# What is the moment of inertia of a pendulum with a mass of 3 kg that is 12 m from the pivot?

Jan 20, 2018

$432 K g . {m}^{2}$

#### Explanation:

Consider the bob of the pendulum to be a point mass as radius is not given.

So moment of inertia of a mass $m$ at a distance of $l$ from a pivoted point is $m {l}^{2}$

Given, $m = 3 K g$ and $l = 12 m$

So,its moment of inertia will be 3×(12)^2 Kg.m^2 or $432 K g . {m}^{2}$