Does rigid body physics support rotations? I would expect this thing to start to during a free fall rotate due to its non-symmetrical mass distribution which would give a larger torque on the left. However all I get is a translation downwards.
$\begingroup$
$\endgroup$
2
-
3$\begingroup$ It sounds like you want to simulate aerodynamic drag. $\endgroup$– Mutant BobJun 15, 2015 at 19:09
-
$\begingroup$ Do I really need drag for a non-zero torque? I should do a manual calculation of that. $\endgroup$– user877329Jun 16, 2015 at 7:26
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
If you neglect air drag and other forces, basic mechanics tells you that there is no net torque due to the weight of the object.
Edit: An example can be this: Imagine a point mass m attached via a massless bar to a point mass 2m. Assume that the bara is 1 meter long. Then the center of mass will be 1/3 meters from the heavier mass.
Now the gravitational force acting on the two masses is mg and 2mg, respectively. 
The torque due to gravity around the center of mass from the left mass is mg*(2/3 meters), where 2/3 meters is the distance from the mass to the center of gravity. This torque tries to rotate the bar counter clockwise. Now the torque from the gravity force acting on the heaver mass is 2mg*(1/3 meters) = mg*(2/3 meters), but this torque tries to rotate the bar clockwise. Hence they two torques exactly cancels and there will be no rotation.
This can be proven also in the general case which is done in most beginners university courses on classical mechanics.
-
$\begingroup$ Is it because any pivot point would accelerate together with the body that makes things different. $\endgroup$ Jun 27, 2015 at 17:21
-
$\begingroup$ This is correct, a gravitational force would not cause torque/rotation on a rigid body, because force on each point is proportional to mass (or density). In theory one could instead use a non-gravitational wind force, where each point receives the same force, giving an overall torque. However, the way forces are applied in Blender on rigid bodies makes this a bit tricky because only the center of mass is actually affected. $\endgroup$– lukas_tOct 2, 2015 at 6:45
$\begingroup$
$\endgroup$
Here is an example for working around the "center-of-mass only" forces on rigid bodies:
- the shape of
MainBodyis irrelevant to the sim, it just illustrates the behavior HelperMassis attached toMainBodywith a simple fixed constraint- a "Wind" force is used instead of gravity, because gravity's effect is independent of mass
- Layers are used for controlling force effects, note that
HelperMassdoes not share a layer with the wind field! That way it only shifts the center-of-mass away from the wind force attack point (the object center)
