$M$ and $R$ are the mass and radius of a uniform disc. A small disc of radius $R/3$ is removed from the bigger disc as shown in the figure (the small disc touches the rim and its centre lies on a diameter, at distance $2R/3$ from O). The moment of inertia of the remaining part of the bigger disc about an axis AB passing through the centre O and perpendicular to the plane of the disc is $\dfrac4xMR^2$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 9
The removed disc has $\tfrac19$ of the area, so its mass is $\dfrac M9$. Its moment of inertia about the axis through O (parallel axis theorem):
$$I_2 = \frac12\cdot\frac M9\left(\frac R3\right)^2 + \frac M9\left(\frac{2R}{3}\right)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18}$$
$$I = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{4MR^2}{9}$$
So $x = 9$.
Solution by Sreeraj P, M.Sc Physics