mjbWorld

Wow! I just realised that this was the result that was originally posted by Martin.

Quote "
I think its linear motion is simple, since F = ma = m dv/dt this gives:
Change in velocity = dv = (F/m) dt = impulse/m
So the linear motion will be in the same direction as the impulse, even if the impulse is applied at a point offset from the centre of mass, this seems counter intuitive and this is what I have been trying to get to grips with in the various examples on the website. I think the is correct however, I think that this does not correspond to our intuition because in a practical collision the impulse is not necessarily in the direction of the impact but will have a component which is normal to the surface of the colliding objects.

We then need to calculate the rotation:
I think we could have an equivalent to Newtons 3rd law for impulses, in other words: every impulse has an equal and opposite impulse. If we apply the impulse to an object which is not attached to anything else then I think this equal and opposite reaction would act on the centre of mass.

Perhaps we could have an angular equivalent of impulse as follows:
Linear impulse = integral(Force) dt = change of linear momentum
Angular impulse = integral(Torque) dt = change of angular momentum
However I think we have to be careful here because in the linear case:
F=ma where m=mass which is a scalar quantity
but in the angular case we have:
T = [I] angAcc where [I] = inertia tensor which is a 3x3 matrix

So is it valid to say that:
Change in angular velocity = [I]-1 angular impulse
Where:
* [I]-1 = the inverse of the inertia tensor in global coordinates (which will change as the object rotates)
* angular impulse = Linear impulse x d
* d = distance between c of m and point where impulse is applied (which is a 3D vector quantity)
* x = vector cross product
* Linear impulse = 3D vector quantity

"End Quote

Now I think I understand it a bit more.