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 measures an object's resistance to rotational acceleration about a specific axis.
For a continuous mass distribution, it is calculated using the integral \( I = \int r^2 dm \).

Step 2: Key Formula or Approach:
The approach involves taking an elemental mass \( dm \) at a distance \( x \) from the axis and integrating over the entire length of the rod.
The relevant formula is \( I = \int x^2 dm \), where the limits of integration are from \( -L/2 \) to \( L/2 \).

Step 3: Detailed Explanation:
Consider a thin uniform rod of mass \( M \) and length \( L \).
Let the axis of rotation pass through its center and be perpendicular to its length.
We take an elemental mass \( dm \) of length \( dx \) at a distance \( x \) from the center.
The linear mass density is \( \lambda = M/L \), so the mass of the element is \( dm = (M/L) dx \).
The moment of inertia of this elemental mass is \( dI = dm \cdot x^2 \).
Integrating from \( x = -L/2 \) to \( x = L/2 \) gives the total moment of inertia:
\[ I = \int_{-L/2}^{L/2} x^2 \left(\frac{M}{L}\right) dx \]
\[ I = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-L/2}^{L/2} \]
\[ I = \frac{M}{3L} \left( \frac{L^3}{8} - \left( -\frac{L^3}{8} \right) \right) \]
\[ I = \frac{M}{3L} \left( \frac{2L^3}{8} \right) = \frac{M}{3L} \left( \frac{L^3}{4} \right) = \frac{ML^2}{12} \]

Step 4: Final Answer:
The correct option is (B) as the calculated moment of inertia is \( ML^2/12 \).