A uniform rod of mass m and lengh I makes a constant angle `theta` with an axis of rotation which passes through one end of therod. Find the moment of inertia about this axis.

Explanation

(a) `theta`- angle that rod makes with the axis (vertical ). Linear mass density `lambda = (m)/(t)`

Consider dx.the part of rod
Mass dx, the part of rod `dm = lambda.dx`
`dm= (m)/(l)dx`. The M.I of the rod about 00'
`I= int r^2 dm`
Where is the perpendicular distance from the elementary mass dm.
`sin theta = (r )/(x) " ":. r=x.sin theta`
`I= int_(-(l)/(2))^(l/2) x^2 sin^2 theta. (m)/(l). dx`
`=(m)/(l)*sin theta int_(-(l)/(2))^(l/2) xdx = (m)/(l)*sin^2 theta [(x^2)/(2)]_(-l/2)^(l/2)`
`=(m)/(l)*sin^2theta (2l^3)/(24)= m.sin^2 theta*(l^2)/(12)`
M.I about the axis `00'= (l)/(12)ml^2.sin^2 theta`.