Something I can't figure out about Angular Momentum!

1. Nov 25, 2003

davesbit

I've been reading some pages on the Internet talking about how to implement rotational dynamics in computer games using Angular Momentum and Linear Momentum, however there's one major thing which really doesn't seem to make sense...

Let's say I have a pencil lying horizontally - if I hit it with a quick force (well, Impulse) in the middle of F Newtons, it gets an acceleration of a=F/m

Now if I hit it off-centre with the same force (/Impulse), it also gets an acceleration of a=F/m, however it's now got angular acceleration too. So it's got MORE energy! It's moving and it's spinning

Where the hell did the extra energy come from? I applied the same force in both cases?

This is messing with my head... where the hell did the extra energy come from?

2. Nov 25, 2003

jcsd

Here's a quick and easy answer, perform the experiment yorself on a flat surface, once heating your pencil dead centre and once hitting it right on the end, which one has more linear accelration along the axis you hit it and which one has more angular momentum?

If we idealize this experiment, you find that they both have the same energy.

3. Nov 25, 2003

davesbit

Just had a go... problem is I can't tell, it's hard to hit the pencil with the same force each time!

I think, and I might be wrong, the pencil has less linear momentum when hit on the edge, but friction might have a factor... I still don't know for sure.

Please put me out of my misery - it's driving me mad trying to figure it out... here are the supposed 'facts' I've been taught about physics:

1) If you hit something with a Force F ANYWHERE on the object, it will accelerate linearly with a=F/m
2) If you hit something with a Force F a distance r from the centre, it with accelerate angularly with r*F
3) Energy cannot be created nor destroyed

So surely the pencil thing suggests that one of these is wrong... and I'm guessing from your answer your saying (1) is wrong? Is that right?

4. Nov 25, 2003

jcsd

If you can reduce nfriction to a minimum you find that the pencil hit in the middle will move further along the axis it is hit.

5. Nov 25, 2003

NateTG

Don't worry so much about precision, just try flicking a pencil in the middle, and at the end. You'll see the difference.

6. Nov 25, 2003

davesbit

Hmmm - so does that mean F != ma for a pencil hit on the end?

7. Nov 25, 2003

LURCH

That's correct. The energy you impart to the pencil will be devided up between angular and linear momentum (minus the standard fee for entropy, of course). If you hit it on the end, it will tend to rotate a lot and not go very far (spinning in place). If you hit it smack on the center of gravity, it will go farther and not rotate at all. Add up the angular and the linear momentum together, and it will equal the amount of momentum put in.

8. Nov 25, 2003

davesbit

But I found this code on the internet from a rigid body Dynamics program which is meant to illustrate this process...

void simulation_world::ResolveCollisions( int ConfigurationIndex )
{
rigid_body &Body = aBodies[CollidingBodyIndex];
rigid_body::configuration &Configuration =
Body.aConfigurations[ConfigurationIndex];

vector_2 Position =
Configuration.BoundingBox.aVertices[CollidingCornerIndex];

vector_2 CMToCornerPerp = GetPerpendicular(Position -
Configuration.CMPosition);

vector_2 Velocity = Configuration.CMVelocity +
Configuration.AngularVelocity * CMToCornerPerp;

real ImpulseNumerator = -(r(1) + Body.CoefficientOfRestitution) * DotProduct(Velocity,CollisionNormal);

float PerpDot = DotProduct(CMToCornerPerp,CollisionNormal);

real ImpulseDenominator = Body.OneOverMass + Body.OneOverCMMomentOfInertia * PerpDot * PerpDot;

real Impulse = ImpulseNumerator / ImpulseDenominator;

Configuration.CMVelocity += Impulse * Body.OneOverMass * CollisionNormal;

Configuration.AngularVelocity += Impulse * Body.OneOverCMMomentOfInertia * PerpDot;
}

As you can see by the last two lines, the Impulse isn't split between Linear and Angular... it's added wholly to both. Is this example wrong then?

9. Nov 25, 2003

NateTG

No. Linear and angular momentum are both conserved during colisions. Energy is often not conserved.

10. Nov 25, 2003

davesbit

Actually (forgive the double post!) I've realised I can rephrase my question in a little more exact way...

I have a completely frictionless table and/or I am in space!
If I have a stationary pencil of length l metres and mass m kg and I apply an impulse of N to the end of it at 90 degrees to the direction the pencil is pointing. It will spin and move (presumably).

Does anyone know what is the linear momentum and the angular momentum?

11. Nov 25, 2003

NateTG

If you're talking about colisions (as implied by impulse) then you've got to give more information about the object that's hitting the pencil. Even then, I'm not sure that there is enough information:

Conservation of linear momentum gives 2, conservation of angular momentum gives another, but there are six unknowns (x and y velocities and angluar velocity for each object).

If you assume, for example that the colision is perfectly inelastic, then it's pretty easy to determine the motion since there are now 3 equations and 3 unknowns.

12. Nov 25, 2003

davesbit

Okay understood, in that case let's redefine slightly to a bit more defined situation... say that the pencil is floating with constant velocity of 0.1ms towards a fixed bar, like this:

o <- the bar

/\ 0.1ms
|
|
<======pencil=======

So it starts off with an angular momentum of zero, and a linear momentum of 0.1m/s * mass (say 0.1kg), so that'll be a linear momentum of 0.01 kgm/s

Then it hits the bar exactly on the left edge of the pencil at exactly 90 degrees, and because the bar is fixed it doesn't move (consider it to have infinite mass in this context). And lets assume the collision is completely Elastic this time. Presumably the pencil will start spinning and may keep moving in a linear direction too?

After the collision, what is the new angular momentum and linear momentum?
You said conservation of linear momentum gives 2 equations not one... can you tell me what they are?

Last edited: Nov 25, 2003
13. Nov 25, 2003

Integral

Staff Emeritus

14. Nov 26, 2003

davesbit

"The force acting through the CM will contribut to translational motion. The Component acting perpendicular to the line from point of application to the CM will contribute to rotation."

Is this correct? Is this the answer to my question then?
In the case of hitting a pencil on the end at 90 degrees, all of the direction is *perpendicular* to the CM and none is acting through.

That suggests it ALL goes into angular momentum?

But that would mean that if you flick a pencil on the end it should spin on the spot? ... Actually if I flick a pencil very lightly and carefully on the edge I can *almost* get it to spin on the spot...

Is that right?

[Actually I guess the pencil experiment is a bit flawed anyway because you can never apply an instant force with your finger, as soon as it starts rotating you are still applying the force only a component *is* going through the CM.]

I'm very confused now :( does anyone have any more pointers to similar problems like this? I need to work this out further!

UPDATE: no hang on, that can't be right - look at the example at the bottom of this page:
http://electron9.phys.utk.edu/phys135d/modules/m8/angular.htm [Broken]
It says that a tangential Force contributes to both linear AND angular momentum... hmmm

Okay, I'm just as confused as when I started off :)

Last edited by a moderator: May 1, 2017
15. Nov 26, 2003

NateTG

Conservation of linear momentum effectively gives one equation per (linear) dimension of the system. When you have objects on a table, that means equations for the x and y components.

The example you asked me about involves a fixed bar, which means that momentum is not conserved. (Consider the linear case where the same is true.)

Let's say instead that we are shooting clay pellets at a pencil that starts at rest

===============

^
|
|
o

And that the pellet sticks to the pencil afterwards, then we have:
mv=m0v0
and
I&omega;=L0
which will give you the resulting linear and angular velocity of the resulting glob respectively.

If I've done my math right, and the two have equal masses, then the final velocity is 1/2 v0 and the final angular speed is 3v0/2l where v0 is the initial velocity of the glob, and l is the lenght of the pencil.

16. Nov 27, 2003

davesbit

Thanks - the concepts in this thread are starting to make sense to me a little now.
I also found this Java applet:
http://www.myphysicslab.com/collision.html
and had a little play around with it

If you look at the source code:
http://www.myphysicslab.com/source/Thruster5.java [Broken]
It does seem to suggest again (like the other source code) that an impulse from a collision is applied to both the linear and angular velocity (depending on the moment of the impulse of course).

It's this bit in the code which applies the impulse (j) to both the linear and angular:
// v2 = v1 + j n / m = new linear velocity
velo[1+offsetA] += j*nx/ma;
velo[3+offsetA] += j*ny/ma;
velo[1+offsetB] += -j*nx/mb;
velo[3+offsetB] += -j*ny/mb;
// w2 = w1 + j(r x n)/I = new angular velocity
velo[5+offsetA] += j*(-ray*nx + rax*ny)/Ia;
velo[5+offsetB] += -j*(-rby*nx + rbx*ny)/Ib;

And it has the exact same equation for working out the size of the impulse of the collision. So I'm thinking it *must* be correct, I was just thinking about it in the wrong way when I got confused about the impulse giving the object 'more' energy if it caused a rotation. A force doesn't really 'carry' energy does it?

Last edited by a moderator: May 1, 2017