question_answer

The Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is\[1/2M{{R}^{2}}\]. Derive the values of moment of inertia of the disc about its diameter and about an axis tangential to the disc lying on its plane?

Answer:

                M.I. of a disc about any diameter. In Fig. AB and CD are two mutually perpendicular diameters in the plane of the disc. Applying the theorem of perpendicular axes, we get \[{{I}_{AB}}+{{I}_{CD}}={{I}_{YY'}}\] or \[{{I}_{D}}+{{I}_{D}}=\frac{1}{2}M{{R}^{2}}\] or \[{{I}_{D}}=\frac{1}{4}M{{R}^{2}}\] M.I. of a disc about a tangent in its plane. Let \[{{I}_{T}}\] the moment of inertia of the disc about a tangent EBF in the plane of the disc. This tangent is parallel to the diameter CD of the disc. Applying the theorem of parallel axes, we get \[{{I}_{T}}=Moment\,of\,inertia\,of\,disc\,about\,CD+M{{R}^{2}}\]or \[{{I}_{T}}=\frac{1}{4}M{{R}^{2}}+M{{R}^{2}}=\frac{5}{4}M{{R}^{2}}\]