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

To find the value of \( x \), we need to calculate the moment of inertia of the remaining part of the larger disc after the smaller disc is removed. The moment of inertia \( I \) of a disc about an axis through its center and perpendicular to its plane is given by \( I = \frac{1}{2}MR^2 \). For a smaller disc with radius \( \frac{R}{3} \), its mass \( m \) is \( \frac{M}{9} \) because mass is proportional to the area (and hence proportional to the square of the radius). The moment of inertia of this smaller disc is \( I_{\text{small}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 = \frac{1}{2}\left(\frac{M}{9}\right)\frac{R^2}{9} = \frac{MR^2}{162} \). The remaining disc's moment of inertia \( I_{\text{remaining}} \) is the moment of inertia of the whole disc minus that of the smaller disc removed: \( I_{\text{remaining}} = \frac{1}{2}MR^2 - \frac{MR^2}{162} \). Simplifying this gives:

\( I_{\text{remaining}} = \frac{1}{2}MR^2 - \frac{MR^2}{162} = \frac{81}{162}MR^2 - \frac{1}{162}MR^2 = \frac{80}{162}MR^2 = \frac{40}{81}MR^2 \).

According to the problem, this is also given as \( \frac{4}{x}MR^2 \), so:

\(\frac{4}{x}MR^2 = \frac{40}{81}MR^2 \)

On comparing these, the \( MR^2 \) terms cancel out and we get:

\(\frac{4}{x} = \frac{40}{81} \)

Solving for \( x \), we multiply both sides by \( x \) and then multiply both sides by 81:

\( 4 \cdot 81 = 40x \)

\( 324 = 40x \)

\( x = \frac{324}{40} \)

\( x = 8.1 \)

Given the range is 9,9, it suggests there is an expectation error since our calculation of \( 8.1 \) seems consistent without errors. The theoretical steps are validated by standard moments of inertia operations and simplifications assuming uniform density and separation consistency.

Answer 2

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: \[ x = 9 \]