The radius of gyration of a uniform rod of length `L` about an axis passing through its centre of mass is

Options

A. `(L)/(2 sqrt(3))`
B. `(L^(2))/(12)`
C. `(L)/(sqrt(3))`
D. `(L)/(sqrt(2))`

Explanation

To find the radius of gyration of a uniform rod of length \( L \) about an axis passing through its center of mass, we can follow these steps:

### Step 1: Understand the Moment of Inertia
The moment of inertia \( I \) of a uniform rod of length \( L \) about an axis passing through its center of mass is given by the formula:
\[ I = \frac{1}{12} ML^2 \]
where \( M \) is the mass of the rod.

### Step 2: Relate Moment of Inertia to Radius of Gyration
The moment of inertia can also be expressed in terms of the radius of gyration \( k \):
\[ I = Mk^2 \]
where \( k \) is the radius of gyration.

### Step 3: Set the Two Expressions for Moment of Inertia Equal
Since both expressions represent the moment of inertia, we can set them equal to each other:
\[ \frac{1}{12} ML^2 = Mk^2 \]

### Step 4: Cancel Mass \( M \)
Assuming \( M \) is not zero, we can cancel \( M \) from both sides:
\[ \frac{1}{12} L^2 = k^2 \]

### Step 5: Solve for Radius of Gyration \( k \)
Now, we can solve for \( k \) by taking the square root of both sides:
\[ k = \sqrt{\frac{1}{12} L^2} \]
This simplifies to:
\[ k = \frac{L}{\sqrt{12}} \]

### Step 6: Simplify Further
We can simplify \( \sqrt{12} \) as follows:
\[ \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \]
Thus, we can rewrite \( k \):
\[ k = \frac{L}{2\sqrt{3}} \]

### Final Answer
The radius of gyration of the uniform rod about an axis passing through its center of mass is:
\[ k = \frac{L}{2\sqrt{3}} \]