What will be the moment of inertia (a) about the diameter, (b) about the tangent perpendicular to the plane of a uniform circular ring of mass 2 kg and diameter 60 m.

Answer

To solve the problem of finding the moment of inertia of a uniform circular ring, we will break it down into two parts: (a) about the diameter and (b) about the tangent perpendicular to the plane of the ring.

### Given:
- Mass of the ring, \( m = 2 \, \text{kg} \)
- Diameter of the ring, \( d = 60 \, \text{m} \)
- Radius of the ring, \( r = \frac{d}{2} = \frac{60}{2} = 30 \, \text{m} \)

### (a) Moment of Inertia about the Diameter

1. **Formula for Moment of Inertia about the Diameter**:
The moment of inertia \( I \) of a circular ring about an axis through its center and perpendicular to its plane is given by:
\[ I = m r^2 \]
where \( m \) is the mass and \( r \) is the radius.

2. **Calculate Moment of Inertia**:
Substituting the values:
\[ I = 2 \, \text{kg} \times (30 \, \text{m})^2 \]
\[ I = 2 \times 900 = 1800 \, \text{kg m}^2 \]

### (b) Moment of Inertia about the Tangent Perpendicular to the Plane

1. **Using the Perpendicular Axis Theorem**:
According to the perpendicular axis theorem, the moment of inertia about an axis perpendicular to the plane of the ring (let's call this \( I_z \)) is equal to the sum of the moments of inertia about two perpendicular axes in the plane of the ring (let's call them \( I_x \) and \( I_y \)):
\[ I_z = I_x + I_y \]
For a circular ring, \( I_x = I_y = \frac{1}{2} I_z \).

2. **Calculate Moment of Inertia about the Tangent**:
The moment of inertia about the tangent \( I_t \) can be calculated using the parallel axis theorem:
\[ I_t = I_z + m d^2 \]
where \( d \) is the distance from the center of the ring to the tangent line. In this case, \( d = r = 30 \, \text{m} \).

3. **Substituting Values**:
We already calculated \( I_z = 1800 \, \text{kg m}^2 \):
\[ I_t = 1800 \, \text{kg m}^2 + 2 \, \text{kg} \times (30 \, \text{m})^2 \]
\[ I_t = 1800 \, \text{kg m}^2 + 2 \times 900 = 1800 + 1800 = 3600 \, \text{kg m}^2 \]

### Final Answers:
- (a) Moment of Inertia about the diameter: \( 1800 \, \text{kg m}^2 \)
- (b) Moment of Inertia about the tangent perpendicular to the plane: \( 3600 \, \text{kg m}^2 \)