A massless equilateral triangle EFG of side ' | vedclass

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(A) Let the vertices be $E(0, 0)$,$G(a, 0)$,and $F(a/2, a\sqrt{3}/2)$.The line $EX$ is the $y$-axis (perpendicular to $EG$ at $E$).The distances of th

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Let the vertices be $E(0, 0)$,$G(a, 0)$,and $F(a/2, a\sqrt{3}/2)$.The line $EX$ is the $y$-axis (perpendicular to $EG$ at $E$).The distances of the particles at $E, G, F$ from the $y$-axis are $r_E = 0$,$r_G = a$,and $r_F = a/2$.The moment of inertia $I$ about the $y$-axis is given by $I = \sum m_i r_i^2$.$I = m(0)^2 + m(a)^2 + m(a/2)^2$$I = 0 + ma^2 + \frac{ma^2}{4} = \frac{5}{4} ma^2$.We are given $I = \frac{N}{20} ma^2$.Equating the two expressions: $\frac{5}{4} ma^2 = \frac{N}{20} ma^2$.$\frac{5}{4} = \frac{N}{20} \implies N = \frac{5 	imes 20}{4} = 25$.

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