question_answer

Three particles, each of mass m grams situated at the vertices of an equilateral triangle ABC of side 1 cm (as shown in the figure). The moment [AIPMT (S) 2004]

of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram\[-c{{m}^{2}}\] units will be:

A)            \[(3/4)\,m{{l}^{2}}\]

B) \[2\,m{{l}^{2}}\]

C) \[(5/4)\,\,m{{l}^{2}}\]

D) \[(3/2)\,\,m{{l}^{2}}\]

Correct Answer: C

Solution :

Moment of inertia of the system about AX is given by

\[MI={{m}_{A}}r_{A}^{2}+{{m}_{B}}r_{B}^{2}+{{m}_{C}}r_{C}^{2}\]

\[MI=m{{(0)}^{2}}+m{{(l)}^{2}}+m{{(l\sin {{30}^{0}})}^{2}}\]

\[=m{{l}^{2}}+\frac{m{{l}^{2}}}{4}=\frac{5}{4}m{{l}^{2}}\]

Alternative: Moment of inertia of a system about a line OC perpendicular to AB, in the plane of ABC is

\[{{I}_{CO}}=m\times 0+m\times {{\left( \frac{1}{2} \right)}^{2}}+m\times {{\left( \frac{1}{2} \right)}^{2}}\]

\[\therefore \]      \[{{I}_{CO}}=\frac{m{{l}^{2}}}{4}+\frac{m{{l}^{2}}}{4}=\frac{m{{l}^{2}}}{2}\]

According to parallel-axis theorem

\[{{I}_{AX}}={{I}_{CO}}+M{{x}^{2}}\]

where \[x=\] distance of \[AX\] from CO, \[M=\] total mass of system

\[{{I}_{AX}}=\frac{m{{l}^{2}}}{2}+3m\times {{\left( \frac{l}{2} \right)}^{2}}\]

\[{{I}_{AX}}=\frac{m{{l}^{2}}}{2}+\frac{3m{{l}^{2}}}{4}=\frac{5}{4}m{{l}^{2}}\]