# Mechanics - Angular momentum not about C.O.M. with translational motion...

by boumas
Tags: angular momentum, mechanics, translational
 P: 7 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 :)
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hi boumas! welcome to pf!
 Quote by boumas 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...
about any point, angular momentum = angular momentum about centre of mass plus angular momentum of centre of mass: $\mathbf{L}_P\ =\ I_{c.o.m.}\,\mathbf{\omega}\,+\, \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}$ …
parallel axis theorem not needed
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## Mechanics - Angular momentum not about C.O.M. with translational motion...

hi boumas!
 Quote by boumas 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 parallel axis theorem always correctly gives you the moment of inertia, I

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 rc.o.m x mv happens to equal mr2 times the angular velocity only if rc.o.m is measured from the centre of rotation)
so you only use the parallel axis theorem if either
i] your point is the centre of rotation of the whole body, or
ii] your point is the centre of mass of the whole body, but the body is irregular, and you're finding the moment of inertia by splitting it into parts, each with a different centre of mass
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hi boumas!

(have an omega: ω )
 Quote by boumas 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 parallel axis theorem always correctly gives you the moment of inertia, I

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 rc.o.m x mv happens to equal mr2ω only if rc.o.m is measured from the centre of rotation)
so you only use the parallel axis theorem if either
i] your point is the centre of rotation of the whole body, or
ii] your point is the centre of mass of the whole body, but the body is irregular, and you're finding the moment of inertia by splitting it into parts, each with a different centre of mass
 P: 7 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!
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hi boumas!

(try using the B and X2 icons just above the Reply box, or for latex use itex tags )
 Quote by boumas 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...
the angular momentum for the translational motion is the parallel axis adjustment …

rc.o.m x mvc.o.m = mr2ω

(because vc.o.m = ω x rc.o.m)
 P: 7 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...
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 Quote by boumas … 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...
sorry, you can't have it both ways

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!
 P: 7 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... :)
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hi boumas!
 Quote by boumas 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...
yup!
 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...
if z is fixed (even if only instantaneously), then the angular momentum about z can be found either as Izω or as Icω + zc x mvc (they're the same in this case)
 Also if you were calculating the angular momentum as such but from a moving inertial frame?
not following what you're trying to do here
P: 7
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

 Quote by tiny-tim sorry, you can't have it both ways … 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!
by taking it from the point of view of a moving inertial frame so it would appear to that observer to be both moving and rotating around that axis...

Perhaps though that would bring torque into it???