Physics - Rotation Question with Solution | TestHub

Let the three masses be placed at the vertices A, B, and C of the equilateral triangle. Let the axis of rotation be along the side AB.

The perpendicular distance of the masses at A and B from the axis of rotation is zero, so their contribution to the moment of inertia is zero.

The perpendicular distance of the mass at C from the side AB is the altitude of the equilateral triangle, given by $r=L\sin (60^{\circ })=\frac{\sqrt{3}L}{2}$.

The moment of inertia of the system about the side AB is:

$I=m_A{r}_A^2+m_B{r}_B^2+m_C{r}_C^2$

Substituting the values $m_A=m_B=m_C=2$ kg, $r_A=0$, $r_B=0$, and $r_C=\frac{\sqrt{3}L}{2}$:

$I=2(0{)}^2+2(0{)}^2+2{\left(\frac{\sqrt{3}L}{2}\right)}^2$

$I=2\left(\frac{3L^2}{4}\right)=\frac{3L^2}{2}$

Answer: $\frac{3L^2}{2}$