# Basic Dynamics on "2D" Vacuum • ### Question

• Team:

I'm trying to simulate objects in a vacumm for a simple 2D game; unfortunately, my understanding of dynamics is a little lacking right now.  I understand how the acceleration, velocity and position vectors all play together, but I am having trouble summing all of the forces to achieve the final resulting acelleration vector on a uniformly distributed mass.  In my 2D model to be, all acceleration, velocity and position vectors will have three components: x, y and theta.  A final acelleration after all force summations are done should be expressible in these three "hopefully" orthogonal basis.  Please see picture below for an example of what is confusing me and see if you can explain what acceleration should take place in each case: Thanks,

Shawn

Saturday, October 1, 2011 6:18 PM

• I ran across a Torque in Space answer by Lauren Scott here: http://helios.gsfc.nasa.gov/qa_sp_ms.html

So my initial assumptions about rotation "may" have been wrong. NASA has a lot of good information on their site that make it easy for kids (and people like me) to understand, but sometimes finding that information might be an issue.

So if I read Ms. Scott's explanation correctly, all I need to do is:

1. Sum all forces into a single force vector
2. Computer linear acceleration by projecting the final force onto my x-y plane.
3. Sum all torques into a single torque "T" (Force*Distance from Center of Mass) vector
4. Compute angular acceleration by using T = I*alpha 

In regards to step#4, this is just a basic video game so the variable "I" can be approximated with a constant for each body. For example, very spread out objects like a very long board can have a high moment of inertia but tightly wound objects like spheres might have a small movement of inertia constant. Then alpha or my angular acceleration just becomes

theta = T/I (Torque divided by Moment of Inertia)

after all forces have been summed. This makes the problem much simpler. Again, since this is just a video game, the "I" (moment of inertia) constants for each object can be estimated and the players "if any" probably won't know the difference (if they even care about 2D games anymore...).

So back to the picture above:

Assumption#1 should be correct since the only force is acting on the center of mass so there should be no rotation.

Assumption#2 should be correct since there is no net x-y force, but there is a torque equal to FD + FD = 2FD where D is the tangential length of each of the two forces from the center of mass of the board.

Assumption#3 should be correct since there is no net force and no net torque.

Case#4: There should be a force of F in +Y and a rotation in +theta where torque is FD given that D is (Length of Board / 2)

Case#5: There should be a force of 2F+2F - 1F = 3F in +Y and a torque of 2FD+FD-2FD=FD so there will be a rotation in +theta.

Case#6: There should be a force of F in -Y and a force of F in -X and a torque of FD+(F*0) = FD in positive theta because the force in -X applied very near the center of mass so it has no torque.

So I "guess" I can start on my game again assuming someone doesn't come by and correct me. Again this is just a game, so pretty much anything goes. I just thought it would be neat to see things fly around a bit on impact :-)

 - This formula is the Same as Scott's formula with out the division by 2. I don't need the division by 2 because will be measuring the tangential distance of my forces from the center of gravity. Scott used L/2 because L was the length of the board, but in my situation I would rather L be the tangential distance from the center of mass.

Thanks,

Shawn

I'm starting to remember dynamics (I got a "C" in that class and I don't remember doing anything in outer space. Maybe if my instructor would have covered outer space, I might have paid attention...)

Saturday, October 1, 2011 7:34 PM