Question

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 \(\alpha\):

• A. 109
• B. 119
• C. 99
• D. 128

Hint:

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.

Answer 1

The Correct Option is A

To solve this problem, we need to find the moment of inertia of the remaining disc after two smaller discs have been removed. The following steps demonstrate the process: 

1. Determine the mass of the smaller discs:
   • The total mass of the large disc is \( M \).
   • The two smaller discs have a radius of \( \frac{R}{4} \). The area of one smaller disc is \( \pi \left(\frac{R}{4}\right)^2 = \frac{\pi R^2}{16} \).
   • Assuming uniform density, the entire disc's area is \( \pi R^2 \). Thus, the mass of each smaller disc is \( \frac{M}{\pi R^2} \times \frac{\pi R^2}{16} = \frac{M}{16} \).
2. Calculate the moment of inertia of one smaller disc:
   • The moment of inertia of a disc about its own center is \( \frac{1}{2}mr^2 \). For one smaller disc: \( m = \frac{M}{16} \), \( r = \frac{R}{4} \).
   • \( I_{\text{small}} = \frac{1}{2} \times \frac{M}{16} \times \left(\frac{R}{4}\right)^2 = \frac{1}{2} \times \frac{M}{16} \times \frac{R^2}{16} = \frac{MR^2}{512} \).
3. Apply the parallel axis theorem:
   • The distance of each smaller disc’s center from the center of the larger disc is \( R - \frac{R}{4} = \frac{3R}{4} \).
   • By the parallel axis theorem, the moment of inertia of one smaller disc about the center of the large disc is \[ I_{\text{parallel}} = I_{\text{small}} + md^2 \], where \( d = \frac{3R}{4} \).
   • \[ I_{\text{parallel}} = \frac{MR^2}{512} + \frac{M}{16} \left(\frac{3R}{4}\right)^2 = \frac{MR^2}{512} + \frac{M}{16} \times \frac{9R^2}{16} = \frac{MR^2}{512} + \frac{9MR^2}{256} \].
   • \[ I_{\text{parallel}} = \frac{MR^2}{512} + \frac{18MR^2}{512} = \frac{19MR^2}{512} \].
4. Calculate the moment of inertia of the two small discs and subtract from the larger disc:
   • Moment of inertia of two smaller discs: \[ 2 \times \frac{19MR^2}{512} = \frac{38MR^2}{512} = \frac{19MR^2}{256} \].
   • The moment of inertia of the larger disc about its center is \( \frac{1}{2}MR^2 \).
   • The moment of inertia of the remaining disc: \[ \frac{1}{2}MR^2 - \frac{19MR^2}{256} = \frac{128MR^2}{256} - \frac{19MR^2}{256} = \frac{109MR^2}{256} \].

Thus, \(\alpha = 109\) is the correct answer.