Four point masses 2 kg, 3 kg, 4 kg and 5 kg are respectively located at the four corners A, B, C and D of a square of side 1 m as shown in the following figure. Calculate the moment of inertia of the system about
(i) an aris coinciding with side BC
(ii) an axis coinciding with diagonal BD
(iii) an axis passing through A and perpendicular to the plane of the square.

Explanation

Given `AB=BC=CD=DA=1m`
(i) Moment of inertia about an axis coinciding with BC is
`I=2(AB)^(2)+3(0^(2))+4(0^(2))+5(DC)^(2)`
`=2xx1^(2)+5xx1^(2)=7kgm^(2)`
(ii) Moment of inertia about an axis coinciding with BD is
`I=2(OA)^(2)+4(OC)^(2)`
Here, `AC=sqrt(1^(2)+1^(2))=sqrt2m`
`:.OA=OC=(sqrt2)/(2)=(1)/(sqrt(2))m`
`:.I=2xx((1)/(sqrt2))^(2)+4xx((1)/(sqrt2))^(2)`
`=1+2=3kgm^(2)`
Moment of inertia of an axis through A and perpendicular to the plane of the square is
`I=2(0)^(2)+3(BA)^(2)+4(CA)^(2)+5(DA)^(2)`
`=0+3xx1^(2)+4xx(sqrt2)^(2)+5xx1^(2)`
`=3+8+5=16kgm^(2)`