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 ____.
Show 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.
To determine the value of \( x \), we must compute the moment of inertia for the larger disc after the smaller disc has been excised. The moment of inertia \( I \) for a disc about an axis through its center and perpendicular to its plane is \( I = \frac{1}{2}MR^2 \). For a smaller disc with radius \( \frac{R}{3} \), its mass \( m \) is \( \frac{M}{9} \) due to the proportionality of mass to area (and thus 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 moment of inertia of the remaining portion, \( I_{\text{remaining}} \), is the difference between the moment of inertia of the entire disc and that of the removed smaller disc: \( I_{\text{remaining}} = \frac{1}{2}MR^2 - \frac{MR^2}{162} \). Simplifying this expression yields:
The problem states this is equivalent to \( \frac{4}{x}MR^2 \), leading to the equation:
\(\frac{4}{x}MR^2 = \frac{40}{81}MR^2 \)
Upon canceling the \( MR^2 \) terms from both sides, we obtain:
\(\frac{4}{x} = \frac{40}{81} \)
Solving for \( x \), we multiply both sides by \( x \) and then by 81:
\( 4 \cdot 81 = 40x \)
\( 324 = 40x \)
\( x = \frac{324}{40} \)
\( x = 8.1 \)
The provided range of 9,9 suggests a potential discrepancy, as our calculated value of \( 8.1 \) appears consistent with standard principles of moments of inertia, assuming uniform density and consistent geometric assumptions.
