The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and a circular ring of the same radius about a tengential axis in the plane of the ring is

Options

A. `2 : 3`
B. `2 : 1`
C. `sqrt(5) : sqrt(6)`
D. `1 : sqrt(2)`

Answer

To solve the problem of finding the ratio of the radii of gyration of a circular disc and a circular ring about a tangential axis in their respective planes, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Moment of Inertia**:
- The moment of inertia \( I \) can be expressed in terms of the radius of gyration \( k \) as:
\[ I = m k^2 \]
- Here, \( m \) is the mass of the object, and \( k \) is the radius of gyration.

2. **Moment of Inertia of the Disc**:
- The moment of inertia of a circular disc about a tangential axis in its plane is given by:
\[ I_d = \frac{5}{4} m r^2 \]
- Setting this equal to the expression in terms of the radius of gyration:
\[ \frac{5}{4} m r^2 = m k_d^2 \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{5}{4} r^2 = k_d^2 \]

3. **Moment of Inertia of the Ring**:
- The moment of inertia of a circular ring about a tangential axis in its plane is given by:
\[ I_r = \frac{3}{2} m r^2 \]
- Setting this equal to the expression in terms of the radius of gyration:
\[ \frac{3}{2} m r^2 = m k_r^2 \]
- Again, dividing both sides by \( m \):
\[ \frac{3}{2} r^2 = k_r^2 \]

4. **Finding the Ratio of Radii of Gyration**:
- Now we have:
\[ k_d^2 = \frac{5}{4} r^2 \quad \text{and} \quad k_r^2 = \frac{3}{2} r^2 \]
- Taking the ratio of \( k_d^2 \) to \( k_r^2 \):
\[ \frac{k_d^2}{k_r^2} = \frac{\frac{5}{4} r^2}{\frac{3}{2} r^2} \]
- The \( r^2 \) cancels out:
\[ \frac{k_d^2}{k_r^2} = \frac{5/4}{3/2} = \frac{5}{4} \cdot \frac{2}{3} = \frac{5}{6} \]

5. **Taking the Square Root**:
- To find the ratio of the radii of gyration:
\[ \frac{k_d}{k_r} = \sqrt{\frac{5}{6}} = \frac{\sqrt{5}}{\sqrt{6}} \]

### Final Answer
The ratio of the radii of gyration of the circular disc to the circular ring is:
\[ \frac{k_d}{k_r} = \frac{\sqrt{5}}{\sqrt{6}} \]