Concept of Moment of Inertia and its limits of integration

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a rotating rod and a solid cylinder, specifically addressing the limits of integration. For the rod, the integration limits are correctly set from L/2 to -L/2. In contrast, the cylinder's limits should start from 0 to R, with the angular coordinate varying from 0 to 2π to cover the entire volume. The integration accounts for the cylindrical shape by incorporating the volume element, which includes the angular component. This clarification helps in understanding the integration process for both shapes.
Yam
Messages
32
Reaction score
1

Homework Statement



I am trying to work the moment of inertia for

a) rotating rod, axis through the centre of the rod
http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#irod3

b) Solid cylinder
http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html#icyl2
[/B]

Homework Equations


I = R^2 dM

The Attempt at a Solution



I understand all of the steps, except the limits of integration in terms of R or L.

For the rod, the limit of integration is from L/2 to -L/2

Why is the limit of integration for the cylinder from R to 0?
Should it be from the end to end? R to -R?[/B]
 
Physics news on Phys.org
You need to fix your integration limits such that you integrate over the entire body once. In the case of the cylinder, you integrate from zero, but you allow the angular coordinate to vary between zero and 2pi, thus covering the full lap.
 
Orodruin said:
but you allow the angular coordinate to vary between zero and 2pi
I don't quite understand this part, when did i allow the angular coordinate to vary between zero and 2pi? What is meant by the angular coordinate?
 
It is implicit when you multiply by 2pi in ##dV = 2\pi L r dr##. The integral is really a volume integral, but the direction along the cylinder axis and the angular direction are trivial since the integrand does not depend on them, thus they are replaced by the size of their integration ranges, ##L## and ##2\pi##, respectively.
 
I understand now, thank you for your help! cheers.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top