question_answer

The moment of inertia of a uniform circular disc about a tangent in its own plane is \[\frac{5}{4}M{{R}^{2}},\] where M is mass and R is the radius of the disc. Find its moment of inertia about an axis through its centre and perpendicular to its plane.

Answer:

                If \[{{I}_{D}}\] is the M.I. of the disc about its diameter, then from the theorem of parallel axes, \[{{I}_{T}}={{I}_{D}}+M{{R}^{2}}\]                 or            \[{{I}_{D}}={{I}_{T}}-M{{R}^{2}}=\frac{5}{4}M{{R}^{2}}\]               \[-M{{R}^{2}}=\frac{1}{4}M{{R}^{2}}\] By the theorem of perpendicular axes, the M.I. about an axis through the centre and perpendicular to the plane of the disc, \[I=\] Sum of moments of inertia about two perpendicular diameters \[={{I}_{D}}+{{I}_{D}}=2\times \frac{1}{4}M{{R}^{2}}\] or            \[I=\frac{1}{2}M{{R}^{2}}\].