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

1. Jun 30, 2011

### boumas

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. Jun 30, 2011

### tiny-tim

welcome to pf!

hi boumas! welcome to pf!
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

3. Jul 1, 2011

### boumas

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=(r-x)w

L = Summation{(d)(x)(r-x)(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. Jul 1, 2011

### tiny-tim

hi boumas!
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

5. Jul 1, 2011

### tiny-tim

hi boumas!

(have an omega: ω )
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

6. Jul 1, 2011

### 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...

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. Jul 1, 2011

### tiny-tim

hi boumas!

(try using the B and X2 icons just above the Reply box, or for latex use itex tags )
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)

8. Jul 1, 2011

### boumas

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. Jul 1, 2011

### 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!

10. Jul 4, 2011

### 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...

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. Jul 4, 2011

### tiny-tim

hi boumas!
yup!
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)
not following what you're trying to do here

12. Jul 4, 2011

### boumas

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

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...
:tongue:

Perhaps though that would bring torque into it???

13. Jul 4, 2011

### tiny-tim

if it's rotating about an axis, then it isn't moving

(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. Jul 4, 2011

### boumas

Cool...

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

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