The ratio of the radii of gyration of a circular disc and a circular ring of the same radii about a tangential axis perpendicular to plane of disc or ring is

Options

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

Explanation

To solve the problem of finding the ratio of the radii of gyration of a circular disc and a circular ring of the same radius about a tangential axis perpendicular to the plane of the disc or ring, we can follow these steps:

### Step 1: Understand the Concept of Radius of Gyration
The radius of gyration (k) is defined as \( k = \sqrt{\frac{I}{m}} \), where \( I \) is the moment of inertia and \( m \) is the mass of the object.

### Step 2: Calculate the Moment of Inertia for the Disc
For a circular disc, the moment of inertia about an axis through its center and perpendicular to its plane is given by:
\[ I_{d} = \frac{1}{2} m r^2 \]
Using the parallel axis theorem, the moment of inertia about a tangential axis (distance \( r \) away from the center) is:
\[ I_{d, \text{tangential}} = I_{d} + m r^2 = \frac{1}{2} m r^2 + m r^2 = \frac{3}{2} m r^2 \]

### Step 3: Calculate the Radius of Gyration for the Disc
Now, we can find the radius of gyration for the disc:
\[ k_{d} = \sqrt{\frac{I_{d, \text{tangential}}}{m}} = \sqrt{\frac{\frac{3}{2} m r^2}{m}} = \sqrt{\frac{3}{2}} r \]

### Step 4: Calculate the Moment of Inertia for the Ring
For a circular ring, the moment of inertia about an axis through its center and perpendicular to its plane is:
\[ I_{r} = m r^2 \]
Using the parallel axis theorem, the moment of inertia about a tangential axis is:
\[ I_{r, \text{tangential}} = I_{r} + m r^2 = m r^2 + m r^2 = 2 m r^2 \]

### Step 5: Calculate the Radius of Gyration for the Ring
Now, we can find the radius of gyration for the ring:
\[ k_{r} = \sqrt{\frac{I_{r, \text{tangential}}}{m}} = \sqrt{\frac{2 m r^2}{m}} = \sqrt{2} r \]

### Step 6: Find the Ratio of the Radii of Gyration
Now we can find the ratio of the radii of gyration of the disc to the ring:
\[ \frac{k_{d}}{k_{r}} = \frac{\sqrt{\frac{3}{2}} r}{\sqrt{2} r} = \frac{\sqrt{\frac{3}{2}}}{\sqrt{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Conclusion
Thus, the ratio of the radii of gyration of the circular disc and the circular ring about a tangential axis perpendicular to their planes is:
\[ \frac{k_{d}}{k_{r}} = \frac{\sqrt{3}}{2} \]

### Final Answer
The correct option is \( \frac{\sqrt{3}}{2} \).

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