Question

Three particles, each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertial of the system about an axis
(a) joining two of the particles and
(b) passing through one of the particles and perpendicular to the plane of the particles.

Answer 1

(a) The distance of mass at A from the axis passing through side BC,
$\left(\mathrm{AD}\right)=\frac{\sqrt{3}}{2}\times 10=5\sqrt{3}\mathrm{cm}$



Therefore, we have:
Moment of inertia of mass about the axis BC,
$$\begin{gathered} l=m{r}^2=200\times {\left(5\sqrt{3}\right)}^2 \\ =200\times 25\times 3 \\ =15000\mathrm{gm}-{\mathrm{cm}}^2 \\ =1.5\times {10}^{-3}\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$

(b) Now, let the axis of rotation passes through A and is perpendicular to the plane of triangle.
Therefore, we have:
Net moment of inertia,
$$\begin{gathered} l=m{r}^2+m{r}^2 \\ =2m{r}^2 \\ =2\times 200\times {10}^2 \\ =400\times 100 \\ =40000\mathrm{gm}-{\mathrm{cm}}^2 \\ =4\times {10}^{-3}\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$