# What is the moment of inertia of a pendulum with a mass of 8 kg that is 16 m from the pivot?

Therefore,moment of inertia of a point mass $m$ located at a distance of $r$ from a point is m×r^2
Given,$m = 8 K g$ & $r = 16 m$
So,the value becomes $2048 K g . {m}^{2}$