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 (( 3 ) /(8)) l `
B. ` ( 1 ) /(4) ` l
C. ` sqrt(( 7 ) /(48)) ` l
D. ` ( 1 ) /(8) ` 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 need to find the radius of gyration \( K \) about an axis that is not the center of mass of the rod. The radius of gyration is defined as the distance from the axis of rotation to a point where the entire mass of the body can be assumed to be concentrated without changing the moment of inertia.

### Step 2: Identify the moment of inertia about the center of mass
The moment of inertia \( I_{CM} \) of a thin uniform rod about its center of mass is given by:
\[ I_{CM} = \frac{1}{12} M L^2 \]

### Step 3: Use the parallel axis theorem
To find the moment of inertia \( I_{AB} \) about the new axis (which is at a distance \( D = \frac{L}{4} \) from the center of mass), we use the parallel axis theorem:
\[ I_{AB} = I_{CM} + M D^2 \]
Substituting the values:
\[ I_{AB} = \frac{1}{12} M L^2 + M \left(\frac{L}{4}\right)^2 \]

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

### Step 5: Substitute \( D^2 \) into the equation
Now substitute \( D^2 \) into the equation for \( I_{AB} \):
\[ I_{AB} = \frac{1}{12} M L^2 + M \cdot \frac{L^2}{16} \]

### Step 6: Find a common denominator
To add the two fractions, find a common denominator (which is 48):
\[ I_{AB} = \frac{4}{48} M L^2 + \frac{3}{48} M L^2 = \frac{7}{48} M L^2 \]

### Step 7: Relate moment of inertia to radius of gyration
The moment of inertia can also be expressed in terms of the radius of gyration \( K \):
\[ I_{AB} = M K^2 \]
Setting the two expressions for \( I_{AB} \) equal to each other gives:
\[ M K^2 = \frac{7}{48} M L^2 \]

### Step 8: Solve for \( K^2 \)
Dividing both sides by \( M \) (assuming \( M \neq 0 \)):
\[ K^2 = \frac{7}{48} L^2 \]

### Step 9: Take the square root to find \( K \)
Taking the square root gives:
\[ K = \sqrt{\frac{7}{48}} L \]

### Final Answer
The radius of gyration \( K \) for the rod about the specified axis is:
\[ K = \sqrt{\frac{7}{48}} L \]