Playing with physics stuff; with came up. Any ideas of where to go to explore this? Given an object represented as a point mass M floating in free space, imagine a force F is applied to the point mass from a position relative to it''s coordinates R. If the point mass was fixed in space, the torque on the mass is given by R x F. However, the mass is floating in free space. There is going to be some component of the force that goes toward moving the entire body rather than just contributing to the rotational force on the point mass. The question is, how the heck do you calculate it. Obviously there is some dynamic balance given by the magnitude of R; as R->0 the component given to bulk motion approaches 1.0, while as R->infinity the component to bulk motion approaches 0.0, while the component to rotation motion approaches 1.0. I''m just lost as to where to go from there. Everything I''ve looked at so far appear to deal with torque and bulk motion are two distinct things which are not related (except for a snippet on calculating shuttle motion, which was more concered with the thrust ratios, etc than the "trivial" physics of a body in free space). Is there some subsection of rigid body physics I just don''t have the name of to look at or something? Thanks. =)

I don''t know much about rigid body dynamics, so I''m sure many will flame me for just opening my big mouth, but I''d like to take a stab at it anyway.

The direction the the point mass (center of mass as well, maybe) moves will have to do with the angle of incidence of the force and the surface it is being applied. In actuality, it wouldn''t be force, but momentum, so for the rest of the example let''s say a ball hit a surface of this body for a certain amount of time, transfering momentum to the object. If the momentum was applied at a 0° angle of incidence to the surface (parallel to the normal), then I think the point mass of the body will move in the direction of the object that struck it. * those more knowledgable can flame me now * The rotational force wouldn''t affect the direction the point mass travels. You would now know the direction the body would travel in, but now you''d have to figure out the velocity. The amount of momentum transfered to the body would have to do with the intertia of the body, which has something to do with the mass (look at me, I''m explaining rigid body dynamics and I don''t know how to calculate intertia You can shoot me now).

The less trivial example would be if the angle of incidence isn''t 0°. In that case, there would be two momentum directions. One of them would be parallel to the normal, and the other would actually maintain the direction of the force applied. If we can calculate how much kinetic energy the ball had before the collision, how much kinetic energy the ball has after the collision (using momentum stuff), and how much energy is absorbed along the normal (woudl use velocity parallel to normal in that case), then you might be able to figure out how much kinetic energy is left for the collision in the ball''s direction, and finally figure out the momentum. After all that, I believe you might just be able to add these two momentum "vectors" together to get a complete and final momentum, complete with direction and magnitude.

I hope I''m at least someone in the ballpark. All flames are appreciated, because I can always learn from them

The relationships between force and acceleration, and force and torque, are as simple as your research suggests. The trick is to go beyond the force and look at work and energy to see how the two interact.

E.g. if you apply a force to a free object at it''s centre of mass ir accelerates. If you apply the same force off-centre it accelerates and rotates. You seem to be getting something for nothing as the same force gives you two different outcomes, one with more energy.

But it requires more energy in the second case. The rate of doing work = force x velocity. When you push something at the centre it moves with a certain velocity. When you apply the same force off-centre it moves with the same velocity but also rotatates away from the force, and so moves away more quickly.

So in the second case the force is the same but the speed of the point of application of the force is greater. So the rate of work, and the total energy required, to rotate and translate it, is greater. Unsuprisingly when you work out the maths energy is conserved in both cases.

In practical situations the force is often non-constant. E.g. many engines (including humans doing work) can only do so much work, and if asked to work faster will decrease the effort/force they generate as they do so.

I think there are several things missing here. First, Shadow Mint is talkinga about a point object. You simply don''t have point objects rotating. In fact, you can''t even start to consider rotation of objects until you consider it to be an exteneded object. To think about rotation of point objects, which lack important things such as radius vectors and moments of inertia, is completely pointless. So, now that that is out of the way. . . As already mentioned, the torque on an object is defined as radius vector cross force vector. Torque is analogous to force, so torque = moment of inertia times roational acceleration. From there you will be able to solve for your rotational accel. and perform all the operations you need. To get the amount of force that is parallel to the radius vector, take the dot product of the two vectors and that quanity is the resulting force that causes an acceleration linerarly. Hope this helps.
Brendan

Um... ok. You''re right, a point mass is meaningless, imagine
it as a small sphere for the purposes of moment of intertia.

What you say kind of makes sense but imagine that a very large
force is applied perpendicular to the radial vector. I assume
some kind of energy loss occurs or something?

I don''t think all of energy is going to go into the rotational
energy... doesn''t make sense even on a think-about-it scale...

Now, I''ve also been told:

Calculate the initial energy of the object.

Use the moment of interia to find the find how much of the (new)
energy goes toward rotational energy.

The left over is added to Ek which will let you figure out the
next velocity of the object.

It''s that second step which is getting me. I''ll have to look
up moments of interia, but anyway. In the last step, if you do
it this way, how are you going to find the new direction of the
velocity?

cheers.

Hmm.... I know energy isn''t going to just disappear. Think about gyroscope that you use a string for. When you pull the string, the only force that you are feeling is the string putting a torque on the g''scope in the direction of the string, but the center of mass isn''t trying to go anywhere. Simply put, if you have your force vector perpendicular to the surface, there will be no for trying to nudge it along horizontally. When people are talking about there being more energy, there isn''t. It''s always going to be the same amount of energy, as long as the work done is constant. If you move the object more linearly, then it wont be spinning as much, and vice-versa. No linear energy will be added when the force is applied perpendicular to the radius vector. No rotational energy will be added when it is applied in the direction of the radius vector. Look up in some physics texts if you aren''t convinced by this argument, but I am almost certain that you can''t accelerate an object linearly when applying a force perp. to the radius vector. Maybe this changes under non-spherical objects, but I don''t think so.
Brendan

No. I got a confirmation of this from my uni teacher. Think about it this way: A force is a variation of linear momentum. If the object is initially at rest, it has 0 linear momentum (p). The force applied will inevitably increase it. The rotation of the object does not affect p because the linear momentum of the mass on one side of the object is cancelled by the mass on the other side of the object (because p is a vectorial quantity).

Cédric

Applying the same net force to an object will always produce the same linear acceleration (Newton''s 2nd law).

Applying the same net moment or torque to an object will always produce the same angular acceleration.

A given force applied to an object will produce the same linear acceleration wherever you apply it, but depending on the source, the duration for which the force is applied may change (giving something a shove), or the direction of the force may vary (firing attached thrusters) giving a different change in linear motion.

I believe you can calculate instantaneous linear and angular accelerations independently from the same applied force - in most real world examples, both the force and the accelerations will change in an interdependent fashion over time.

I used to know how to calculate center of moment for an object, but I can''t recall details at the moment. For a uniform sphere, it''ll be at the center of the body.

Of course, for a spherical body, collisions are only going to impart torque through friction, so tangential collisions aren''t going to have any effect under the simplest friction model (max friction is proportional to normal reaction force at point of contact)

What I wrote earlier still stands, but I''ll just comment on a few of the points you''ve mentioned.

> I assume some kind of energy loss occurs or something?

There''s no energy loss from such calculations. Energy loss can occur from friction, can be dispersed (e.g. as sound) in a collision, can be lost through inefficiencies in the motor, but if these are ruled out energy can and should be conserved.

But this is of limited use: energy is a very blunt tool, as it is only a scalar, with no directional information, and so does not tell you very much. It''s worth calculating energy before and after to confirm you are conserving it, but apart from this there''s not much you can do with it.

> I don''t think all of energy is going to go into the rotational
> energy... doesn''t make sense even on a think-about-it scale...

No. You can have almost all energy going into the rotational energy but some will always go into the linear energy.

> Now, I''ve also been told:
> Calculate the initial energy of the object.
> Use the moment of interia to find the find how much of the
> (new) energy goes toward rotational energy....

As you''ve found this is not much use as knowing the energy does not give you enough information.

You can just apply the force seperately to the linear and rotational effects. To account for the distribution of effect between linear and angular use a variant of the collision equation, e.g. Figure 4 in Chris Hecker''s 4th GD article. This article goes into a lot of detail on the concepts involved so is well worth downloading.

Check out this thread:

http://www.llamasoft.co.uk/forum/viewtopic.php?t=2085

[teamonkey]