A non uniform rod AB of mass M and length 2l. the mass per unit lenght of the rod is mx.At a point of the rod distance x from A.

Find the moment of inertia of this rod.About an axis perpendicular to the rod through A

Dear bharath,

This is a very simple problem of Integration.

Length of the rod is 2l. Let 0 be its centre. Take any small element dx on the rod.

Then Moment of inertia Integral mx²dx and integrate it from -l to +l.

So we get 2ml³/3 as the required moment of inertia...

Sagar Singh

B.Tech, Chemical Engg

IIT Delhi

let the rod of length 2L is placed in +ve x axis with its end A is at origin.....

consider a small element dx at a distance x from origen.......

mass of that element is dm=(linear mass density).dx

                                      dm =mxdx

now moment of inertial of this small element about end A is

          dI=dmx^2

             =(mxdx)x^2

           dI  =mx^3dx

now integrating this expression limit from 0 to 2L

          I=16mL^4/4

           I=4mL^4 units

 approve my ans if u like