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

Concept: The moment of inertia is the rotational analogue of mass in linear motion. It represents the resistance of a body to rotational acceleration about a given axis. For a rigid body, it depends on both the mass of the body and the distribution of that mass relative to the axis of rotation. Mathematically, the moment of inertia is defined as: \[ I = \sum m_i r_i^2 \] or for a continuous body, \[ I = \int r^2 \, dm \] where $r$ is the perpendicular distance of the mass element from the axis of rotation. For a uniform thin rod, the mass is evenly distributed along its length.

Step 1: Consider a small element of the rod. Let a small element of length $dx$ at a distance $x$ from the center have mass \[ dm = \frac{M}{L}dx \] because the mass per unit length is constant.

Step 2: Apply the definition of moment of inertia. \[ I = \int x^2 dm \] Substituting $dm$, \[ I = \int x^2 \left(\frac{M}{L}\right) dx \] The limits of integration are from $-L/2$ to $L/2$ since the axis passes through the center. \[ I = \frac{M}{L} \int_{-L/2}^{L/2} x^2 dx \]

Step 3: Evaluate the integral. \[ I = \frac{M}{L}\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} \] \[ I = \frac{M}{L}\left(\frac{L^3}{24} + \frac{L^3}{24}\right) \] \[ I = \frac{ML^2}{12} \] Thus, the moment of inertia of a uniform thin rod about an axis through its center and perpendicular to its length is: \[ I = \frac{ML^2}{12} \]