Moment of inertia of a ring of mass M and radius R about an axis passing through the centre and perpendicular to the plane is

Options

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

Answer

To find the moment of inertia of a ring of mass \( M \) and radius \( R \) about an axis passing through the center and perpendicular to the plane of the ring, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ring and Axis**:
- Consider a ring of mass \( M \) and radius \( R \).
- The axis of rotation is perpendicular to the plane of the ring and passes through its center.

2. **Defining a Small Mass Element**:
- Let’s take a small mass element \( dm \) on the ring.
- The distance of this mass element from the axis of rotation is \( R \) (the radius of the ring).

3. **Calculating the Moment of Inertia of the Mass Element**:
- The moment of inertia \( dI \) of the small mass element \( dm \) about the axis is given by:
\[ dI = r^2 \cdot dm \]
- Since the distance \( r \) is constant and equal to \( R \) for all mass elements on the ring, we can substitute \( R \) into the equation:
\[ dI = R^2 \cdot dm \]

4. **Integrating Over the Entire Ring**:
- To find the total moment of inertia \( I \) of the ring, we integrate \( dI \) over the entire mass of the ring:
\[ I = \int dI = \int R^2 \, dm \]
- Since \( R^2 \) is constant, it can be factored out of the integral:
\[ I = R^2 \int dm \]

5. **Evaluating the Integral**:
- The integral \( \int dm \) gives the total mass \( M \) of the ring:
\[ I = R^2 \cdot M \]

6. **Final Expression for Moment of Inertia**:
- Therefore, the moment of inertia of the ring about the specified axis is:
\[ I = M R^2 \]

### Conclusion:
The moment of inertia of a ring of mass \( M \) and radius \( R \) about an axis passing through the center and perpendicular to the plane is:
\[ I = M R^2 \]