Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $A B C$ of side $\ell \mathrm{~cm}$ (as shown in the figure). The moment of inertia of the system about a line $A X$ perpendicular to $A B$ and in the plane of $A B C$, in gram $-\mathrm{cm}^2$ units…

Question

Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $A B C$ of side $\ell \mathrm{~cm}$ (as shown in the figure). The moment of inertia of the system about a line $A X$ perpendicular to $A B$ and in the plane of $A B C$, in gram $-\mathrm{cm}^2$ units will be

Aicharya · Accepted Answer

$\mathrm{I}_{\mathrm{AX}}=\mathrm{m}(\mathrm{AB})^{2}+\mathrm{m}(\mathrm{OC})^{2}=\mathrm{m} l^{2}+\mathrm{m}\left(l \cos 60^{\circ}\right)^{2}=m l^{2}+\frac{\mathrm{m} l^{2}}{4}=\frac{5}{4} \mathrm{~m} l^{2}$

Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $A B C$ of…

Solution