The moment of inertia of a disc of mass `M` and radius `R` about an axis. Which is tangential to sircumference of disc and parallel to its diameter is.

Options

A. `(5)/(4)MR^(2)`
B. `(2)/(3)MR^(2)`
C. `(3)/(2)MR^(2)`
D. `(1)/(2)MR^(2)`

Answer

To find the moment of inertia of a disc of mass \( M \) and radius \( R \) about an axis that is tangential to the circumference of the disc and parallel to its diameter, we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the moment of inertia of a disc about a specific axis. The axis is tangential to the circumference of the disc and parallel to its diameter.

### Step 2: Moment of Inertia about the Center
The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is given by the formula:
\[ I_{CM} = \frac{1}{2} M R^2 \]

### Step 3: Use the Perpendicular Axis Theorem
According to the perpendicular axis theorem, for a planar body, the moment of inertia about an axis perpendicular to the plane (z-axis) is equal to the sum of the moments of inertia about two perpendicular axes (x-axis and y-axis) lying in the plane of the body:
\[ I_Z = I_X + I_Y \]
For a disc, since the moment of inertia about the x-axis and y-axis is the same (due to symmetry):
\[ I_X = I_Y = I_D \]
Thus,
\[ I_Z = 2 I_D \]
From the earlier step, we know:
\[ I_Z = \frac{1}{2} M R^2 \]
So,
\[ 2 I_D = \frac{1}{2} M R^2 \implies I_D = \frac{1}{4} M R^2 \]

### Step 4: Apply the Parallel Axis Theorem
Now, we need to find the moment of inertia about the tangential axis. We can use the parallel axis theorem, which states:
\[ I = I_{CM} + Md^2 \]
where \( d \) is the distance from the center of mass axis to the new axis. In this case, \( d = R \) (the radius of the disc).

### Step 5: Calculate the Moment of Inertia about the Tangential Axis
Substituting the values we have:
\[ I = I_D + M R^2 \]
Substituting \( I_D = \frac{1}{4} M R^2 \):
\[ I = \frac{1}{4} M R^2 + M R^2 \]
\[ I = \frac{1}{4} M R^2 + \frac{4}{4} M R^2 = \frac{5}{4} M R^2 \]

### Final Answer
Thus, the moment of inertia of the disc about the tangential axis is:
\[ I = \frac{5}{4} M R^2 \]