Physics - Rotation Question with Solution | TestHub

Let the three masses be placed at the vertices A, B, and C of the equilateral triangle of side $L$.

The axis of rotation passes through vertex A and is parallel to the side BC.

The perpendicular distance of vertex A from the axis is zero.

The perpendicular distance of vertices B and C from the axis is equal to the altitude of the equilateral triangle.

Altitude of the equilateral triangle, $h=L\sin (60^{\circ })=\frac{\sqrt{3}}{2}L$.

The moment of inertia of the system about the given axis is:

$I=I_A+I_B+I_C$

$I=m(0{)}^2+m(h{)}^2+m(h{)}^2$

$I=2m{h}^2$

Substituting the value of $h$:

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

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

$I=\frac{3}{2}m{L}^2$

Answer: $\frac{3{\mathrm{mL}}^2}{2}$