Question

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

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

Answer 1

The correct option is D $\frac{M{R}^2}{4}+\frac{M{l}^2}{12}$

Let, XX' be the axis of symmetry and YY' be the axis perpendicular to XX'.

Let us consider a circular disc S of width dx at a distance x from YY' axis.

Mass per unit length of the cylinder is $\frac{M}{l}$.

Thus, the mass of disc is $\frac{M}{l}dx$
Moment of inertia of this disc about the diameter of the rod $=\left(\frac{M}{l}dx\right)\frac{R^2}{4}$.
Moment of inertia of disc about YY' axis given by parallel axes theorem is $=\left(\frac{M}{l}dx\right)\frac{R^2}{4}+\left(\frac{M}{l}dx\right)x^2$

Moment of inertia of cylinder,

$I={\int }_{-1/2}^{1/2}\left(\frac{M}{l}dx\right)\frac{R^2}{4}+{\int }_{-1/2}^{1/2}\left(\frac{M}{l}dx\right)x^2$

$=\frac{M}{l}\left[{\int }_{-1/2}^{1/2}(\frac{R^2}{4}+x^2)dx\right]=\frac{M}{l}{[\frac{R^{2}x}{4}+\frac{x^3}{3}]}_{-1/2}^{1/2}$

$\Rightarrow I=M[\frac{R^2}{4}+\frac{l^2}{12}]$