# What is the moment of inertia of a pendulum with a mass of 8 kg that is 2 m from the pivot?

$I = m \cdot {r}^{2} = 8 \cdot {2}^{2} = {2}^{5} = 32$
A pendulum you can treat as point mass of 8 kg, with this in mind you can simply use the straight forward formula $I = m {r}^{2}$
In general the mass moment of inertia ${I}_{m} = \int {r}^{2} \mathrm{dm}$
$I = m \cdot {r}^{2} = 8 \cdot {2}^{2} = {2}^{5} = 32$