Question

A solid cylinder has diameter $D$ and length $L$ . Find its moment of inertia about an axis perpendicular to its length and passing through the centre of gravity.

• A. $M\left[\frac{D^2}{4}+\frac{L^2}{12}\right]$
• B. $M\left[\frac{D^2}{16}+\frac{L^2}{12}\right]$
• C. $M\left[\frac{D^2}{8}+\frac{L^2}{16}\right]$
• D. $M\left[\frac{D^2}{4}+\frac{L^2}{6}\right]$

Answer 1

Answer: (B)

Let us consider an element of the cylinder, which is a disc of thickness $dx$ at a distance $x$ from the center.
Density of cylinder, $\rho =\frac{\text{ Mass }}{\text{ Volume }}$
$\begin{array}{l}\Rightarrow \rho =\frac{M}{\pi {\left(\frac{D}{2}\right)}^{2}L}\\ \Rightarrow \rho =\frac{4M}{\pi D^{2}L}\end{array}$
Mass of the element, $dm=\rho dV$
$\begin{array}{l}\Rightarrow dm=\rho \left(\frac{\pi D^2}{4}dx\right)\\ \Rightarrow dm=\frac{M}{L}dx\end{array}$
Moment of inertia of a disc of mass $m$ and radius $r$ about an axis passing through its center and lying in its plane is $I_{Disc}=\frac{m{r}^2}{4}$
Parallel axis theorem is given by: $I=I_{cm}+m{d}^2$
Using these concepts we can say that moment of inertia of the given element of the cylinder about an axis passing through the center of the cylinder is:
$\begin{array}{l}dI=\frac{(dm)R^2}{4}+(dm)x^2\\ \Rightarrow dI=\frac{(dm)D^2}{16}+(dm)x^2\\ \Rightarrow dI=\frac{M{D}^2}{16L}dx+\frac{M}{L}{x}^{2}dx\end{array}$
Moment of inertia of the cylinder is, $I=\int dI$
$\begin{array}{l}\Rightarrow I={\int }_{x=\frac{-L}{2}}^{x=\frac{L}{2}}\frac{M{D}^2}{16L}dx+{\int }_{x=\frac{-L}{2}}^{x=\frac{L}{2}}{x}^{2}dx\\ \Rightarrow I={\frac{M{D}^2}{16}x|}_{-\frac{L}{2}}^{\frac{L}{2}}+{\frac{M}{3L}{x}^3|}_{\frac{-L}{2}}^2\\ \Rightarrow I=\frac{M{D}^2}{16}+\frac{M{I}^2}{12}\end{array}$