Mass moment of inertia — thin rod comparison How does the mass moment of inertia of a thin uniform rod about one end compare with that about its midpoint?

Introduction / Context:
Designers frequently switch reference axes using parallel-axis relations. For a slender rod, knowing standard mass moments of inertia allows quick dynamic and strength calculations for beams, pendulums, and linkages.

Given Data / Assumptions:

• Uniform thin rod of length L and mass m.
• Axis 1: through the centroid (midpoint) and perpendicular to the rod.
• Axis 2: through one end and perpendicular to the rod.

Concept / Approach:
Standard results: I_centroid = (1/12) * m * L^2. Using the parallel-axis theorem to shift to the end: I_end = I_centroid + m * (L/2)^2 = (1/12) m L^2 + (1/4) m L^2 = (1/3) m L^2. The ratio I_end / I_centroid = (1/3) / (1/12) = 4.

Step-by-Step Solution:

Verification / Alternative check:
Dimensional consistency: both are in kg·m^2 and differ only by the dimensionless factor 4, confirming correctness.

Why Other Options Are Wrong:

• Same/twice/thrice/half: Do not match the computed factor from standard formulas and the parallel-axis theorem.

Common Pitfalls:
Forgetting to square the shift distance in the parallel-axis term m * d^2 or using area moments instead of mass moments.

Final Answer:
Four times