Three identical rods,each of length l and mas | vedclass

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(B) Let the three rods be $AB$,$BC$,and $CA$. We want the moment of inertia $(I)$ about an axis perpendicular to the plane passing through vertex $A$.

Vedclass · Accepted Answer

$\frac{l}{\sqrt{2}}$

Vedclass · Answer

(B) Let the three rods be $AB$,$BC$,and $CA$. We want the moment of inertia $(I)$ about an axis perpendicular to the plane passing through vertex $A$.$1$. For rod $AB$: The axis passes through one end of the rod and is perpendicular to its length. $I_1 = \frac{Ml^2}{3}$.$2$. For rod $AC$: The axis passes through one end of the rod and is perpendicular to its length. $I_2 = \frac{Ml^2}{3}$.$3$. For rod $BC$: The axis is parallel to the axis passing through the center of mass of rod $BC$ and perpendicular to its length. The distance of the center of mass of $BC$ from vertex $A$ is $d = l \sin(60^\circ) = \frac{\sqrt{3}l}{2}$. Using the parallel axis theorem: $I_3 = I_{cm} + Md^2 = \frac{Ml^2}{12} + M\left(\frac{\sqrt{3}l}{2}\right)^2 = \frac{Ml^2}{12} + \frac{3Ml^2}{4} = \frac{Ml^2 + 9Ml^2}{12} = \frac{10Ml^2}{12} = \frac{5Ml^2}{6}$.Total moment of inertia $I = I_1 + I_2 + I_3 = \frac{Ml^2}{3} + \frac{Ml^2}{3} + \frac{5Ml^2}{6} = \frac{2Ml^2}{3} + \frac{5Ml^2}{6} = \frac{4Ml^2 + 5Ml^2}{6} = \frac{9Ml^2}{6} = \frac{3}{2}Ml^2$.The total mass of the system is $3M$. The radius of gyration $k$ is defined by $I = (3M)k^2$.$\frac{3}{2}Ml^2 = 3Mk^2 \implies k^2 = \frac{l^2}{2} \implies k = \frac{l}{\sqrt{2}}$.

Three identical rods,each of length $l$ and mass $M$,are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is

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