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 intertia 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 from the axis

(AD)=$\frac{\sqrt{3}}{2}\times =5\sqrt{3}$

Therefore moment of inertia about the axis BC will be,

$l=m{r}^2=200\times (5\sqrt{3}{)}^2$

$=200\times 25\times 3$

$=15000\text{gm}-{\text{cm}}^2$

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

(b) Therefore, the torque will be produced by mass B and C.

Therefore, Net moment of inertia

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

$=2m{r}^2$

$=2\times 200\times {10}^2$

$=400\times 100$

$=40000{\text{gm-cm}}^2$

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