What is the moment of inertia of a pendulum with a mass of $9 kg$ that is $10 m$ from the pivot?

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Answer 1 · 2018-08-14T02:53:41

$900 \ "kg m"^2$

Moment of inertia of a simple pendulum is given by the equation:

$I=mr^2$

where:

$m$ is the mass of the pendulum's bob in kilograms

$r$ is the distance of the pendulum's bob from the pivot

So, we get:

$I=9 \ "kg"*(10 \ "m")^2$

$=9 \ "kg"*100 \ "m"^2$

$=900 \ "kg m"^2$

What is the moment of inertia of a pendulum with a mass of 9 kg9 kg that is 10 m10 m from the pivot?