Three point masses,each of mass m,are placed | vedclass

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(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. The masses are placed at $A, B,$ and $C$. The axis of rotation passes through sid

Vedclass · Accepted Answer

$\frac{3}{4}m a^2$

Vedclass · Answer

(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. The masses are placed at $A, B,$ and $C$. The axis of rotation passes through side $AB$.$1$. The perpendicular distance of mass at $A$ from the axis $AB$ is $r_A = 0$.$2$. The perpendicular distance of mass at $B$ from the axis $AB$ is $r_B = 0$.$3$. The perpendicular distance of mass at $C$ from the axis $AB$ is the altitude $x$ of the equilateral triangle.Using the Pythagorean theorem in the triangle formed by the altitude:$x^2 + (a/2)^2 = a^2$$x^2 = a^2 - a^2/4 = 3a^2/4$$x = \frac{\sqrt{3}}{2}a$The moment of inertia $I$ of the system about the axis $AB$ is given by:$I = \sum m_i r_i^2 = m(r_A^2) + m(r_B^2) + m(r_C^2)$$I = m(0)^2 + m(0)^2 + m(x)^2$$I = m \left( \frac{\sqrt{3}}{2}a \right)^2 = m \left( \frac{3}{4}a^2 \right) = \frac{3}{4}m a^2$

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $a$. The moment of inertia of this system about an axis passing along one side of the triangle is:

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