Question

Find the moment of the inertia of a uniform rod of length L , mass M about an axis passing through its center and perpendicular to its length .

Answer 1

$$\begin{gathered} \mathrm{Moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{a}\mathrm{rod}\mathrm{about}\mathrm{its}\mathrm{centre}\mathrm{of}\mathrm{mass}\mathrm{and}\mathrm{perpendicular}\mathrm{to}\mathrm{length}: \\ \mathrm{I}={\int }_0^{\mathrm{M}}{\mathrm{r}}^{2}d\mathrm{m} \\ \mathrm{where} \\ \mathrm{M}=\mathrm{Total}\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{rod} \\ \mathrm{r}=\mathrm{Distane}\mathrm{from}\mathrm{centre}\mathrm{of}\mathrm{mass} \\ \mathrm{dm}=\mathrm{Point}\mathrm{mass} \\ \mathrm{L}=\mathrm{Total}\mathrm{length} \\ \mathrm{Expressing}\mathrm{the}\mathrm{mass}\mathrm{element}\mathrm{interms}\mathrm{of}\mathrm{length}: \\ \mathrm{I}={\int }_{-\frac{\mathrm{L}}{2}}^{\frac{\mathrm{L}}{2}}{\mathrm{r}}^2\frac{\mathrm{M}}{\mathrm{L}}\mathrm{dr} \\ =\frac{\mathrm{M}}{\mathrm{L}}{\int }_{-\frac{\mathrm{L}}{2}}^{\frac{\mathrm{L}}{2}}{\mathrm{r}}^2\mathrm{dr} \\ =\frac{\mathrm{M}}{\mathrm{L}}{\left[\frac{{\mathrm{r}}^3}{3}\right]}_{-\frac{\mathrm{L}}{2}}^{\frac{\mathrm{L}}{2}} \\ =\frac{\mathrm{M}}{3\mathrm{L}}\left\{{\left[\frac{\mathrm{L}}{2}\right]}^3-{\left[-\frac{\mathrm{L}}{2}\right]}^3\right\} \\ =\frac{\mathrm{M}}{3\mathrm{L}}\left[\frac{{\mathrm{L}}^3}{8}+\frac{{\mathrm{L}}^3}{8}\right] \\ =\frac{\mathrm{M}}{3\mathrm{L}}\times \frac{{\mathrm{L}}^3}{4} \\ \mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}{\mathbf{ML}}^{\mathbf{2}} \end{gathered}$$