Question

A square frame ABCD is formed by 4 identical rods each of mass \(m\) and length \(\ell\). This frame is in XY plane such that side AB coincides with x-axis and side AD along y-axis. The moment of inertia of the frame about x-axis is

• A. \(\frac{5m\ell ^{2}}{3}\)
• B. \(\frac{2m\ell ^{2}}{3}\)
• C. \(\frac{4m\ell ^{2}}{3}\)
• D. \(\frac{m\ell ^{2}}{12}\)

Hint:

Add the moments of inertia of the four rods about the x-axis. One lies on the axis, one is parallel to it, and two are perpendicular with one end on it.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The moment of inertia of a frame is the sum of those of its rods about the same axis.

Step 2: Rod AB
AB lies along the x-axis, so \(I_{AB}=0\).

Step 3: Rod CD
CD is parallel to the x-axis at distance \(\ell\), so \(I_{CD}=m\ell^{2}\).

Step 4: Rods AD and BC
Each is perpendicular to the axis with one end on it, so \(I=\frac{m\ell^{2}}{3}\) for each.

Step 5: Total
\(I=0+m\ell^{2}+\frac{m\ell^{2}}{3}+\frac{m\ell^{2}}{3}=\frac{5m\ell^{2}}{3}\).

Final Answer:
The moment of inertia of the frame about the x-axis is 5 m l squared over 3. \[ \boxed{\text{(A) }\dfrac{5m\ell^{2}}{3}} \]