What is the moment of inertia of a pendulum with a mass of $8 k g$ $8 kg$ that is $4 m$ $4 m$ from the pivot?

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Answer 1 · 2017-03-31T05:46:50

The moment of inertia is $= 128 k g m^{2}$ $=128kgm^2$

The moment of inertia is

$I = m \cdot r^{2}$ $I=m*r^2$

mass, $m = 8 k g$ $m=8kg$

radius, $r = 4 m$ $r=4m$

$I = 8 \cdot 4^{2} = 128 k g m^{2}$ $I=8*4^2=128kgm^2$

What is the moment of inertia of a pendulum with a mass of 8kg8 kg that is 4m4 m from the pivot?