Rigid body physics question
Hi there,
I have a rigid body made up of 2 spheres connected by a thin bar (the mass of the bar is negligible). Sphere1 has mass m1 and Sphere2 has mass m2. The body is falling towards a horizontal plane. At the point of impact with the plane, the body has linear velocity v and rotational velocity w. From this I can calculate the instantaneous velocity at impact of each sphere, say v1 and v2 (I know how to calculate these quantities).
How do I calculate the impulsive force applied from the collision in the following cases:
Only one sphere collides (one impulse);
Both spheres collide. (two impulses)
I''m a bit confused as to how the mass distribution affects the force at each collision point.
Cheers,
Mikey.
Possibly ... possibly not. Really I''m looking for a solution in the general case.
If they were not rotating then the dot product between the of the second sphere''s momemtum and the direction of the bar would say how much of the momemtum of the second sphere was transferred to the first through the bar. It seems like all of the rotational mememtum would be transferred.
Keys to success: Ability, ambition and opportunity.
Yeah, I thought of that. But if you take the more general case where n spheres are each interconnected it gets even morer complicated (especially if the spheres can have different radii).
Imagine, for example, having the spheres setup as before, but now there is a 3rd sphere directly inbetween. This sphere is connected to each of the other 2 spheres, but has half there radius. The 2 bigger spheres both collide similtaneously. How does the momentum of the smaller sphere get transferred now?
Assuming the small sphere is at the centre of gravity of the object, it will have no rotational momentum, so it''s momentum will be entirely downwards. The dot product of the momentum in the direction of the bars would be zero.
(This example may seem contrived, but another way of thinking of it is to let the bar from the previous example now have mass, i.e. a dumbell).
Cheers,
Mikey.
Imagine, for example, having the spheres setup as before, but now there is a 3rd sphere directly inbetween. This sphere is connected to each of the other 2 spheres, but has half there radius. The 2 bigger spheres both collide similtaneously. How does the momentum of the smaller sphere get transferred now?
Assuming the small sphere is at the centre of gravity of the object, it will have no rotational momentum, so it''s momentum will be entirely downwards. The dot product of the momentum in the direction of the bars would be zero.
(This example may seem contrived, but another way of thinking of it is to let the bar from the previous example now have mass, i.e. a dumbell).
Cheers,
Mikey.
I don''t know. I can''t decide if the entire momemtum of the object would be divided by the number of points of impact or if the mass of the individual spheres would determine the force of impact of the seperate spheres.
Keys to success: Ability, ambition and opportunity.
I see what you''re getting at, and that''s the problem I''m facing. I initially tried just calculating an impulse based on the mass of the individual spheres, but that doesn''t work. If you take the case when the object is resting on the plane, then the total impulse produced (ignoring restitution) should be enough to counteract gravity. If the sum of the masses of the resting spheres is not equal to the total mass of the object, then the total impulse produced would not be enough (since the force is divided by mass to give acceleration). The result is that the object slowly sinks into the ground.
I think probably the best way is to spread the mass amongst the resting spheres, but I was hoping there was a better solution. This way seems like it will get very fiddly when dealing with an arbitary collection of spheres.
Thanks anyway ... you''ve helped me get a couple of things straight in my mind.
Mikey.
I think probably the best way is to spread the mass amongst the resting spheres, but I was hoping there was a better solution. This way seems like it will get very fiddly when dealing with an arbitary collection of spheres.
Thanks anyway ... you''ve helped me get a couple of things straight in my mind.
Mikey.
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