Question

The moment of inertia of a thin uniform rod of mass ' (M) ' and length ' (L) ', about an axis perpendicular to length and at a distance ' (L/4) ' from one end is

• A. (\frac{ML^2}{6})
• B. (\frac{ML^2}{12})
• C. (\frac{7ML^2}{24})
• D. [suspicious link removed]

Hint:

$I = \frac{ML^2}{3}$ about one end and $\frac{ML^2}{12}$ about the center.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the moment of inertia (\(I\)) of a rod about a specific axis.
The axis is perpendicular to the rod's length and is located at \(L/4\) from one end.
Step 2: Key Formula or Approach:
1) Moment of inertia about the center of mass (COM) for a rod: \(I_{\text{COM}} = \frac{ML^2}{12}\).
2) Parallel Axis Theorem: \(I = I_{\text{COM}} + Mh^2\), where \(h\) is the distance between the COM and the required axis.
Step 3: Detailed Explanation:
The center of mass of a uniform rod of length \(L\) is at its midpoint, i.e., \(L/2\) from either end.
The required axis is at a distance \(L/4\) from one end.
The distance between the COM and the required axis is:
\[ h = \frac{L}{2} - \frac{L}{4} = \frac{L}{4} \]
Now, using the Parallel Axis Theorem:
\[ I = \frac{ML^2}{12} + M\left(\frac{L}{4}\right)^2 \]
\[ I = \frac{ML^2}{12} + \frac{ML^2}{16} \]
To add these, find a common denominator (48):
\[ I = \frac{4ML^2}{48} + \frac{3ML^2}{48} \]
\[ I = \frac{7ML^2}{48} \]
Step 4: Final Answer:
The moment of inertia is \(\frac{7ML^2}{48}\).