The radius of gyration of a disc about an axis coinciding with a tangent in its plane is

Options

A. `(R )/(2)`
B. `(R )/(sqrt(2))`
C. `sqrt(2)R`
D. `(sqrt(5))/(2)R`

Explanation

To find the radius of gyration of a disc about an axis coinciding with a tangent in its plane, we can follow these steps:

### Step 1: Understand the Concept of Radius of Gyration
The radius of gyration (k) is defined as the distance from the axis of rotation at which the entire mass of the body can be assumed to be concentrated for the purpose of calculating rotational inertia (moment of inertia).

### Step 2: Moment of Inertia of the Disc
The moment of inertia (I) of a solid disc of mass \( M \) and radius \( R \) about an axis through its center and perpendicular to its plane is given by the formula:
\[ I_{center} = \frac{1}{2} M R^2 \]

### Step 3: Use the Parallel Axis Theorem
To find the moment of inertia about an axis that is tangent to the disc, we can use the Parallel Axis Theorem. This theorem states that:
\[ I_{tangent} = I_{center} + Md^2 \]
where \( d \) is the distance from the center of mass to the new axis. For a tangent line, \( d = R \).

### Step 4: Calculate the Moment of Inertia about the Tangent
Substituting the values into the equation:
\[ I_{tangent} = \frac{1}{2} M R^2 + M R^2 \]
\[ I_{tangent} = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Step 5: Calculate the Radius of Gyration
The radius of gyration \( k \) is related to the moment of inertia by the equation:
\[ I = M k^2 \]
Setting \( I = I_{tangent} \):
\[ \frac{3}{2} M R^2 = M k^2 \]
Dividing both sides by \( M \):
\[ \frac{3}{2} R^2 = k^2 \]
Taking the square root of both sides:
\[ k = R \sqrt{\frac{3}{2}} = R \cdot \frac{\sqrt{3}}{\sqrt{2}} = \frac{R \sqrt{3}}{\sqrt{2}} \]

### Final Answer
Thus, the radius of gyration of the disc about an axis coinciding with a tangent in its plane is:
\[ k = \frac{R \sqrt{3}}{\sqrt{2}} \]