Question

Find the moment of inertia of the system formed using two identical rods about the given axis of rotation as shown in the figure. Each rod has mass \( M \) and length \( L \). v

• A. \( \dfrac{17}{12}ML^2 \)
• B. \( \dfrac{13}{12}ML^2 \)
• C. \( \dfrac{2}{3}ML^2 \)
• D. \( \dfrac{3}{4}ML^2 \)

Hint:

For composite rigid bodies:
Split the system into simple components
Use standard moment of inertia formulas
Apply the parallel axis theorem whenever the axis does not pass through the centre of mass

Answer 1

The Correct Option is A

Step 1: The system consists of two identical rods:
One vertical rod of length \( L \) and mass \( M \), with axis passing through its upper end \( O \)
One horizontal rod of length \( L \) and mass \( M \), attached at the lower end of the vertical rod The axis of rotation passes through point \( O \) and is perpendicular to the plane of the rods.
Step 2: Moment of inertia of the vertical rod about one end: \[ I_1 = \frac{1}{3}ML^2 \]
Step 3: Moment of inertia of the horizontal rod: About its own centre (axis perpendicular to the rod): \[ I_{\text{cm}} = \frac{1}{12}ML^2 \] Distance of its centre from axis \( O \) is \( L \), so by parallel axis theorem: \[ I_2 = I_{\text{cm}} + ML^2 = \frac{1}{12}ML^2 + ML^2 = \frac{13}{12}ML^2 \]
Step 4: Total moment of inertia of the system: \[ I = I_1 + I_2 = \frac{1}{3}ML^2 + \frac{13}{12}ML^2 = \frac{4}{12}ML^2 + \frac{13}{12}ML^2 = \frac{17}{12}ML^2 \]