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

In summary: I said, I'm just a summarizer of content and do not provide answers or responses. Sorry for any confusion.
  • #1
boumas
7
0
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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  • #2
welcome to pf!

hi boumas! welcome to pf! :smile:
boumas said:
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, https://www.physicsforums.com/library.php?do=view_item&itemid=313" = angular momentum about centre of mass plus angular momentum of centre of mass: [itex]\mathbf{L}_P\ =\ I_{c.o.m.}\,\mathbf{\omega}\,+\, \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}[/itex] …

parallel axis theorem not needed :wink:
 
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  • #3
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
hi boumas! :smile:
boumas said:
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) :wink:

(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
hi boumas! :smile:

(have an omega: ω :wink:)
boumas said:
I didn't feel like using parallel axis theorem to find the https://www.physicsforums.com/library.php?do=view_item&itemid=31" 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) :wink:

(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
 
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  • #6
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! :redface:
 
  • #7
hi boumas! :smile:

(try using the B and X2 icons just above the Reply box, or for latex use itex tags :wink:)
boumas said:
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 [itex]\mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}[/itex] 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) :wink:
 
  • #8
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...
:smile:
 
  • #9
boumas said:
… 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 :redface:

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! :smile:
 
  • #10
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
hi boumas! :smile:
boumas said:
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! :biggrin:
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) :wink:
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 :confused:
 
  • #12
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! :biggrin:

I was trying to get around the

tiny-tim said:
sorry, you can't have it both ways :redface:

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! :smile:

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
boumas said:
… 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:

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 https://www.physicsforums.com/library.php?do=view_item&itemid=632" wheel, the axis of rotation has moved to the bottom of the wheel)
 
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  • #14
Cool...

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

:smile:
 

1. What is angular momentum?

Angular momentum is a physical quantity that measures the rotational motion of an object around an axis. It is defined as the product of an object's moment of inertia and its angular velocity.

2. How is angular momentum different from linear momentum?

Angular momentum is different from linear momentum in that it is a measure of rotational motion, while linear momentum is a measure of translational motion. Angular momentum also takes into account an object's moment of inertia, which is not a factor in linear momentum.

3. What are the units of angular momentum?

The SI unit of angular momentum is kilogram-meter squared per second (kg·m^2/s), while the imperial unit is pound-foot squared per second (lb·ft^2/s). It can also be expressed in terms of joule-seconds (J·s).

4. How is angular momentum conserved?

Angular momentum is conserved in a closed system, meaning that it remains constant as long as there are no external torques acting on the system. This is known as the law of conservation of angular momentum.

5. What are some real-life applications of angular momentum?

Angular momentum has many applications in various fields, such as in the design of vehicles and machinery, the study of celestial bodies, and the analysis of fluid dynamics. It is also used in sports, such as figure skating and gymnastics, to perform rotational movements with precision and control.

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