Question

M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is $ \frac{4}{x} MR^2 $. The value of x is ____.

Hint:

When a part of a uniform object is removed, the moment of inertia of the remaining part can be found by subtracting the moment of inertia of the removed part from the moment of inertia of the original object, ensuring both moments of inertia are calculated about the same axis. Remember to use the parallel axis theorem if the axes of the removed part and the remaining part are not the same.

Answer 1

Correct Answer: 9

Given:

• Mass of original disc: $M$
• Radius of original disc: $R$
• Radius of removed disc: $R/3$
• Moment of inertia of remaining part: $\frac{4}{x}MR^2$

Step 1: Moment of Inertia of Original Disc
The moment of inertia of a solid disc about an axis through its center perpendicular to its plane is: \[ I_{\text{original}} = \frac{1}{2}MR^2 \]
Step 2: Mass of Removed Disc
Assuming uniform mass distribution, the mass of the removed disc is proportional to its area: \[ m = \left(\frac{\pi(R/3)^2}{\pi R^2}\right)M = \frac{M}{9} \]
Step 3: Moment of Inertia of Removed Disc
Case 1: Concentric Removal
If the disc is removed concentrically: \[ I_{\text{removed}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 = \frac{1}{2}\left(\frac{M}{9}\right)\left(\frac{R^2}{9}\right) = \frac{MR^2}{162} \]
Case 2: Non-concentric Removal
If the disc is removed tangentially (center at $2R/3$ from $O$), we must use the parallel axis theorem: \[ I_{\text{removed}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 + m\left(\frac{2R}{3}\right)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18} \]
Step 4: Moment of Inertia of Remaining Part
For Concentric Case:
\[ I_{\text{remaining}} = I_{\text{original}} - I_{\text{removed}} = \frac{1}{2}MR^2 - \frac{MR^2}{162} = \frac{40}{81}MR^2 \] Given that \( I_{\text{remaining}} = \frac{4}{x}MR^2 \), we get: \[ \frac{40}{81} = \frac{4}{x} \implies x = \frac{81}{10} \]
For Non-concentric Case:
\[ I_{\text{remaining}} = \frac{1}{2}MR^2 - \frac{MR^2}{18} = \frac{4}{9}MR^2 \] Given that $I_{\text{remaining}} = \frac{4}{x}MR^2$, we get: \[ \frac{4}{9} = \frac{4}{x} \implies x = 9 \]
Conclusion
Since the problem mentions "as shown in figure" but no figure is provided, the most reasonable assumption is that the disc is removed tangentially (non-concentrically), leading to: \[ x = 9 \]