The moment of inertia of a meter scale of mas | vedclass

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(B) The mass of the meter scale is $M = 0.6\, kg$ and its length is $L = 1\, m$.The center of mass of the scale is at the $50\, cm$ mark.The axis of r

Vedclass · Accepted Answer

$0.104$

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(B) The mass of the meter scale is $M = 0.6\, kg$ and its length is $L = 1\, m$.The center of mass of the scale is at the $50\, cm$ mark.The axis of rotation is at the $20\, cm$ mark.The distance $r$ between the center of mass and the axis of rotation is $r = |50\, cm - 20\, cm| = 30\, cm = 0.3\, m$.Using the parallel axis theorem,$I = I_{CM} + Mr^2$,where $I_{CM} = \frac{ML^2}{12}$.Substituting the values: $I = \frac{0.6 	imes (1)^2}{12} + 0.6 	imes (0.3)^2$.$I = 0.05 + 0.6 	imes 0.09$.$I = 0.05 + 0.054 = 0.104\, kg\, m^2$.

The moment of inertia of a meter 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).

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