# What is the moment of inertia of a pendulum with a mass of 5 kg that is 11m from the pivot?

Apr 27, 2018

$605 \setminus {\text{kg m}}^{2}$

#### Explanation:

Moment of inertia $\left(I\right)$ of a simple pendulum is given by:

$I = m {r}^{2}$

• $m$ is the mass of the pendulum bob

• $r$ is the distance of the pendulum from the pivot

So, inputting our given values, we get:

I=5 \ "kg"*(11 \ "m")^2

$= 5 \setminus {\text{kg"*121 \ "m}}^{2}$

$= 605 \setminus {\text{kg m}}^{2}$