Given the moment of interia of a disc of mass M and radii R about any of its diameters to be `(MR^(2))/4` . What is its moment of interia about an axis normal to the disc and passing through a point on its edge .

Options

A. `3/2 MR^(2)`
B. `(MR^(2))/2`
C. `MR^(2)`
D. `2MR^(2)`

Answer

To find the moment of inertia of a disc of mass \( M \) and radius \( R \) about an axis normal to the disc and passing through a point on its edge, we can follow these steps:

### Step 1: Moment of Inertia about Diameter
The moment of inertia of the disc about any of its diameters is given as:
\[ I_{xx} = \frac{MR^2}{4} \]

### Step 2: Use Perpendicular Axis Theorem
According to the perpendicular axis theorem, the moment of inertia about an axis perpendicular to the plane of the disc (let's call it \( I_{zz} \)) is the sum of the moments of inertia about two perpendicular axes in the plane of the disc (which are the diameters):
\[ I_{zz} = I_{xx} + I_{yy} \]
Since \( I_{xx} = I_{yy} \) (because of symmetry), we can write:
\[ I_{zz} = 2I_{xx} = 2 \left( \frac{MR^2}{4} \right) = \frac{MR^2}{2} \]

### Step 3: Apply Parallel Axis Theorem
Now, we need to find the moment of inertia about an axis that is normal to the disc and passes through a point on its edge. We can use the parallel axis theorem, which states:
\[ I = I_{zz} + Md^2 \]
where \( d \) is the distance from the center of mass to the new axis. In this case, the distance \( d \) is equal to the radius \( R \) of the disc. Therefore:
\[ I = I_{zz} + MR^2 \]
Substituting the value of \( I_{zz} \):
\[ I = \frac{MR^2}{2} + MR^2 = \frac{MR^2}{2} + \frac{2MR^2}{2} = \frac{3MR^2}{2} \]

### Final Answer
Thus, the moment of inertia of the disc about an axis normal to the disc and passing through a point on its edge is:
\[ I = \frac{3MR^2}{2} \]