Step 1: Understanding the Concept:
The moment of inertia of a complex or hollowed-out object can be calculated using the principle of superposition.
We treat the object as a full, uniform disc and then subtract the moment of inertia of the portion that has been removed (the hole).
Crucially, mass is distributed uniformly across the area.
Therefore, the mass of the removed portion is proportional to its area relative to the original disc.
In this problem, the phrasing implies that \(M\) is the mass of the *original* solid disc before the hole was cut.
Key Formula or Approach:
1. Moment of inertia of a solid disc: \(I = \frac{1}{2}mr^2\)
2. Mass per unit area (surface density) \(\sigma = \frac{\text{Mass}}{\text{Area}}\)
3. Total Inertia \(I_{final} = I_{full} - I_{removed}\)
Step 2: Detailed Explanation:
Let \(M\) be the mass of the original solid disc of radius \(R\).
Area of the original disc \(A_1 = \pi R^2\).
Area of the hole \(A_2 = \pi \left( \frac{R}{2} \right)^2 = \frac{\pi R^2}{4}\).
1. Finding the Mass of the Removed Portion (\(m\)):
Since the disc is uniform, mass is proportional to area.
\[ \frac{m}{M} = \frac{\text{Area of hole}}{\text{Area of full disc}} = \frac{\pi R^2 / 4}{\pi R^2} = \frac{1}{4} \]
So, \(m = \frac{M}{4}\).
2. Calculating Individual Moments of Inertia:
Inertia of the full disc: \(I_1 = \frac{1}{2} M R^2\).
Inertia of the removed disc (hole) about the center:
\[ I_2 = \frac{1}{2} m r^2 = \frac{1}{2} \left( \frac{M}{4} \right) \left( \frac{R}{2} \right)^2 \]
\[ I_2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{1}{32} M R^2 \]
3. Finding Net Inertia:
Subtract the hole's inertia from the full disc's inertia:
\[ I_{net} = I_1 - I_2 = \frac{1}{2} M R^2 - \frac{1}{32} M R^2 \]
Convert to common denominator:
\[ I_{net} = \frac{16}{32} M R^2 - \frac{1}{32} M R^2 = \frac{15}{32} M R^2 \]
Step 3: Final Answer:
The moment of inertia of the remaining disc is \(\frac{15}{32} MR^2\).