# What is the moment of inertia of a pendulum with a mass of 4 kg that is 6 m from the pivot?

Sep 21, 2017

=144$k g \cdot {m}^{2}$

#### Explanation:

For this question we will assume that the pendelum has no friction and is only being acted upon by gravity. In conditions like this, the moment of inertia for the pendulum is equal to
$I = m {r}^{2}$, where $I$ is the moment of inertia in kilograms times metres squared, $m$ is the mass of the pendulum in kilograms, and $r$ is the length of the pendulum from its pivot, in metres.

We know that $m = 4$, and $r = 6$. Plug these into the formula to get
$I = 4 \cdot {6}^{2}$
$I = 4 \cdot 36$
$I = 144 k g \cdot {m}^{2}$, the moment of inertia.

https://en.wikipedia.org/wiki/Moment_of_inertia

I hope that helped!