Question

find the moment of inertia of thin uniform rod mass 3kg,length 150cm rotating about an axis perpendicular to its length and (1)passing through its centre. (2) one of its ends

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Mass}\mathrm{of}\mathrm{rod},\mathrm{M}=3\mathrm{kg} \\ \mathrm{Length},\mathrm{L}=150\mathrm{cm}=1.50\mathrm{m} \\ \mathrm{Moment}\mathrm{of}\mathrm{inertia}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{for}\mathrm{the}\mathrm{two}\mathrm{cases}, \\ \left(1\right)\mathrm{Passing}\mathrm{through}\mathrm{centre}, \\ \mathrm{I}=\frac{{\mathrm{ML}}^2}{12} \\ =\frac{3\times 1.{50}^2}{12}=0.5625\mathrm{kg}{\mathrm{m}}^2 \\ \left(2\right)\mathrm{One}\mathrm{of}\mathrm{its}\mathrm{ends} \\ \mathrm{I}=\frac{{\mathrm{ML}}^2}{3}=\frac{3\times 1.{50}^2}{3}=2.25\mathrm{kg}{\mathrm{m}}^2 \end{gathered}$$