question_answer

Three point masses, each of mass M are placed at the comers of an equilateral triangle of side L. The moment of inertia of this system about an axis along one side of the triangle is

A) \[\frac{1}{3}M{{L}^{2}}\]                                             

B) \[\frac{3}{2}M{{L}^{2}}\]

C) \[\frac{3}{4}M{{L}^{2}}\]                                             

D) ML2

Correct Answer: C

Solution :

From the triangle BCD   \[{{(CD)}^{2}}={{(BC)}^{2}}-{{(BD)}^{2}}\] \[={{L}^{2}}-{{\left( \frac{L}{2} \right)}^{2}}\] \[{{x}^{2}}=\frac{3{{L}^{2}}}{4}\] \[\Rightarrow \]   \[x=\frac{\sqrt{3}}{2}L\] Moment of inertia of system along the side AB \[{{I}_{sys.}}={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\] \[=m\times {{(0)}^{2}}+m\times {{(x)}^{2}}+m\times {{(0)}^{2}}\] \[=m{{x}^{2}}=m{{\left( \frac{\sqrt{3L}}{2} \right)}^{2}}\] \[=\frac{3m{{L}^{2}}}{4}\]