- #1
shotgunshogun
- 11
- 0
Homework Statement
A thin rod is exactly 1.9 meters long. The density in this rod varies in a peculiar manner. If we call the left-hand end of the rod x=0 and the right-hand end of the rod x=L , then the linear density can be expressed in units of kilograms per meter as http://coswebhost.rit.edu/webwork2_files/tmp/equations/6b/d06d55bce8c4f91789b960b87d21021.png
You grab the RIGHT-hand end of the rod and prepares to swing the rod this end. What is the moment of inertia of the rod around this end?
The Attempt at a Solution
I tried to intergrate from -L to 0 since your changing the axis of rotation and i got 1.16 kgm^2. You integrate the linear density times x^2 dx if you revolve from the left end , the bounds would be from 0 to L (1.85 kgm^2). The right side though... it doesn't work just the same, what needs to be changed to fufill the question, change of bounds, change of the density equation?
Last edited by a moderator: