Derive Linear and Angular Acceleration from A Force and a Point of application

1. Jun 7, 2004

Perseus

Setting the Stage:

A uniform sphere has a mass of 100 kilograms and a radius of 10 meters.
It is located at [0,0,0]
It is in static equilibrium.
We are in constant time.
We are in a vacuum, there is no gravity, friction, or any other forces to act upon our sphere.
We are using a coordinate system compatible with OpenGL; right-hand.
You are looking down the negative Z axis and can see the sphere in front of you.

Z', A force of [1,0,0] Newtons, is applied at point, P', [0,0,10].

Stating the Obvious for Clarity:
This will create a counter-clockwise rotation.
Z' is tangential to the surface of the sphere.

The Question:

Finding Torque is no problem. A cross product and some work with the moment of inertia. That is not the question.

Q1) How does one find the resulting linear acceleration?
Q2) Is the resulting linear acceleration inversely proportional to the distance from the center of mass the Force is applied?

2. Jun 7, 2004

Perseus

Progress, Perhaps.

I have derived a formula for solving this. Based on my guesswork of how a free-body in the described environment would work.

[1-(r-d/r)]F' = L' (1)

Such that L' is the new scaled Force that takes into consideration how far away it is from the center of mass.

This should approximate what would happen if you did this. Its close enough for a 3D game engine. Perhaps not for rocket science. But I would just like confirmation from anyone out there that this seems plausable.

Thanks!

3. Jun 7, 2004

Perseus

To expand to the 3D environment I created a vector from the center of mass to the point of impact. I then took the inner product (dot product) of these and used that as a scale factor in the linear force.

4. Jun 7, 2004

baffledMatt

I've never been happy with this type of problem because it is not at all clear what you mean by applying a force at point P' once P' starts to move. I think the problem is better posed if you ask what happens if you apply an impulse of, say $$\underline {p_0}$$.

Nevertheless, my feeble brain is still unable to get round this problem. But here is how I might go about it:

You supply an impulse $$\underline {p_0} = [1,0,0]$$ to point [0,0,10]. This supplies momentum of $$\underline {p_0}$$ to the sphere as well as an angular momentum $$\underline L = [0,10,0]$$. Thus the sphere will move linearly such that it has this much linear momentum (v = p / m) and rotate such that it has this much angular momentum (you have to look up the moment of inertia of a sphere). To turn this into a force you then consider supplying a certain amount of momentum (an impulse) each timestep. So, if you want a force of 10N with a timestep of 0.01 seconds, you supply an impulse of 0.1 each step.

Not sure if any of this is correct though (at the very least check my math!!!), but should make for some interesting simulations.

Matt

5. Jun 7, 2004

Perseus

Hmmm

I could have sworn I explicitly stated exact details in my original problem.

In order for you to apply force in a 3D simulation you must "pick a point". It is not enough to know the direction and magnitude of the force being applied. The point P' doesn't "move" its merely a position in 3D space telling you where that force is being applied on the sphere.

The moment of inertia for a sphere is known in my existing torsion functions. Finding torsion and/or using F = ma has never been a problem.

But its incorrect that a free object in this environment stated above would use F = ma if a force were applied a point totally to the far extremes away from its center of mass. It would have less of a net translation than if you had applied that force orthogonal to its center of mass.

The formula I came up with is as follows:

F', force
O', origin vector of Sphere
P', point of impact that F' is applied
r, scalar, bounding radius of sphere
f, scale factor in the range [0,1]
%, dot product
D', origin to point of impact normal
S', the scaled down force

D' = normalize(O' - F')
f = |(|F'| % D')|
S' = F' * f

This works in my simulation. There is still linear translation if you hit the sphere at or near its edges but not as much as if you hit it at the dead center. This gives a very realistic simulation of asteroids in space to be shot at.

6. Jun 7, 2004

Perseus

What is really interesting is that I have not found a single shred of information on the internet that covers explicitly what I have done here.

Calculating torque finds the proper ammount of torque based on the distance from the axis of rotation. It should be no different for a free body with no forces of gravity or wind resistance acting against it.

Or is my observation incorrect?
Do you simply use F = ma, regardless of where the rigid body was impacted?

In that case, I overcomplicated the simulation by orders of magnitude.

I'll just ask my original (and most important question) again:

Is the resulting linear acceleration inversely proportional to the distance from the center of mass the Force is applied?

7. Jun 7, 2004

baffledMatt

No.

Ok, I've been scratching my head for the last half hour over this and I think (aka hope!!) I have the explanation. I will attempt to answer this in two stages. First of all I will consider the case when you are considering an impulse such as you might get with bouncing an object off your sphere (which is quite apt to your application I think). Then I will try and convince you that in order to have a well defined problem you have to take the method of the impulse into account - you can't just say "lets apply a force here" and hope to get the correct answer.

Right.

Let's say that we have supplied a momentum $$\mathbf{\Delta p}$$ to our sphere with the coordinates used earlier, such that the angular momentum supplied is $$r \Delta p$$, where r is the radius of the sphere. Now, let's look at the resulting spinning/moving sphere. We can find the total linear momentum of the sphere by integrating $$\mathbf{v}dM$$ throughout the sphere (dM is an element of mass) and you should be able to convince yourself that due to symmetry all the contribution from $$\mathbf{v}$$ which is purely rotational will cancel. hence the only linear momentum of the sphere is exactly its linear velocity times its mass. Since we supplied it with $$\Delta p$$ momentum its mass, M, and linear velocity, $$\mathbf{V}$$ must satisfy $$M\mathbf{V} = \mathbf{\Delta p}$$.

A similar argument goes for the angular momentum. So, you see that nomatter where this momentum is supplied the linear momentum of the sphere will be the same.

however... we have considered this problem by saying "we supply an impulse". What we want to do is consider what happens if this momentum is supplied by a collision (bouncing a smaller object off the sphere). This is easily done. If the collision is elastic you have three equations which equate energy and momentum before and after the collision

notation:
m = mass of smaller object
v = initial velocity of smaller object
v' = final velocity of smaller object
M = mass of sphere
V = linear velocity of sphere after collision
$$\omega$$ = angular momentum of sphere after collision

$$m \mathbf{v} = m \mathbf{v'} + M\mathbf{V}$$

$$\frac{1}{2}m v^2 = \frac{1}{2}m v'^2 + \frac{1}{2}MV^2 + \frac{1}{5}M r^2 \omega^2$$

$$\frac{2}{5}M r^2 \omega = mvr$$

now, we have 3 equations for 3 unknowns (v, v' and V - remember that $$\omega$$ is a function of V) so we can solve this problem.

Now, when I first tried this I thought to myself 'imagine the smaller object was absorbed into the larger one?' If we do this we have v'=0 and so too many equations for our unknowns... or do we? Of course not because the collision is no longer elastic so conservation of KE no longer applies. however, we will get a different answer so we see that in order to have a well defined problem we need to specify the nature of the collision.

I think this is much easier to consider than saying "what happens if we supply a force", and certainly more applicable to your simulation.

Does that clarify things?

Matt

8. Jun 7, 2004

baffledMatt

By the way:

This is a common problem and I think it's due to the fact that this is textbook stuff, but the sort of thing you find near the end of a textbook. So it is too difficult/obscure to go on any website giving a basic introduction to mechanics but way too easy to be found in any research work.

Matt

9. Jun 7, 2004

TALewis

It's as simple as this:

$$\vec{F} = m\vec{a}_G$$

At any given instant in time, the sum of the forces acting on a rigid body causes a linear acceleration of the body's mass center in the same direction as the resultant force. The beauty of this is that you can consider the linear effect of the forces independently of any rotational effect the forces may also cause. It doesn't even matter where the force is applied, if you are considering only the linear effect.

\newcommand{\unit}[1]{\mathrel{\mathrm{#1}}} \begin{align*} \vec{F} &= \langle 1,0,0 \rangle \unit{N}\\ \vec{a}_G &= \frac{\vec{F}}{m}\\ &= \frac{\langle 1,0,0 \rangle \unit{N}}{100\unit{kg}}\\ \end{align*}

Keep in mind that's only valid for the instant in time when F is initially applied. It's up to you to consider how long and in what orientation this force is applied to the body and how that affects its linear and angular acceleration.

Last edited: Jun 7, 2004
10. Jun 7, 2004

baffledMatt

This is exactly why I felt it easier to deal with impulses as opposed to forces. If you want the problem to be well posed you need to specify exactly what the forces do after the sphere starts accelerating.

Matt

Last edited: Jun 7, 2004
11. Jun 7, 2004

reilly

Rigid Body Dynamics

It's not clear to me that your problem is well set. That is, how the force behaves over time as the sphere moves is not specified. Is it always applied to the same physical point? Whatever. You have given amongst the most difficult problems in mechanics, provided that you can fully specify the force. You must consider both linear and rotational motion together. While your problem is not as difficult as the free symmetric top, it still requires some fairly high horsepower analysis -- some form of the Euler equations for rigid bodies.I would suggest using a Lagrangian approach with body-centered coordinates. This type of analysis is usually discussed in any text on advanced mechanics -- Goldstein's Classical Mechanics, a great classic, is very good on this topic.

Regards,
R. Atkinson

12. Jun 8, 2004

Perseus

Thank everyone for the replies. The clearest answer I was looking for was that force affects a ridgid body regardless of the point of application. Thank you to baffledMatt and TALewis!

Yes it is quite a simple problem indeed. Really all one has to do for this rigid body is calculate rotation and translation independantly. The kinematics engine takes over from there and handles the time issues and what results occur after this. The result is an incredibly realistic collision for the asteroids. If you shoot the top of the asteroid, it will torque, more than if you had shot it closer to the center. Thanks to all. If you would like to see the eventuality of this work, we will be releasing a demo in the near future.

http://www.universeunlimited.com

As for reilly,

My problem is well set. The only thing I failed to explicitly state was that this was an inelastic collision. Other than that I provide all information that is needed (and then some) to solve the problem. The problem is not complex given the scenario. One simply uses the P', and F' to calculate [tau]/torque, and then uses F' and the mass of the sphere to calculate the resulting linear acceleration. Its elementary physics. I made a simple oversight in my observations of free-bodies, or, spheres that slip and how they move. Since it is a rigid body, accelerating any point of that body accelerates its whole. I had to setup a complex arena to describe a seemily difficult question. My clarity was to provide undisputable context to the question at hand.