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`
`:.` M.I of the element about the axis of rotation is `dm=(m)/(L).dx`
`:. dI=(m)/(L) .dx.(x sin theta)^(2)`
The total M.I of the rod is given by
`int dI= int _(0)^(L) sin^(2) theta. x^(2)dx, I=(mL^(2))/(3) sin^(2) theta`