Question

A solid cylinder has mass ‘$M$’, radius ‘$R$’ and length ‘$l$’. Its moment of inertia about an axis passing through its centre and perpendicular to its own axis is

• A. $\frac{2M{R}^2}{3}+\frac{M{I}^2}{12}$
• B. $\frac{M{R}^2}{3}+\frac{M{I}^2}{12}$
• C. $\frac{3M{R}^2}{4}+\frac{M{I}^2}{12}$
• D. $\frac{M{R}^2}{4}+\frac{M{I}^2}{12}$

Answer 1

Answer: (D)

Let, $X{X}^{\prime }$ be the axis of symmetry and $Y{Y}^{\prime }$ be the axis perpendicular to $X{X}^{\prime }$. Let us consider a circular disc $S$ of width $dx$ at a distance $x$ from $Y{Y}^{\prime }$ axis. Mass per unit length of the cylinder is $\frac{M}{I}.$ Thus, the mass of disc is $\frac{M}{I}dx$

Moment of inertia of this disc about the diameter of the rod
$=\left(\frac{M}{I}dx\right)\frac{R^2}{4}$
Moment of inertia of disc about $Y{Y}^{\prime }$ axis given by parallel axes theorem is
$=\left(\frac{M}{I}dx\right)\frac{R^2}{4}+\left(\frac{M}{I}dx\right)x^2$
$\therefore$ Moment of inertia of cylinder,
$I=\underset{-I/2}{\overset{I/2}{\int }}\left(\frac{M}{I}dx\right)\frac{R^2}{4}+\underset{-I/2}{\overset{I/2}{\int }}\left(\frac{M}{I}dx\right)x^2$
$=\frac{M}{I}\left[\underset{-I/2}{\overset{I/2}{\int }}\left(\frac{R^2}{4}+x^2\right)dx\right]=\frac{M}{I}{\left[\frac{R^{2}x}{4}+\frac{x^3}{3}\right]}_{-I/2}^{I/2}$
$\Rightarrow I=M\left[\frac{R^2}{4}+\frac{I^2}{12}\right]$