A thin rod has a length of 1 m and a mass of | vedclass

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(B) Given: Mass of the rod $m = 0.6 \ kg$,Length $l = 1 \ m$.The center of mass of the rod is at $l/2 = 0.5 \ m$ from one end.The axis is at $20 \ cm

Vedclass · Accepted Answer

$0.104$

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(B) Given: Mass of the rod $m = 0.6 \ kg$,Length $l = 1 \ m$.The center of mass of the rod is at $l/2 = 0.5 \ m$ from one end.The axis is at $20 \ cm = 0.2 \ m$ from one end.The distance $x$ between the center of mass and the given axis is $x = |0.5 \ m - 0.2 \ m| = 0.3 \ m$.Using the parallel axis theorem,$I = I_{cm} + mx^2$.$I = \frac{ml^2}{12} + mx^2 = m \left( \frac{l^2}{12} + x^2 ight)$.Substituting the values: $I = 0.6 \left( \frac{1^2}{12} + (0.3)^2 ight) = 0.6 \left( \frac{1}{12} + 0.09 ight)$.$I = 0.6 \left( 0.0833 + 0.09 ight) = 0.6 	imes 0.1733 = 0.104 \ kg \cdot m^2$.

$A$ thin rod has a length of $1 \ m$ and a mass of $0.6 \ kg$. The moment of inertia of the rod about an axis perpendicular to its length and passing through a point $20 \ cm$ from one end is ................... $kg \cdot m^2$ (the width of the rod is negligible).

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