Question

Three particles, each of mass $200g$ are kept at the corners of an equilateral triangle of side $10cm$. Find the moment of inertia 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

$Mass=200g=0.2kg$

$sides=10cm=0.1m$

perpendicular distance

$AD=\frac{\sqrt{3}}{2}\times 10$

$=5\sqrt{3}cm$

a)

$\text{Joining of the particle}$

$\text{Moment of inertia along BC}$

$I=m{r}^2$

$=0.2(5\sqrt{3}{)}^2\times {10}^{-4}$

$I=1.5\times {10}^{-3}kg{m}^2$

b)

Passing through particle of perpendicular moment of inertia

$I=m{r}^2+m{r}^2$

$=2m{r}^2$

$I=4\times {10}^{-3}kg{m}^2$