The radius of gyration of a uniform disc of radius R, about an axis passing through a point `(R )/(2)` away from the centre of disc, and perpendicular to the plane of disc is:

Options

A. `sqrt((13)/(20))R`
B. `sqrt((23)/(40))R`
C. `sqrt((23)/(20))R`
D. `sqrt((13)/(40))R`

Explanation

To find the radius of gyration \( k \) of a uniform disc of radius \( R \) about an axis passing through a point \( \frac{R}{2} \) away from the center of the disc and perpendicular to the plane of the disc, we can follow these steps:

### Step 1: Define the mass and moment of inertia
Let the mass of the disc be \( m \). The moment of inertia \( I \) of a uniform disc about its center is given by:
\[ I_{\text{center}} = \frac{1}{2} m R^2 \]

### Step 2: Use the parallel axis theorem
To find the moment of inertia about the new axis that is \( \frac{R}{2} \) away from the center, we can use the parallel axis theorem:
\[ I = I_{\text{center}} + m d^2 \]
where \( d \) is the distance from the center to the new axis. Here, \( d = \frac{R}{2} \).

### Step 3: Substitute values into the equation
Substituting the values into the equation:
\[ I = \frac{1}{2} m R^2 + m \left(\frac{R}{2}\right)^2 \]
Calculating \( \left(\frac{R}{2}\right)^2 \):
\[ \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
Thus, we have:
\[ I = \frac{1}{2} m R^2 + m \cdot \frac{R^2}{4} \]

### Step 4: Combine the terms
Now, we combine the terms:
\[ I = \frac{1}{2} m R^2 + \frac{1}{4} m R^2 = \left(\frac{2}{4} + \frac{1}{4}\right) m R^2 = \frac{3}{4} m R^2 \]

### Step 5: Relate moment of inertia to radius of gyration
The moment of inertia can also be expressed in terms of the radius of gyration \( k \):
\[ I = m k^2 \]
Setting the two expressions for \( I \) equal gives:
\[ m k^2 = \frac{3}{4} m R^2 \]

### Step 6: Solve for \( k^2 \)
Dividing both sides by \( m \):
\[ k^2 = \frac{3}{4} R^2 \]

### Step 7: Take the square root to find \( k \)
Taking the square root of both sides:
\[ k = \sqrt{\frac{3}{4}} R = \frac{\sqrt{3}}{2} R \]

### Final Answer
Thus, the radius of gyration \( k \) of the disc about the given axis is:
\[ k = \frac{\sqrt{3}}{2} R \]

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