What is the moment of inertia of a pendulum with a mass of $8 k g$ $8 kg$ that is $16 m$ $16 m$ from the pivot?

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Answer 1 · 2018-01-27T18:53:11

As,here the radius of the bob of the pendulum is not mentioned,so we shall have to consider the bob to be a point mass.

Therefore,moment of inertia of a point mass $m$ $m$ located at a distance of $r$ $r$ from a point is $m \times r^{2}$ $m×r^2$

Given, $m = 8 K g$ $m=8 Kg$ & $r = 16 m$ $r=16m$

So,the value becomes $2048 K g . m^{2}$ $2048 Kg.m^2$

What is the moment of inertia of a pendulum with a mass of 8kg8 kg that is 16m16 m from the pivot?