What is the moment of inertia of a pendulum with a mass of $3 kg$ that is $10 m$ from the pivot?

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Answer 1 · 2018-04-21T09:43:49

$300 \ "kg m"^2"$

Moment of inertia of a simple pendulum is given by:

$I=mr^2$

$m$ is the mass of the object in kilograms

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

And so, we get:

$I=3 \ "kg"*(10 \ "m")^2$

$=3 \ "kg"*100 \ "m"^2$

$=300 \ "kg m"^2$

What is the moment of inertia of a pendulum with a mass of 3 kg3 kg that is 10 m10 m from the pivot?