# What is the moment of inertia of a pendulum with a mass of 2 kg that is 7 m from the pivot?

Dec 18, 2016

The answer is $I = m {r}^{2}$ = 98 $k g {m}^{2}$

#### Explanation:

Assuming you are not being asked to derive the moment of inertia of the pendulum, I will state without proof that this is

$I = m {r}^{2}$

where $m$ is the mass hanging from the pendulum and $r$ is the distance from the pivot to the centre of the mass.

With your values, this comes to

$I = 2 \times {7}^{2}$ = 98 $k g {m}^{2}$