Mechanics  Angular momentum not about C.O.M. with translational motion...by boumas Tags: angular momentum, mechanics, translational 

#1
Jun3011, 05:53 AM

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Perhaps this should be under physics, but my mechanics course is done by the maths department...
I don't actually have a particular problem, just a question. If you have a body (say a rod) with translational motion and rotation about an axis that is not its centre of mass, is there a way of neatly finding the angular momentum. I feel that using Lz=I0w +(RcrossMV)z (sorry for crummy equation writing...) in combination with parallel axis theorem would be wrong, but am not quite sure how else to go about the problem... Sorry for being vague, but I don't really have an example problem to give, and this has had me puzzled and confused for quite a while... Thanks :) 



#2
Jun3011, 10:33 AM

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hi boumas! welcome to pf!
parallel axis theorem not needed 



#3
Jul111, 06:26 AM

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Thanks! :)
Sorry for late response, I left the library early and don't have internet in my flat... (Like living in the dark ages here!) What I think I'm having difficulty understanding is if the body is rotating about an axis that isn't its centre of mass, for instance a uniform rod with translational motion but also rotating about its endpoint. To find the angular momentum about some other point p you'd add the angular momentum of the centre of mass wrt p (which is ok) to the "spin" angular momentum, ie the rods angular momentum about its centre of mass. However, if the rod is rotating about its endpoint, I don't really understand how I'd find its angular momentum wrt its center of mass... I didn't feel like using parallel axis theorem to find the moment of inertia about the endpoint would work as that's just finding the angular momentum about its endpoint, not its centre of mass. The other thought was to find the angular momentum of all the "little pieces" of mass of the rod about its centrepoint as it swings about its endpoint (integrating them in a similar way to calculating moment of inertia). Doing this and set the length of the rod at 2r, angular momentum about c.o.m. came out like so... m=mass of rod d=density x=distance from centre to piece of mass v=velocity of piece of mass w=angular velocity of piece of mass about endpoint L = Summation{(d)(x)(v)(delta x)} but v=(rx)w L = Summation{(d)(x)(rx)(w)(delta x)} Integrating from r to r L = (2r^3wd)/3 = (wmr^2)/3 Is this OK? I guess now this isn't coincidence that this is that this is the moment of inertia of the centre times the angular velocity... It just seems rather weird that the axis of the angular velocity wouldn't matter? Thanks again, I'm just not sure about it and these problems are starting to make my head... uh... spin. ;) 



#4
Jul111, 07:26 AM

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Mechanics  Angular momentum not about C.O.M. with translational motion...
hi boumas!
but that's only helpful if the angular momentum about that point = I times the angular velocity and that is only true for the centre of rotation (the end of the rod, in this case) (and only because r_{c.o.m} x mv happens to equal mr^{2} times the angular velocity only if r_{c.o.m} is measured from the centre of rotation) so you only use the parallel axis theorem if either 



#5
Jul111, 07:29 AM

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hi boumas!
(have an omega: ω ) but that's only helpful if the angular momentum about that point = Iω and that is only true for the centre of rotation (the end of the rod, in this case) (and only because r_{c.o.m} x mv happens to equal mr^{2}ω only if r_{c.o.m} is measured from the centre of rotation) so you only use the parallel axis theorem if either 



#6
Jul111, 08:07 AM

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So when finding the angular momentum about the endpoint (centre of rotation) I can use the parallel axis theorem, but how can I now add the angular momentum for the translational motion?
Adding the \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.} component to it seems wrong as now the axis of rotation for the other component is not the center of mass... Also I'm not quite sure how this would lead to calculating the angular momentum about some other point in space... Thanks Hope I haven't missed the answer to this in your previous post! 



#7
Jul111, 09:23 AM

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hi boumas!
(try using the B and X_{2} icons just above the Reply box, or for latex use itex tags ) r_{c.o.m} x mv_{c.o.m} = mr^{2}ω (because v_{c.o.m} = ω x r_{c.o.m}) 



#8
Jul111, 09:55 AM

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Just to clarify, when I said the translational motion I meant a translational motion independent of the rotation. So for instance, the endpoint of the rod is also moving with velocity u, as well as being the axis for the rotation of the body...
Just to make sure... 



#9
Jul111, 10:03 AM

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if it's (on) the axis for the rotation of the body, then it's stationary if it's moving, then the axis is somewhere else! 



#10
Jul411, 03:58 AM

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Ah, OK, so it's basicly unnatural it say a body rotates about anywhere other than its centre of mass without having an axis of rotation...
However, I'm still unsure how I would find the angular momentum about a point other than the axis of rotation (Taking the case that it is fixed and not moving), say a point z which is at some moment in time at the opposite end of the rod... Also if you were calculating the angular momentum as such but from a moving inertial frame? Maybe that's just silly now... Sorry to pile these on, I appreciate it... :) 



#11
Jul411, 04:20 AM

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hi boumas!




#12
Jul411, 04:30 AM

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Ahhh...
I think my brain just was refusing to believe that you could use Izω when the angular velocity isn't around that point... Neat! I was trying to get around the Perhaps though that would bring torque into it??? 



#13
Jul411, 04:56 AM

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(but you can often find a useful frame in which the axis of rotation has moved to a different point … eg in a frame moving with a rolling wheel, the axis of rotation has moved to the bottom of the wheel) 



#14
Jul411, 05:02 AM

P: 7

Cool...
Thanks very much for clearing that up for me... 


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