Moment of inertia of a uniform circular disc about a diameter is $I$. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be: (1990)
Solution:
Given $I_{\text{diameter}} = \frac{MR^2}{4} = I$, which means $MR^2 = 4I$. Using the parallel axis theorem, the moment of inertia about a perpendicular axis on the rim is $I_{\text{rim}} = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2 = \frac{3}{2} (4I) = 6I$.
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