The moment of inertia of a disc of mass `M` and radius `R` about a tangent to its rim in its plane is

Options

A. `2/3 MR^(2)`
B. `3/2 MR^(2)`
C. `4/5 MR^(2)`
D. `5/4 MR^(2)`

Answer

To find the moment of inertia of a disc of mass \( M \) and radius \( R \) about a tangent to its rim in its plane, we can follow these steps:

### Step 1: Understand the moment of inertia about the center
The moment of inertia \( I \) of a disc about an axis passing through its center and perpendicular to its plane (diametrical axis) is given by the formula:
\[ I_{\text{center}} = \frac{1}{2} M R^2 \]

### Step 2: Use the parallel axis theorem
To find the moment of inertia about an axis that is tangent to the rim of the disc, we can use the parallel axis theorem. The parallel axis theorem states that:
\[ I = I_{\text{cm}} + Md^2 \]
where:
- \( I \) is the moment of inertia about the new axis,
- \( I_{\text{cm}} \) is the moment of inertia about the center of mass,
- \( M \) is the mass of the disc,
- \( d \) is the distance between the center of mass and the new axis.

### Step 3: Identify the distance \( d \)
In this case, the distance \( d \) from the center of the disc to the tangent line is equal to the radius \( R \) of the disc.

### Step 4: Calculate the moment of inertia about the tangent
Now, substituting the values into the parallel axis theorem:
\[ I_{\text{tangent}} = I_{\text{center}} + M R^2 \]
Substituting \( I_{\text{center}} = \frac{1}{2} M R^2 \):
\[ I_{\text{tangent}} = \frac{1}{2} M R^2 + M R^2 \]
\[ I_{\text{tangent}} = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Step 5: Final calculation
Now, we need to add the moment of inertia about the center (which we calculated as \( \frac{1}{2} M R^2 \)) to \( M R^2 \) (the term from the parallel axis theorem):
\[ I_{\text{tangent}} = \frac{1}{2} M R^2 + M R^2 = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{5}{2} M R^2 \]

Thus, the moment of inertia of the disc about a tangent to its rim in its plane is:
\[ I = \frac{5}{2} M R^2 \]

### Final Result
The moment of inertia of the disc about a tangent to its rim in its plane is:
\[ \frac{5}{2} M R^2 \]