What is the moment of inertia of a pendulum with a mass of #5 kg# that is #3 m# from the pivot?

1 Answer
Dec 24, 2017

Answer:

45 # kgm^2#

Explanation:

Just use the definition of moment of inertia
#I = mr^2 = 5 kg* (3m)^2 = 45# # kgm^2#

To appreciate this definition, take a look at its kinetic energy at any given constant,

#K = 1/(2)mv^2#

Since a pendulum swings back and forth, you can measure/observe two kinds of speeds, the speed of the blob (v) and the speed of the angular change over time (#omega#). Their relationship is

# v=romega #

Then # K = 1/(2) m (romega)^2 =1/(2) (m r^2) omega^2 #

Let # I = 1/(2) mr^2# and call it moment of inertia

#K =1/(2) I omega^2#

From an utility point of view, it is much easier to measure the angular change over time (#omega#) then the actual speed (v) of the blob. The trade off is that besides knowing the mass, you also need to know the radius of rotation (length of the pendulum) #I# in order to figure out the kinetic energy of the pendulum.

And of course, you can use the concept of angular momentum to appreciate the definition of #I# as well.