Problem — Moment of inertia of a rod

Problem

A uniform rod of mass $M = 0{,}5\;\mathrm{kg}$ and length $L = 1\;\mathrm{m}$. Calculate $I$ about an axis perpendicular through the centre and then through one end (parallel axis theorem).

Solution

Axis through the centre: $I_\text{centre} = \tfrac{1}{12}ML^2 = \tfrac{1}{12}\cdot 0{,}5\cdot 1^2 \approx \ev{0{,}042\;\mathrm{kg\,m^2}}$ Axis through one end (Huygens–Steiner theorem, $d = L/2$): $I_\text{end} = I_\text{centre} + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{12}ML^2 + \tfrac{1}{4}ML^2 = \tfrac{1}{3}ML^2 = \tfrac{1}{3}\cdot 0{,}5\cdot 1^2 \approx \ev{0{,}167\;\mathrm{kg\,m^2}}$ The moment about the end is four times that about the centre: shifting the axis, the mass rotates on average further away.

Links

Topics: Rotational dynamics Concepts: Moment of inertia Objects: Rod