2D rigid body linear/angular force application

for some force F, and some vector R describing the position of the point of application from the centre of mass, with body mass m

change in linear acceleration is F/m, and change in torque is R×F

Quote:

It makes sense to me that if the force pierces the center of mass of the body, no torque is induced. Likewise for forces that act on the body as a whole, like gravity or buoyancy. Am I right in these assertions?

Yep, you are correct.

The linear acceleration of a body is equal to the sum of the forces that act upon it, divided by its mass. No matter where on the body do these forces act, you just add them, divide by the scalar mass and you have the linear acceleration vector. If you start changing the points on the body where the force is applied, but without changing directions of the forces, the linear acceleration stays the same.

Now the for the angular acceleration you need to find the sum of all the Torques that act on the body and divide by the moment of inertia I of the body (http://en.wikipedia.org/wiki/Moment_of_inertia). This is calculated easily for simple bodies. To find the torque T applied to the body due to a force f, you take the cross product of the force with the vector p - c (like luca-deltodesco said). In the case the line on which f lies crosses the center of mass, f will be co-linear with p - c and the cross product will be zero, meaning this force won't change the angular velocity of the object, like you correctly asserted.

Hope i helped.. I can clarify more if needed :)

Edit: as for the s segmentation, it does not affect the linear acceleration, as said before. But the amount of angular acceleration due to f grows linearly with the distance of the intersection pointe from the center, yes. (|p-c| grows linearly and consequently so does (f x (p - c)) / I).

[Edited by - D_Tr on August 4, 2007 4:35:31 PM]

thanks so much! especially D_Tr for the explanation- the simulation is working to my satisfaction now!

i'm left wondering though how the whole of a force can be applied linearly and also, in part, to angular acceleration... wouldn't this indicate a reaction greater than the causing action??

Thanks again!

You mean Newton's 3rd law about action-reaction? The law says that if you apply a force, you get the negative of this force as a reaction, so it has nothing to do with movement or acceleration. The force just produces more work if it causes rotational motion too, because the velocity of the point of application will in average be greater than if there was not angular motion.