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

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Answer 1 · 2016-03-29T20:49:32

$512 k g . m^{2}$ $512kg.m^2$

$I = \sum j m_{j} r_{p e r p j}^{2}$ $I=sum_jm_jr_(perpj)^2$

$= M R^{2}$ $=MR^2$

$= 8 \times 8^{2}$ $=8xx8^2$

$= 512 k g . m^{2}$ $=512kg.m^2$ .

What is the moment of inertia of a pendulum with a mass of 8 kg8kg8 kg that is 8 m8m8 m from the pivot?