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 $d x$ 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{4 M}{\pi D ^2 L } \end{array} $
Mass of the element, $d m=\rho d V$
$ \begin{array}{l} \Rightarrow d m=\rho\left(\frac{\pi D ^2}{4} d x\right) \\ \Rightarrow d m=\frac{M}{L} d x \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 $I_{D i s c}=\frac{m r^2}{4}$
Parallel axis theorem is given by: $I=I_{c m}+m d^2$
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} d I=\frac{(d m) R^2}{4}+(d m) x^2 \\ \Rightarrow d I=\frac{(d m) D^2}{16}+(d m) x^2 \\ \Rightarrow d I=\frac{M D^2}{16 L} d x+\frac{M}{L} x^2 d x \end{array} $
Moment of inertia of the cylinder is, $I=\int d I$
$ \begin{array}{l} \Rightarrow I=\int_{x=\frac{-L}{2}}^{x=\frac{L}{2}} \frac{M D^2}{16 L} d x+\int_{x=\frac{-L}{2}}^{x=\frac{L}{2}} x^2 d x \\ \Rightarrow I=\left.\frac{M D^2}{16} x\right|_{-\frac{L}{2}} ^{\frac{L}{2}}+\left.\frac{M}{3 L} x^3\right|_{\frac{-L}{2}} ^2 \\ \Rightarrow I=\frac{M D^2}{16}+\frac{M I^2}{12} \end{array} $