What is the moment of inertia of a pendulum with a mass of $8 k g$ $8 kg$ that is $6 m$ $6 m$ from the pivot?

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Answer 1 · 2016-06-01T18:39:12

$288 k g \cdot m^{2}$ $288kg*m^2$

In the most simple form, this is an equation that you may use for this type of problem (simple pendulum):

$I = m r^{2}$ $I=mr^2$ , where

I :

$I:$

k g \cdot m^{2}

$kg*m^2$

m :

$m:$

k g

$kg$

r :

$r:$

m

$m$

$I = m r^{2}$ $I=mr^2$
$I = (8 k g) {(6 m)}^{2}$ $I=(8kg)(6m)^2$
$I = 288 k g \cdot m^{2}$ $I=288kg*m^2$

Best of luck in your studies!

EDIT: the following response to the same question (with different values) has a lovely diagram which may further help enhance your understanding.
https://socratic.org/questions/what-is-the-moment-of-inertia-of-a-pendulum-with-a-mass-of-4-kg-that-is-4-m-from

What is the moment of inertia of a pendulum with a mass of 8kg8 kg that is 6m6 m from the pivot?