# 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>

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

http://hyperphysics.phy-astr.gsu.edu/hbase/intcon.html#intcon
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html
https://en.wikipedia.org/wiki/Moment_of_inertia

Best of luck in your studies!

(EDIT: the following response to the same question (with different values) has a lovely diagram which may further help enhance your understanding.)
https://socratic.org/questions/what-is-the-moment-of-inertia-of-a-pendulum-with-a-mass-of-4-kg-that-is-4-m-from