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```        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```
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										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
```
7 years ago
510 Points
```										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

```
7 years ago
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