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 of each particle m =100 g = 0.1kg
Length of side of square , a= 10cm = 0.1 m is square distance of corner from centre of square `1/2` diagonal = `(sqrt(2)a)/(2) = (a)/(sqrt2)`

`therefore` Total moment of inertia
(a) ` I = m_1r_^2 + m_2r_2^2 + m_3r^3 + m_4r_4^2 = 4mr^2`
`therefore I = 4xx 0.1 ((1)/(sqrt2) xx 0.1)^2 = 4 xx 1/2 xx 0.1^3`
` 2xx 10^(-3) kgm^2`
(d) Radius of gyration k = `sqrt(1/m) = sqrt((2xx10^(-3))/(4xx10^(-1))`
`(1)/(sqrt2) xx 10^(-1)`
` =0.7071 xx 10^(-1) m " or " 7.071 cm`