Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.

Explanation

mass m = 100g
`=100xx10^(-3)kg`
`m=4m=400xx10^(-3)kg`
radius `r=10cm`
`=10xx10^(-2)m`
Moment of inertia `l=mr^(2)`
`=4xx100xx10^(-3)xx[(10)/(sqrt(2))xx10^(-2)]^(2)`
`=(4xx10^(-3))/(2)=2xx10^(-3)kg-m^(2)`

But `I=MK^(2)`, K = radius of gyration
`K=sqrt((I)/(m))=sqrt((2xx10^(-3))/(400xx10^(-3)))=sqrt(2)/(20)=(1.414)/(20)`
`=0.0707m=7.07cm`