Question

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

Answer 1

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.