of inertia \(I_y\) up to the proportionality constant.
- Again open the nut and screw it in the bolt at the center of the plate. Get the time period of twist oscillations. From this find the moment of inertia \(I_z\) up to the proportionality constant.
- Find the value of \(({I_x+I_y})/{I_z}\).