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

Answer

A, B, C, D

Explanation

a. The distance form the axis
`(AD)=sqrt3/2xx10=5sqrt3cm`

Therefore moment of inertial about the axis BC will be
`l=mr^2=200xx(5sqrt3)^2`
`=200xx25xx3`
`=15000 gm-cm^2`
`=1.5xx10&-3kg-m^2`
b. The axis of rotation let pass through A ankd perpendicular to the plane of triangle.
Therefore the torque will be produced by mass B and C.
Therefore Net moment of inertia
`=l=mr^2=mr^2`
`=2mr^2`
` =2xx200xx10^2`
`=400xx100`
`=40000gm-cm^2`
`=4xx10^-3kg-m^2`