A rod PQ of mass m and length L is rotated about an axis through P as shown in figure. Find the moment of inertia of the rod about the axis of rotation.

Explanation

Consider a small element .dx. of the rod which is at a distance .x. from the end .P.. If `.theta.` is the inclinaition of rod w.r.t to the axis of rotation, the radius of the circle in which the element rotates is given by
`sin theta = (r )/(x) " " r = x sin theta`
`therefore` M.I of the element about the axis of rotation is `dI=dm.r^(2)`
where dm is the mass of element `dm=(m)/(L).dx`
`therefore dI=(m)/(L).dx.(x sin theta)^(2)`
The total M.I. of the rod is given by
`int dI=int_(O)^(L)(m)/(L)sin^(2)theta.x^(2)dx, I=(mL^(2))/(3).sin^(2)theta`