Problem — Pivoted rod falling from horizontal

Problem

A uniform rod of mass $M = 0{,}80\;\mathrm{kg}$ and length $L = 1{,}0\;\mathrm{m}$ is pivoted at one end and can rotate in a vertical plane. It starts from the horizontal position, at rest, and is released. Find the angular velocity as it passes through the vertical (rod pointing down). Use $I = \tfrac{1}{3}ML^2$.

Solution

The rod rotates about the pivot: mechanical energy is conserved. The centre of mass, initially at the height of the pivot, drops by $L/2$ when the rod is vertical. The potential energy lost becomes rotational kinetic energy: $Mg\,\frac{L}{2} = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}\left(\tfrac{1}{3}ML^2\right)\omega^2$ The mass cancels; solving for $\omega$: $g\,\frac{L}{2} = \tfrac{1}{6}L^2\omega^2 \quad\Rightarrow\quad \omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3\cdot 9{,}81}{1{,}0}}$ $\ev{\omega \approx 5{,}4\;\mathrm{rad/s}}$ Note that the result does not depend on the mass: as with free fall, only the geometry matters.

Connections

Topics: Dinamica rotazionale Concepts: Conservazione dell’energia meccanica · Energia cinetica rotazionale · Momento d’inerzia Objects: Asta