From a uniform circular disc of radius R and mass 9M, a small disc of radius R3 is removed as shown

Question

From a uniform circular disc of radius R and mass 9M, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :

Accepted Answer

MOI of removed part about axis passing through COM & $⊥$ to plane of disc = I_cm + md²
$= \frac{(m) (R / 3)^{2}}{2} + m [\frac{4 R^{2}}{9}] = \frac{m R^{2}}{2}$
so MOI of remaining portion
= [MOI of whole disc - MOI of removed part]
= (9m) $\frac{R^{2}}{2} - \frac{m R^{2}}{2} = \frac{m R^{2}}{2} [8]$
I_remaining = 4mR²

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