#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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

SAGAR SINGH - IIT DELHI
879 Points
11 years ago

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

509 Points
11 years ago

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