Moment of Inertia of Bent Rod – Rankers Physics

Moment of Inertia of Bent Rod

A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^\circ$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is: (2008)

$\frac{\sqrt{2}ML^2}{24}$
$\frac{ML^2}{24}$
$\frac{ML^2}{12}$
$\frac{ML^2}{6}$

Solution:

Treat the bent rod as two rods of length $L/2$ and mass $M/2$ connected at the origin. Using the moment of inertia formula for a rod about its end $I = \frac{m l^2}{3}$, total $I = 2 \times \frac{(M/2)(L/2)^2}{3} = \frac{ML^2}{12}$.

Leave a Reply

Your email address will not be published. Required fields are marked *