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 \[I=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{M}{{\mathbf{R}}^{\mathbf{2}}}\].
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