Derive the expresssion for moment of inertia of a rod about its centre and perpendicular to the rod.

Explanation

(i) Consider a uniform rod of mass M and length l as shown in figure.

(ii) Let us consider the rod is along the x-axis and the moment of inertia of the rod is found about the axis, which passes through center of mass (here the geometrical center) of the rod 'O'.
(iii) Now the moment of inertia of an infinitesimal small mass 'dm' of length dx of the rod, which is at a distance 'x' from O can be expressed as,
`d I = (dm)x^(2)" "...(1)`
(iv) The moment of inertia(I) of the entire rod can be found by integrating the equation(1) as,
`I = int dI = underset(-(l)/(2))overset(+(1)/(2))(int) (dm)x^(2)" "...(2)`
(v) If `lambda` is linear mass density (i.e. `lambda = (m)/(l)`) the small mass dm can be written as.
`dm = lambda dx = (m)/(l) dx`
(vi) Substituting the 'dm' value in equation(2), we get,
`I = underset(-(l)/(2))overset(+(l)/(2))(int) ((m)/(l)dx) x^(2) = (m)/(l) underset(-(l)/(2))overset(+(l)/(2))(int) x^(2) dx`
Using integration, `I = (M)/(l) underset(-(l)/(2))overset((l)/(2))(int) x^(2) dx rArr (M)/(l) [(x^(3))/(3)]_(-(l)/(2))^(+(l)/(2))`
`I = (M)/(3l)xx [((l)/(2))^(3)+((l)/(2))^(3)] = (M)/(3l) [(2l^(3))/(8)], I = (1)/(12) Ml^(2)`