# What is the moment of inertia of a pendulum with a mass of 5 kg that is 9 m from the pivot?

Jan 2, 2016

$I = {r}^{2} \cdot m = {9}^{2} \cdot 5 k g \cdot {m}^{2} = 405 k g \cdot {m}^{2}$

#### Explanation:

The moment of inertia is defined as the distances of all infinitely small masses distributed across the body's whole mass. As an integral:

$I = \int {r}^{2} \mathrm{dm}$

This is useful for bodies of which geometry can be expressed as a function. However, since you only have one body in a very specific spot, it is simply:

$I = {r}^{2} \cdot m = {9}^{2} \cdot 5 k g \cdot {m}^{2} = 405 k g \cdot {m}^{2}$