Mass Moment — Thin Rod About Its Centre For a thin, uniform rod of mass m and length l, what is the mass moment of inertia about an axis through its centre and perpendicular to its length?

Difficulty: Easy

Correct Answer: m l^2 / 12

Explanation:

Introduction / Context: Standard mass moment formulas for slender members are used in vibration and rotational dynamics. The rod-about-centre result is one of the most cited identities in mechanical design and analysis.

Given Data / Assumptions:

  • Uniform rod, length l, mass m.
  • Axis through the midpoint, perpendicular to the rod’s length.

Concept / Approach: For a slender rod, distributing mass along a line, the integral I = ∫ r^2 dm with symmetric limits gives I_center = (1/12) m l^2. About an end, it is (1/3) m l^2; the two are connected by the parallel-axis theorem.

Step-by-Step Solution (outline):

Place origin at the center; coordinate x runs from −l/2 to +l/2. Let linear density λ = m / l, and r = x for the perpendicular axis. Compute I = ∫(x^2) dm = ∫(x^2) λ dx = λ ∫_{−l/2}^{+l/2} x^2 dx = (m/l) * (l^3 / 12) = (1/12) m l^2.

Verification / Alternative check: Using parallel-axis: I_end = I_center + m (l/2)^2 = (1/12) m l^2 + (1/4) m l^2 = (1/3) m l^2, a known correct endpoint value.

Why Other Options Are Wrong: m l^2 / 4 and m l^2 / 6 overpredict; m l^2 / 8 is not a standard value for any principal rod axis.

Common Pitfalls: Confusing area versus mass moments; mixing the end-axis (1/3 m l^2) with center-axis (1/12 m l^2).

Final Answer: m l^2 / 12.

More Questions from Engineering Mechanics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion