Moment of inertia of a disc about its own axi | vedclass

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(A) The moment of inertia $(I)$ of a disc of mass $M$ and radius $R$ about its central axis perpendicular to its plane is given by $I = \frac{1}{2}MR^

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$\frac{5}{2}I$

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(A) The moment of inertia $(I)$ of a disc of mass $M$ and radius $R$ about its central axis perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.According to the perpendicular axis theorem,the moment of inertia about a diameter $(I_d)$ is $I_d = \frac{I}{2} = \frac{1}{4}MR^2$.To find the moment of inertia about a tangential axis in the plane of the disc,we use the parallel axis theorem: $I_{tangent} = I_d + MR^2$.Substituting $I_d = \frac{I}{2}$ and $MR^2 = 2I$ (since $I = \frac{1}{2}MR^2 \implies MR^2 = 2I$):$I_{tangent} = \frac{I}{2} + 2I = \frac{5}{2}I$.

Moment of inertia of a disc about its own axis is $I$. Its moment of inertia about a tangential axis in its plane is

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