question_answer

What is the moment of inertia of a uniform circular disc of radius R and mass M about an axis passing through its centre and normal to the disc?  The moment of inertia of the disc about any of its diameters is given to be\[\left( \mathbf{1}\text{ }/\text{ }\mathbf{4} \right)\text{ }\mathbf{M}{{\mathbf{R}}^{\mathbf{2}}}\]. 

Answer:

                M.I. of the disc about any diameter, \[{{I}_{D}}=\frac{1}{4}M{{R}^{2}}.\] By the theorem of perpendicular axes, \[\text{M}.I\] of the disc about an axis through the centre and normal to disc \[=\text{ M}.I\] about any diameter\[+\]\[\text{M}.I\] about perpendicular diameter \[I={{I}_{D}}+{{I}_{D}}=2\times \frac{1}{4}M{{R}^{2}}\] or            \[I=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{M}{{\mathbf{R}}^{\mathbf{2}}}\].