Question

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

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

Hint:

Important Standard Results for Moment of Inertia of a Rod: About center (perpendicular to rod): $I = \frac{ML^2}{12}$ About one end (perpendicular to rod): $I = \frac{ML^2}{3}$ These results are commonly used in rotational dynamics problems.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The moment of inertia ($I$) measures an object's resistance to rotational acceleration. It depends on the distribution of mass relative to the axis of rotation. For a continuous object like a rod, we calculate this using integration.
Step 2: Key Formula or Approach:
The general formula is $I = \int r^2 \, dm$. For a uniform rod, the linear mass density is $\lambda = M/L$.
Step 3: Detailed Explanation:
1. Place the center of the rod at the origin $(x=0)$. The rod extends from $-L/2$ to $+L/2$.
2. Take a small element of length $dx$ at distance $x$ from the center. Its mass $dm = (M/L)dx$.
3. Integrate the moment of inertia for this element: \[ I = \int_{-L/2}^{L/2} x^2 \left( \frac{M}{L} \right) dx = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-L/2}^{L/2} \]
4. Substitute the limits: \[ I = \frac{M}{L} \left[ \frac{(L/2)^3}{3} - \frac{(-L/2)^3}{3} \right] = \frac{M}{L} \left[ \frac{L^3}{24} + \frac{L^3}{24} \right] \] \[ I = \frac{M}{L} \left( \frac{2L^3}{24} \right) = \frac{ML^2}{12} \]
Step 4: Final Answer
The moment of inertia is $ML^2/12$.