A small hole is made in a disc of mass `M` and radius `R` at a distance `R//4` from centre. The disc is supported on a horizontal peg through this hole. The moment of inertia of the disc about horizontal peg is

Options

A. `(MR^(2))/9`
B. `5/16MR^(2)`
C. `9/16MR^(2)`
D. `5/4MR^(2)`

Explanation

To find the moment of inertia of a disc about a horizontal peg located at a distance \( \frac{R}{4} \) from the center, we can follow these steps:

### Step 1: Identify the moment of inertia of the disc about its center
The moment of inertia \( I_{cm} \) of a solid disc about its center of mass is given by the formula:
\[ I_{cm} = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the disc and \( R \) is its radius.

### Step 2: Apply the Parallel Axis Theorem
The Parallel Axis Theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is given by:
\[ I = I_{cm} + Md^2 \]
where \( d \) is the distance from the center of mass to the new axis.

### Step 3: Determine the distance \( d \)
In this case, the distance \( d \) is \( \frac{R}{4} \) (the distance from the center of the disc to the peg).

### Step 4: Substitute the values into the Parallel Axis Theorem
Now we can substitute the values into the formula:
\[ I = \frac{1}{2} M R^2 + M \left(\frac{R}{4}\right)^2 \]

### Step 5: Calculate \( Md^2 \)
Calculating \( Md^2 \):
\[ Md^2 = M \left(\frac{R}{4}\right)^2 = M \cdot \frac{R^2}{16} = \frac{M R^2}{16} \]

### Step 6: Combine the terms
Now substituting back:
\[ I = \frac{1}{2} M R^2 + \frac{M R^2}{16} \]
To combine these fractions, we need a common denominator:
\[ I = \frac{8}{16} M R^2 + \frac{1}{16} M R^2 = \frac{9}{16} M R^2 \]

### Final Answer
Thus, the moment of inertia of the disc about the horizontal peg is:
\[ I = \frac{9}{16} M R^2 \]

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