Question

A uniform rod of mass 2 kg and length of 2.0 m is pivoted about an axis perpendicular to the rod and 50 cm from its left end. Find the rotational inertia about this axis.

Answer 1

maggM =2kg

L=2.0m

$I_G=\frac{ML^2}{12}\\ I_0=I_G+Mr^2\\ I_0=\frac{ML^2}{12}+ M(\frac{50}{100})^2=\frac{2\times2^2}{12}+(2\times0.5^2)=\frac{2}{3}+\frac{1}{2}=1.1667kgm^2\\ I_0=1.1667 kgm^2$