A thin uniform rod of mass M and length L. Find the radius of gyration for rotation about an axis passing through a point at a distance of`(L )/(4 )` from centre and perpendicular to rod.

Options

A. `sqrt((7)/(48)L)`
B. `sqrt((5)/(48)L)`
C. `sqrt((7)/(24)L)`
D. `sqrt((19)/(24)L)`

Answer

To find the radius of gyration of a thin uniform rod of mass \( M \) and length \( L \) about an axis that is at a distance of \( \frac{L}{4} \) from the center and perpendicular to the rod, we can follow these steps:

### Step 1: Understand the Problem
We have a thin rod with mass \( M \) and length \( L \). We need to find the radius of gyration \( k \) for rotation about an axis that is located \( \frac{L}{4} \) from the center of the rod.

### Step 2: Moment of Inertia about the Center of Mass
The moment of inertia \( I_{CM} \) of a thin rod about an axis passing through its center of mass and perpendicular to its length is given by:
\[ I_{CM} = \frac{1}{12} ML^2 \]

### Step 3: Apply the Parallel Axis Theorem
To find the moment of inertia about the new axis (which is \( \frac{L}{4} \) away from the center), we use the parallel axis theorem:
\[ I = I_{CM} + Md^2 \]
where \( d \) is the distance from the center of mass to the new axis. Here, \( d = \frac{L}{4} \).

### Step 4: Calculate \( d^2 \)
Calculating \( d^2 \):
\[ d^2 = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16} \]

### Step 5: Substitute into the Parallel Axis Theorem
Substituting \( I_{CM} \) and \( d^2 \) into the equation:
\[ I = \frac{1}{12} ML^2 + M \left(\frac{L^2}{16}\right) \]

### Step 6: Simplify the Moment of Inertia
Now, we need to combine the two terms:
\[ I = \frac{1}{12} ML^2 + \frac{1}{16} ML^2 \]
To combine these fractions, we need a common denominator. The least common multiple of 12 and 16 is 48:
\[ I = \frac{4}{48} ML^2 + \frac{3}{48} ML^2 = \frac{7}{48} ML^2 \]

### Step 7: Find the Radius of Gyration
The radius of gyration \( k \) is defined as:
\[ k = \sqrt{\frac{I}{M}} \]
Substituting \( I \):
\[ k = \sqrt{\frac{\frac{7}{48} ML^2}{M}} = \sqrt{\frac{7}{48} L^2} \]
This simplifies to:
\[ k = L \sqrt{\frac{7}{48}} \]

### Final Answer
Thus, the radius of gyration \( k \) is:
\[ k = L \sqrt{\frac{7}{48}} \]