# Angular momentum of rod

1. Dec 26, 2012

### kuyt

What is the angular momentum of a rod rotating about one end (mass M and angular velocity ш),about its center of mass?

2. Dec 26, 2012

### Staff: Mentor

Is the rod rotating about one end or its center of mass?

3. Dec 26, 2012

### kuyt

4. Dec 26, 2012

5. Dec 26, 2012

### kuyt

But this is something fundamental,calculating angular momentum of a system about arbitary points in space.

6. Dec 26, 2012

In the second link the 'bubbles' are hot links.
Navigate through "Mechanics", then "Rotation", then "Moment of Inertia".
Click on "Common Forms" for Enlightenment!

7. Dec 26, 2012

### schaefera

Find the angular momentum of a differential length (dm*v*r) and integrate from r=0 to L.

8. Dec 26, 2012

### HallsofIvy

Yes, it is, and you can answer it by calculating the angular momentum of individual point about the end of the rod, then integrating them along the length of the rod. The result of doing that would be the formula at http://scienceworld.wolfram.com/physics/MomentofInertiaRod.html
that you could get to following the links you were given in tadchem's response.

9. Dec 26, 2012

### kuyt

it gives moment of inertia not angular momentum! Can we use the general formula(if its correct):angular moment abt any point=angular moment of a fictitious particle (of mass m at the position of COM)abt that point + angular moment of the body abt com ?

10. Dec 26, 2012

### dev70

hello kuyt, i think you should have a look at the angular momentum equation ie., L= r X P and v=rw. So, use r=l/2 where l=length of rod. hope that makes sense..

11. Dec 26, 2012

### kuyt

No,it not that easy I guess.
P.S can anyone giveme the final answer(in terms of angular velocity,mass and length)and ofcourse the explanation,instead of links

12. Dec 26, 2012

### schaefera

It's (1/3)wmL^2 for angular speed w.

13. Dec 26, 2012

### kuyt

^but that about one of the ends ,not com

14. Dec 26, 2012

### BruceW

Ah, so you're trying to find the angular momentum about the com, in a system where the rod is rotating about one end.

15. Dec 26, 2012

### kuyt

yes

16. Dec 26, 2012

### BruceW

hmm. Tricky one. Well, I'm pretty sure that the whole point of saying the angular momentum "about a point" is equivalent to calculating the angular momentum, given that the origin is the point about which we want to find the angular momentum.

So I think the angular momentum about the COM is simply angular momentum, given that our origin is the COM. And in our original reference frame, the rod was rotating around the end. So in a reference frame where the COM stays at the origin, the angular momentum will simply be
$$\omega \frac{mL^2}{12}$$
(in other words, same as what the angular momentum would be for a system where the rod is rotating around it's COM.)

17. Dec 26, 2012

### Astronuc

Staff Emeritus