What is the moment of inertia of a pendulum with a mass of $7 k g$ $7 kg$ that is $3 m$ $3 m$ from the pivot?

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Answer 1 · 2017-03-05T12:10:46

$I = 63 k g m^{2}$ $I = 63kgm^2$

Moment of inertia is the product of mass and distance squared,

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

Here the mass is $7$ $7$ and the distance is $3$ $3$ , so

$I = 7 \times 3^{2} = 7 \times 9 = 63 k g m^{2}$ $I = 7 xx 3^2 = 7 xx 9 = 63kgm^2$

What is the moment of inertia of a pendulum with a mass of 7kg7 kg that is 3m3 m from the pivot?