Question

Consider a disc of radius \(R\) and mass \(M\). A hole of radius \(\dfrac{R}{3}\) is created in the disk such that the centre of the hole is \(\dfrac{R}{3}\) away from the centre of the disk. The moment of inertia of the system along the axis perpendicular to the disc passing through the centre of the disc is

• A. \(\dfrac{MR^2}{2}\)
• B. \(\dfrac{13}{27}MR^2\)
• C. \(\dfrac{1}{3}MR^2\)
• D. \(4MR^2\)

Hint:

For bodies with holes, treat the hole as negative mass. First calculate the moment of inertia of the complete body and then subtract the moment of inertia of the removed part using the parallel axis theorem when required.

Answer 1

The Correct Option is B

Step 1: Understand the concept of removed mass.
The given system can be treated as a full disc of mass \(M\) and radius \(R\) from which a smaller disc is removed.
Radius of the hole is
\[ r=\frac{R}{3} \] Distance of the hole's centre from the main disc centre is
\[ d=\frac{R}{3} \]

Step 2: Find the mass of the removed disc.
Since the surface mass density is uniform, mass is proportional to area.
Mass of removed disc is
\[ m=M\left(\frac{\pi r^2}{\pi R^2}\right) \] Substituting \(r=\dfrac{R}{3}\),
\[ m=M\left(\frac{(R/3)^2}{R^2}\right) \] \[ m=M\left(\frac{1}{9}\right) \] \[ m=\frac{M}{9} \]

Step 3: Moment of inertia of the original full disc.
Moment of inertia of a full disc about an axis perpendicular to the disc through its centre is
\[ I_{\text{full}}=\frac12 MR^2 \]

Step 4: Calculate the moment of inertia of the removed disc about the same axis.
The removed disc has radius \(\dfrac{R}{3}\) and mass \(\dfrac{M}{9}\).
Its own moment of inertia about its centre is
\[ I_c=\frac12 \times \frac{M}{9}\times \left(\frac{R}{3}\right)^2 \] \[ I_c=\frac12 \times \frac{M}{9}\times \frac{R^2}{9} \] \[ I_c=\frac{MR^2}{162} \] Now apply the parallel axis theorem.
Distance between the two centres is
\[ d=\frac{R}{3} \] Therefore,
\[ I_{\text{removed}} = I_c+md^2 \] \[ = \frac{MR^2}{162} + \frac{M}{9}\left(\frac{R}{3}\right)^2 \] \[ = \frac{MR^2}{162} + \frac{MR^2}{81} \] Taking LCM,
\[ = \frac{MR^2}{162} + \frac{2MR^2}{162} \] \[ = \frac{3MR^2}{162} \] \[ = \frac{MR^2}{54} \]

Step 5: Find the moment of inertia of the remaining system.
Subtract the removed portion from the original disc.
\[ I= I_{\text{full}}-I_{\text{removed}} \] \[ = \frac12 MR^2-\frac{MR^2}{54} \] Taking LCM \(54\),
\[ = \frac{27MR^2-MR^2}{54} \] \[ = \frac{26MR^2}{54} \] \[ = \frac{13}{27}MR^2 \]

Step 6: Final conclusion.
Hence, the moment of inertia of the remaining disc is
\[ \boxed{\frac{13}{27}MR^2} \]