# What is the moment of inertia of a pendulum with a mass of 8 kg that is 1 m from the pivot?

Apr 23, 2018

$I = 8 \textcolor{w h i t e}{l} {\text{kg"*"m}}^{2}$

#### Explanation:

Taking the pendulum bulb as a point mass and the pivot as the rotational axis,

${m}_{\text{bulb"=8color(white)(l)"kg}}$
$r = 1 \textcolor{w h i t e}{l} \text{m}$

Moment of inertia of the bulb as a point mass: [1]

$I = m \cdot {r}^{2} = 8 \textcolor{w h i t e}{l} {\text{kg"*"m}}^{2}$

Note that the momentum of inertia expression can be derived using the parallel axis theorem: [2]

$I = {I}_{\text{cm}} + m \cdot {d}^{2}$

Where ${I}_{\text{cm}} = 0$ for a point mass and $d = r$ the axis of rotation distance from the axis passing through the center of mass.

Hence $I = m \cdot {d}^{2} = m \cdot {r}^{2}$
Reference
[1] List of moments of inertia, the English Wikipedia, https://en.wikipedia.org/wiki/List_of_moments_of_inertia
[2] Parallel axis theorem, the English Wikipedia, https://en.wikipedia.org/wiki/Parallel_axis_theorem