The moment of inertia of a thin circular disc | vedclass

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(D) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^

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$\frac{5}{2} I$

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(D) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$. From this, we have $M R^2 = 2 I$. The moment of inertia of the disc about its diameter is $I_d = \frac{1}{4} M R^2$. Using the parallel axis theorem, the moment of inertia about an axis parallel to the diameter and touching the edge of the rim is $I' = I_d + M R^2$. Substituting the values, $I' = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$. Since $M R^2 = 2 I$, we get $I' = \frac{5}{4} (2 I) = \frac{5}{2} I$.

The moment of inertia of a thin circular disc about an axis passing through its centre and perpendicular to its plane is $I$. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is

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