Moment of Inertia, Parallel Axes and Perpendicular Axes Theorems

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.

Theorem of Parallel Axes

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., $I_\mathrm{cm} \leq I_\parallel$.

Theorem of Perpendicular Axes

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.

Solved Problems from IIT JEE

Problem from IIT JEE 2005

From a circular disc of radius $R$ and mass $9M$, a small disc of radius $R/3$ is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is,

1. $4MR^2$
2. $\frac{40}{9}MR^2$
3. $10MR^2$
4. $\frac{37}{9}MR^2$

Solution: The moment of inertia of disc of mass $9M$ and radius $R$ about an axis perpendicular to its plane and passing through its centre O is, \begin{align} I_\text{total}&=\frac{1}{2}(9M)R^2 \\ &=\frac{9}{2}MR^2.\nonumber \end{align} The mass of removed disc is $\frac{9M}{\pi R^2}\frac{\pi R^2}{9}=M$. The parallel axis theorem gives moment of inertia of the removed disc about axis passing through $O$ as, \begin{align} I_\text{removed}&=\frac{1}{2}M\left(\tfrac{R}{3}\right)^{\!2}+Md^2 \\ &=\tfrac{1}{18}MR^2+M\left(\tfrac{2R}{3}\right)^{\!2} \\ &=\tfrac{1}{2}MR^2.\nonumber \end{align} Using, $I_\text{total}=I_\text{remaining}+I_\text{removed}$, we get $I_\text{remaining}=4MR^2$.

Problem from IIT JEE 1997

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 $I_\mathrm{ZZ^\prime}=1.6Ma^2$.

By theorem of perpendicular axes, \begin{align} \label{rob:eqn:1} I_\mathrm{XX^\prime}+I_\mathrm{YY^\prime}=I_{ZZ^\prime}. \end{align} By symmetry, $I_\mathrm{XX^\prime}=I_\mathrm{YY^\prime}$. Substitute in above equation to get \begin{align} I_\mathrm{XX^\prime}&=\frac{1}{2}I_\mathrm{ZZ^\prime} \\ &=0.8Ma^2. \end{align} The theorem of parallel axes gives, \begin{align} I_\mathrm{AB}&=I_\mathrm{XX^\prime}+Md^2 \\ &=I_\mathrm{XX^\prime}+M(2a)^2 \\ &=4.8Ma^2.\nonumber \end{align}