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Angular momentum of rod 
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#1
Dec2612, 11:02 AM

P: 8

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



#3
Dec2612, 12:08 PM

P: 8

about one end



#4
Dec2612, 01:06 PM

P: 70

Angular momentum of rod
A little research is in order.
http://scienceworld.wolfram.com/physics/ http://hyperphysics.phyastr.gsu.edu/hbase/hframe.html "There is no need to know all the answers when you know how to find all the answers that have been found before." 


#5
Dec2612, 01:13 PM

P: 8

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



#6
Dec2612, 01:21 PM

P: 70

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
Dec2612, 01:27 PM

P: 208

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



#8
Dec2612, 01:30 PM

Math
Emeritus
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Thanks
PF Gold
P: 39,338

that you could get to following the links you were given in tadchem's response. 


#9
Dec2612, 01:43 PM

P: 8




#10
Dec2612, 01:52 PM

P: 58

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
Dec2612, 02:04 PM

P: 8

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
Dec2612, 02:06 PM

P: 208

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



#13
Dec2612, 02:09 PM

P: 8

^but that about one of the ends ,not com



#14
Dec2612, 02:57 PM

HW Helper
P: 3,440

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
Dec2612, 03:02 PM

P: 8

yes



#16
Dec2612, 03:54 PM

HW Helper
P: 3,440

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 [tex] \omega \frac{mL^2}{12} [/tex] (in other words, same as what the angular momentum would be for a system where the rod is rotating around it's COM.) 


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