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(A) Given: Mass $M = 1 \, kg$, Length $L = 5 \, m$.The moment of inertia of a thin rod about an axis passing through its center of mass and perpendicu

Vedclass · Accepted Answer

$4.33 \, kg \cdot m^2$

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(A) Given: Mass $M = 1 \, kg$, Length $L = 5 \, m$.The moment of inertia of a thin rod about an axis passing through its center of mass and perpendicular to its length is $I_{CM} = \frac{ML^2}{12}$.$I_{CM} = \frac{1 	imes 5^2}{12} = \frac{25}{12} \approx 2.083 \, kg \cdot m^2$.The axis of rotation is at a distance $x = 1 \, m$ from one end. The center of mass is at the midpoint, which is $L/2 = 2.5 \, m$ from either end.The distance $d$ between the center of mass and the new axis is $d = |2.5 \, m - 1 \, m| = 1.5 \, m$.Using the parallel axis theorem, $I = I_{CM} + Md^2$.$I = 2.083 + (1 	imes 1.5^2) = 2.083 + 2.25 = 4.333 \, kg \cdot m^2$.Thus, the moment of inertia is $4.33 \, kg \cdot m^2$.

Consider a thin metal strip of mass $1 \, kg$ and length $5 \, m$. Calculate its moment of inertia about an axis perpendicular to the strip and located at $100 \, cm$ on the strip from one of its ends. (Assume the breadth of the strip is negligible.)

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