Question

A uniform circular disc of radius \( R \) and mass \( M \) is rotating about an axis perpendicular to its plane and passing through its center. 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.

• A. \( \frac{7}{32} MR^2 \)
• B. \( \frac{9}{32} MR^2 \)
• C. \( \frac{17}{32} MR^2 \)
• D. \( \frac{13}{32} MR^2 \)

Hint:

To calculate the moment of inertia of the remaining part of a disc, subtract the moment of inertia of the removed portion from the original disc's moment of inertia.

Answer 1

The Correct Option is D

The moment of inertia of the original disc is: \[ I_{\text{original}} = \frac{MR^2}{2} \] \[ I = \frac{MR^2}{2} - \left[ \frac{\frac{M}{4} \left( \frac{R}{2} \right)^2}{2} + \frac{M}{4} \left( \frac{R}{2} \right)^2 \right] \] 

Explanation: The first term, $\frac{MR^2}{2}$, represents the moment of inertia of a solid disc about an axis perpendicular to the disc and passing through its center. The terms inside the square brackets represent the moment of inertia of a part of the disc that has been removed. The expression $\frac{M}{4} \left( \frac{R}{2} \right)^2$ relates to the mass and radius of a removed section. The division by 2 in the first term inside the square brackets is related to the moment of inertia of the removed section about its own center. 

Step 1: Simplify the terms inside the square brackets \[ \frac{\frac{M}{4} \left( \frac{R}{2} \right)^2}{2} = \frac{M}{4} \cdot \frac{R^2}{4} \cdot \frac{1}{2} = \frac{MR^2}{32} \] \[ \frac{M}{4} \left( \frac{R}{2} \right)^2 = \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{16} \] 

Explanation: This step simplifies the fractions and squares within the square brackets to make the expression easier to work with. 

Step 2: Combine the terms inside the square brackets \[ \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2}{32} + \frac{2MR^2}{32} = \frac{3MR^2}{32} \] Explanation: This step combines the simplified terms within the square brackets into a single fraction. 

Step 3: Substitute the combined term back into the original equation \[ I = \frac{MR^2}{2} - \frac{3MR^2}{32} \] 

Explanation: This step substitutes the result from Step 2 back into the original equation for I. 

Step 4: Find a common denominator and simplify \[ I = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32} \] 

Explanation: This step finds a common denominator for the two fractions and subtracts them to find the final expression for I. 

Final Result: \[ I = \frac{13}{32} MR^2 \]

Answer 2

Moment of inertia of the original (whole) disc about its centre: \[ I_{\text{full}}=\tfrac{1}{2}MR^{2}. \]

1. Mass of the removed small disc (area proportional): \[ m' = M\cdot\frac{\pi (R/2)^2}{\pi R^2} = M\cdot\frac{1}{4}=\frac{M}{4}. \]
2. Moment of inertia of the small disc about its own centre: \[ I'_{\text{cm}}=\tfrac{1}{2}m'\left(\frac{R}{2}\right)^2 =\tfrac{1}{2}\cdot\frac{M}{4}\cdot\frac{R^2}{4} =\frac{M R^2}{32}. \]
3. Use the parallel-axis theorem to get the small disc’s moment about the big disc’s centre. The distance between centres is \(d=\tfrac{R}{2}\). Thus \[ I'_{\text{about origin}}=I'_{\text{cm}}+m' d^2 =\frac{M R^2}{32}+\frac{M}{4}\cdot\frac{R^2}{4} =\frac{M R^2}{32}+\frac{M R^2}{16} =\frac{3M R^2}{32}. \]
4. Moment of inertia of the remaining part \(=\) full disc \( - \) removed piece: \[ I_{\text{remaining}}=I_{\text{full}}-I'_{\text{about origin}} =\frac{1}{2}MR^2-\frac{3}{32}MR^2 =\frac{16}{32}MR^2-\frac{3}{32}MR^2 =\frac{13}{32}MR^2. \]

Answer

\(\displaystyle I_{\text{remaining}}=\frac{13}{32}\,MR^{2}.\) (Option 4)