question_answer

What is the moment of inertia of a rod of mass M, length \[l\] about an axis perpendicular to it through one end? Given the moment of inertia about the centre of mass is \[\frac{1}{12}M{{l}^{2}}\]  

Answer:

                By using theorem of parallel axes,                 \[{{I}_{end}}={{I}_{CM}}+M{{\left( \frac{l}{2} \right)}^{2}}=\frac{1}{12}M{{l}^{2}}+\frac{1}{4}M{{l}^{2}}=\frac{1}{3}M{{l}^{2}}\]