Question

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its centre and perpendicular to its length is:

• A. $ML^2/3$
• B. $ML^2/12$
• C. $ML^2/6$
• D. $ML^2/2$

Hint:

Memorize standard moments of inertia for common shapes.
For a rod about its center, it is \( ML^2/12 \), and about its end, it is \( ML^2/3 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The moment of inertia of a rigid body depends on its mass distribution relative to the axis of rotation.
For a thin uniform rod, the axis passing through its centre and perpendicular to its length is the perpendicular bisector.
Step 2: Key Formula or Approach:
The moment of inertia \( I \) can be derived using the integral:
\[ I = \int_{-L/2}^{L/2} x^2 dm \]
Step 3: Detailed Explanation:
Let the mass of the rod be \( M \) and its length be \( L \).
The linear mass density of the rod is \( \lambda = \frac{M}{L} \).
Consider a small element of length \( dx \) at a distance \( x \) from the centre.
The mass of this element is \( dm = \lambda dx = \frac{M}{L} dx \).
The moment of inertia of the entire rod is:
\[ I = \int_{-L/2}^{L/2} x^2 \left( \frac{M}{L} \right) dx \]
\[ I = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-L/2}^{L/2} \]
\[ I = \frac{M}{3L} \left( \left( \frac{L}{2} \right)^3 - \left( -\frac{L}{2} \right)^3 \right) \]
\[ I = \frac{M}{3L} \left( \frac{L^3}{8} + \frac{L^3}{8} \right) \]
\[ I = \frac{M}{3L} \left( \frac{2L^3}{8} \right) = \frac{M L^2}{12} \]
Step 4: Final Answer:
The correct moment of inertia is \( M L^2/12 \).