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

Mar 29, 2016

$512 k g . {m}^{2}$

#### Explanation:

$I = {\sum}_{j} {m}_{j} {r}_{p e r p j}^{2}$

$= M {R}^{2}$

$= 8 \times {8}^{2}$

$= 512 k g . {m}^{2}$.