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

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Answer 1 · 2018-04-23T14:32:58

$I=8color(white)(l)"kg"*"m"^2$

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

$m_"bulb"=8color(white)(l)"kg"$
$r=1color(white)(l)"m"$

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

$I=m*r^2=8color(white)(l)"kg"*"m"^2$

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

$I=I_"cm"+m*d^2$

Where $I_"cm"=0$ for a point mass and $d=r$ the axis of rotation distance from the axis passing through the center of mass.

Hence $I=m*d^2=m*r^2$
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

What is the moment of inertia of a pendulum with a mass of 8 kg8 kg that is 1 m1 m from the pivot?