What is the moment of inertia of a pendulum with a mass of $5 kg$ that is $2 m$ from the pivot?

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Answer 1 · 2018-04-30T01:14:14

$20 \ "kg m"^2$

Moment of inertia of a simple pendulum is given by,

$I=mr^2$

$m$ is the mass of the pendulum

$r$ is the distance of the pendulum from the pivot point

And so, plugging in our given values, we get,

$I=5 \ "kg"*(2 \ "m")^2$

$=5 \ "kg"*4 \ "m"^2$

$=20 \ "kg m"^2$

What is the moment of inertia of a pendulum with a mass of 5 kg5 kg that is 2 m2 m from the pivot?