# What is the moment of inertia of a pendulum with a mass of 5 kg that is 12 m from the pivot?

Jun 1, 2016

$720 k g \cdot {m}^{2}$

#### Explanation:

In the most simple form, this is an equation that you may use for this type of problem (simple pendulum):

$I = m {r}^{2}$, where
<ul>
$I :$ moment of inertia, $k g \cdot {m}^{2}$<br>
$m :$ mass of object, $k g$<br>
$r :$ radius from pivotal point to the mass, $m$
</ul>

$I = m {r}^{2}$
$I = \left(5 k g\right) {\left(12 m\right)}^{2}$
$I = 720 k g \cdot {m}^{2}$