A disc of mass (M, R) is given. Two discs of radius \(\frac{R}{4}\) are cut from this, whose centers are at 135° angle. Their peripheries touch the larger disc as shown. If moment of inertia of remaining disc about the center is \(\frac{\alpha}{256}MR^2\), find \(\alpha\):
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For problems involving objects with holes or removed parts, the "subtraction method" is very effective.
Always apply the parallel axis theorem carefully, ensuring that \(d\) is the perpendicular distance between the axis through the center of mass of the object and the axis of rotation.
Step 1: Understanding the Question:
We need to find the moment of inertia of a large disc after two smaller discs have been removed from it. We can do this by using the principle of superposition: the moment of inertia of the remaining part is the moment of inertia of the original full disc minus the moments of inertia of the two removed parts (holes), all calculated about the same axis (the center of the large disc). Step 2: Key Formula or Approach:
1. Moment of inertia of a solid disc about its center: \(I_{cm} = \frac{1}{2} m r^2\).
2. Parallel Axis Theorem: The moment of inertia about an axis parallel to the center of mass axis is \(I = I_{cm} + m d^2\), where \(d\) is the distance between the two axes.
3. Surface Mass Density: \(\sigma = \frac{\text{Mass}}{\text{Area}}\). Step 3: Detailed Explanation: Part A: Properties of the Discs
- Original Disc: Mass = M, Radius = R. Moment of inertia about its center is \(I_{total} = \frac{1}{2}MR^2\).
- Surface Mass Density: \(\sigma = \frac{M}{\pi R^2}\).
- Removed Discs (Holes): Radius \(r = \frac{R}{4}\). The mass of each removed disc is:
\[ m = \sigma \times (\text{Area of small disc}) = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{4}\right)^2 = \frac{M}{16} \]
- Position of Holes: The peripheries of the small discs touch the larger disc. This means the centers of the small discs are at a distance \(d = R - r = R - \frac{R}{4} = \frac{3R}{4}\) from the center of the large disc. The 135° angle between their positions does not affect the calculation of the moment of inertia for each hole about the center. Part B: Moment of Inertia of Removed Discs
We need the moment of inertia of one removed disc about the center of the large disc. We use the parallel axis theorem.
- Moment of inertia of a small disc about its own center: \(I_{hole, cm} = \frac{1}{2}mr^2 = \frac{1}{2} \left(\frac{M}{16}\right) \left(\frac{R}{4}\right)^2 = \frac{MR^2}{512}\).
- Using the parallel axis theorem with \(d = \frac{3R}{4}\):
\[ I_{hole} = I_{hole, cm} + md^2 = \frac{MR^2}{512} + \left(\frac{M}{16}\right) \left(\frac{3R}{4}\right)^2 \]
\[ I_{hole} = \frac{MR^2}{512} + \frac{9MR^2}{256} = \frac{MR^2 + 18MR^2}{512} = \frac{19MR^2}{512} \]
Part C: Moment of Inertia of Remaining Disc
The moment of inertia of the remaining part is the total inertia minus the inertia of the two removed holes.
\[ I_{rem} = I_{total} - 2 \times I_{hole} \]
\[ I_{rem} = \frac{1}{2}MR^2 - 2 \times \left(\frac{19MR^2}{512}\right) = \frac{MR^2}{2} - \frac{19MR^2}{256} \]
\[ I_{rem} = \left(\frac{128}{256} - \frac{19}{256}\right)MR^2 = \frac{109}{256}MR^2 \]
Step 4: Final Answer:
Comparing the result \(I_{rem} = \frac{109}{256}MR^2\) with the given form \(\frac{\alpha}{256}MR^2\), we find that \(\alpha = 109\).
