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

Dec 24, 2017

45 $k g {m}^{2}$

#### Explanation:

Just use the definition of moment of inertia
$I = m {r}^{2} = 5 k g \cdot {\left(3 m\right)}^{2} = 45$ $k g {m}^{2}$

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

$K = \frac{1}{2} m {v}^{2}$

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

$v = r \omega$

Then $K = \frac{1}{2} m {\left(r \omega\right)}^{2} = \frac{1}{2} \left(m {r}^{2}\right) {\omega}^{2}$

Let $I = \frac{1}{2} m {r}^{2}$ and call it moment of inertia

$K = \frac{1}{2} I {\omega}^{2}$

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.