The moment of inertia of a uniform horizontal | vedclass

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Question

(D) The moment of inertia of a solid cylinder about an axis passing through its center of mass and perpendicular to its longitudinal axis is given by:

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$\frac{49 M R^2}{4}$

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(D) The moment of inertia of a solid cylinder about an axis passing through its center of mass and perpendicular to its longitudinal axis is given by:$I_{XY} = M \left( \frac{l^2}{12} + \frac{R^2}{4} \right)$Using the parallel axis theorem, the moment of inertia $I_{AB}$ about an axis passing through its edge and perpendicular to the longitudinal axis is:$I_{AB} = I_{XY} + M \left( \frac{l}{2} \right)^2$Substituting the expression for $I_{XY}$:$I_{AB} = M \left( \frac{l^2}{12} + \frac{R^2}{4} \right) + M \frac{l^2}{4}$$I_{AB} = M \left( \frac{l^2}{3} + \frac{R^2}{4} \right)$Given that the length $l = 6R$, we substitute this into the equation:$I_{AB} = M \left[ \frac{(6R)^2}{3} + \frac{R^2}{4} \right]$$I_{AB} = M \left[ \frac{36R^2}{3} + \frac{R^2}{4} \right]$$I_{AB} = M \left[ 12R^2 + \frac{R^2}{4} \right]$$I_{AB} = M \left( \frac{48R^2 + R^2}{4} \right) = \frac{49 M R^2}{4}$

The moment of inertia of a uniform horizontal solid cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder, when its length is $6$ times its radius $R$, is:

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