The moment of inertia of a metre scale of mass $0.6\ kg$ about an axis perpendicular to the scale and located at the $20\ cm$ position on the scale in $kg\ m^{2}$ is (Breadth of the scale is negligible)

Question

Dr. Abubacker · Accepted Answer

(b) $I = I_{cm} + mx^{2} = \frac{ml^{2}}{12} + mx^{2} = m\left( \frac{(1)^{2}}{12} + (0.3)^{2} \right)$
$= 0.6\left( \frac{1}{12} + 0.09 \right) = 0.104\ kg - m^{2}$

Q. The moment of inertia of a metre scale of mass $0.6\ kg$ about an axis perpendicular to the scale and located at the $20\ cm$ position on the scale in $kg\ m^{2}$ is (Breadth of the scale is negligible)

Explanation

Reviewed by Dr. Abubacker

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