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)  \[M{{L}^{2}}\]

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 \[{{I}_{sys.}}={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\]                 \[=M\times {{(0)}^{2}}+M\times {{(x)}^{2}}+M\times {{(0)}^{2}}\]                 \[=M{{x}^{2}}=M{{\left( \frac{\sqrt{3}L}{2} \right)}^{2}}=\frac{3M{{L}^{2}}}{4}\]