The moment of inertia of a hoop of radius R and mass M, about any tangent

Options

A. `[3mR^(2)]//2`
B. `[mR^(2)]//2`
C. `MR^(2)`
D. `[MR^(2)]//4`

Answer

To find the moment of inertia of a hoop of radius \( R \) and mass \( M \) about any tangent, we can follow these steps:

### Step 1: Understand the Hoop's Moment of Inertia
The moment of inertia of a hoop about an axis passing through its center and perpendicular to the plane of the hoop is given by:
\[ I_{center} = M R^2 \]

### Step 2: Use the Parallel Axis Theorem
To find the moment of inertia about a tangent to the hoop, we can use the Parallel Axis Theorem. This theorem states that if you know the moment of inertia about a parallel axis through the center of mass, you can find the moment of inertia about another parallel axis by adding the product of the mass and the square of the distance between the two axes.

The formula is:
\[ I = I_{cm} + M h^2 \]
where:
- \( I \) is the moment of inertia about the new axis,
- \( I_{cm} \) is the moment of inertia about the center of mass,
- \( M \) is the mass of the object,
- \( h \) is the distance between the two axes.

### Step 3: Identify the Values
In our case:
- \( I_{cm} = M R^2 \) (moment of inertia about the center),
- \( M = M \) (mass of the hoop),
- \( h = R \) (the distance from the center to the tangent).

### Step 4: Substitute the Values into the Formula
Now we substitute these values into the Parallel Axis Theorem:
\[ I_{tangent} = I_{cm} + M h^2 \]
\[ I_{tangent} = M R^2 + M R^2 \]
\[ I_{tangent} = M R^2 + M R^2 = 2 M R^2 \]

### Step 5: Add the Contribution from the Tangent
Since we are finding the moment of inertia about a tangent, we need to add the additional \( M R^2 \) (the contribution from the distance \( R \)):
\[ I_{tangent} = M R^2 + M R^2 = 3 M R^2 \]

### Final Result
Thus, the moment of inertia of the hoop about any tangent is:
\[ I_{tangent} = \frac{3}{2} M R^2 \]

### Conclusion
The correct answer is option A: \( \frac{3}{2} M R^2 \).

---