A uniform circular disc of radius ‘R’ and mass ‘M’ is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius R/2 is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above : (1) (27/32) MR^2 (2) (29/32) MR^2 (3) (17/32) MR^2 (4) (13/32) MR^2

1. Key Concepts

1. Moment of Inertia (MOI)
   The rotational equivalent of mass in linear motion. For a uniform solid disc of mass $M$ and radius $R$ about its central axis:
   $I_{\text{disc}} = \frac{1}{2} M R^2$

2. Mass Density ($\sigma$)
   For a uniform sheet, $\sigma = \frac{\text{mass}}{\text{area}} = \frac{M}{\pi R^2}$. Keeping $\sigma$ constant lets us relate masses of pieces to their areas.

3. Parallel‐Axis Theorem
   If the MOI about a body’s own centre is $I_c$, then about a parallel axis a distance $d$ away:
   $I = I_c + m d^2$

4. ‘Negative Mass’ Trick
   When a part is removed, treat it like adding a negative mass with the same geometry, then add moments algebraically.

2. Logical Chain to Solve

1. Start with the original full disc: we already know $I_{\text{full}} = \tfrac12 MR^2$.
2. Compute mass of the removed piece:
   • Radius of hole: $\tfrac{R}{2}$
   • Area of hole: $\pi (\tfrac{R}{2})^2 = \tfrac{\pi R^2}{4}$
   • Mass of hole: $m = \sigma \times \text{area} = \frac{M}{\pi R^2} \times \frac{\pi R^2}{4} = \frac{M}{4}$
3. Find MOI of the hole about the original centre:
   (a) MOI about its own centre:
   $I_{c\,(\text{hole})} = \frac12 m \left(\frac{R}{2}\right)^2 = \frac{m R^2}{8}$
   (b) Distance between centres ($d$): $\tfrac{R}{2}$
   (c) Use parallel-axis:
   $I_{\text{hole about main axis}} = I_{c\,(\text{hole})} + m d^2 = \frac{m R^2}{8} + \frac{m R^2}{4} = \frac{3 m R^2}{8}$
4. Subtract the hole’s MOI from the full disc’s MOI:
   $I_{\text{remain}} = I_{\text{full}} - I_{\text{hole}}$
   Substitute $m=\tfrac{M}{4}$:
   $I_{\text{remain}} = \frac12 M R^2 - \frac{3}{8}\left(\frac{M}{4}\right) R^2 = \frac{16 M R^2}{32} - \frac{3 M R^2}{32} = \boxed{\frac{13}{32} M R^2}$
5. Pick the option: matches option (4).

That step-by-step path is what any student should follow: identify densities, compute masses, use the parallel-axis theorem, then subtract.

Imagine a Big Spinning Pizza

1. Big Pizza: You have a huge, perfectly round pizza. It spins on a stick poked right through its middle so every part goes around evenly.
2. Cut a Small Slice Out: Now you use a cookie-cutter to punch out a smaller round mini-pizza near the middle edge (but not right at the edge). That mini-pizza has half the radius of the big one.
3. What Happens to Spinning? Taking some mass away makes the whole thing easier to spin. We want to know exactly how much easier.
4. Two Simple Moves:
   • First, write down how hard it was to spin the whole pizza before cutting (its “moment of inertia”).
   • Second, figure out how hard the tiny round piece would have been to spin about the same stick, then subtract that amount.
5. Answer Pops Out: Do the subtraction carefully and you get $\frac{13}{32} MR^2$. That’s choice (4).

That’s all—remove, subtract, done!

Step-by-Step Calculation

1. Surface Density
   $\sigma = \frac{M}{\pi R^2}$

2. Removed Disc
   Radius: $\frac{R}{2}$
   Mass:
   $m = \sigma \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$ Distance of its centre from main centre:
   $d = \frac{R}{2}$

3. MOI of Removed Disc about Its Own Centre
   $I_{c\,(\text{small})} = \frac12 m \left(\frac{R}{2}\right)^2 = \frac{m R^2}{8}$

4. Shift to Main Axis (Parallel-Axis)
   $I_{\text{hole}} = I_{c\,(\text{small})} + m d^2 = \frac{m R^2}{8} + m \left(\frac{R}{2}\right)^2 = \frac{m R^2}{8} + \frac{m R^2}{4} = \frac{3 m R^2}{8}$

5. MOI of Original Full Disc
   $I_{\text{full}} = \frac12 M R^2$

6. MOI of Remaining Part
   $I_{\text{remain}} = I_{\text{full}} - I_{\text{hole}}$ Substitute $m = \tfrac{M}{4}$:

   $I_{\text{remain}} = \frac12 M R^2 - \frac{3}{8}\left(\frac{M}{4}\right) R^2 = \frac{16}{32} M R^2 - \frac{3}{32} M R^2 = \frac{13}{32} M R^2$

7. Final Answer
   $\boxed{\frac{13}{32} M R^2}\quad\text{(Option 4)}$