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

1 Answer
Jun 1, 2016

Answer:

#720kg*m^2#

Explanation:

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

#I=mr^2#, where
<ul>
#I:# moment of inertia, #kg*m^2#<br>
#m:# mass of object, #kg#<br>
#r:# radius from pivotal point to the mass, #m#
</ul>

#I=mr^2#
#I=(5kg)(12m)^2#
#I=720kg*m^2#

The following resources may help you further understand the concepts:
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