Rigid body, impulse based, Rolling and friction questions
Author
I'm trying to understand the concepts of a sphere hitting the ground with friction, but within a rigid body, impulse based physics sim.
Theres a few things I'm trying to find find out.
1) When the sphere hits a flat surface, would it slide at first, or would you immediately calculate it to be rolling (eg translational vel = -tangential vel) as I am using a rigid body and impulses?
2) How is kinetic energy lost?
Like say the sphere is spinning on the collision normal axis, and it is being held onto the ground by gravity (series of impulses every few milliseconds) how does it lose kinetic energy as the collision point velocity would be 0, so no friction force.
Same with a rolling sphere, when the tangential and translation velocities are the same but moving in opposite directions, the collision point's velocity would be 0, so how does it lose energy there too?
Thanks.
Quote:Original post by johnnyBravo
1) When the sphere hits a flat surface, would it slide at first, or would you immediately calculate it to be rolling (eg translational vel = -tangential vel) as I am using a rigid body and impulses?
Unless the ball is already spinning, it won't roll automatically, and even then it's not guaranteed. When the ball hits an incline, assuming no other forces at work, gravity will pull down on it. The normal force of the incline will push up on it, and there will be a net force of motion in some direction (for instance, down the incline). Without friction, the ball would simply slide down like any other object, however if friction is there then it provides a small frictional force on the surface of the sphere that contacts the table. From the sphere's perspective, this is the torque which starts rotation.
Quote:
2) How is kinetic energy lost?
This is the time to start thinking of the objects as lattices of atoms held together with electromagnetic bonds. When two bodies collide, their atoms are forced to come closer than they would be at equilibrium, and they seek to expand from that point, vibrating back and forth like a spring, and sending shockwaves through the objects. A second result is that some of the energy is imparted to the electrons, and they become more excited. Eventually the excited electrons calm down, tossing off a photon. The vibration and the emission will continue, if not prevented, into the surrounding environment, effectively dissipating the heat, which had been the kinetic energy.
Or at least that's the 4:00 AM version :)
Author
Also how is the friction force calculated when the sphere is spinning on the normal axis of a surface?
So far i've only read 2D equivalent things on calculating friction.
So far i've only read 2D equivalent things on calculating friction.
I think friction would still be calculated the same as in a 2d version. When calculating the friction, think less of the sphere and more of the contact patch. In a truly mathematical sense the contact patch between a sphere and a plane would be a single point. That single point will be moving in some direction and it should have some normal force applied on it from the ground plane. Friction on that point, and thus the entire sphere, should be a simple Ff = uN where u is the frictional coefficient and N is the normal force. The direction of the force would be the opposite of the direction of the point's velocity vector.
Ok, let's define friction force as a force acting on two sliding surfaces.
Assuming that the plane is horizontal and that the sphere has been initially in rotation (but above the plane, with no velocity). After it falls, it will hit the plane.
Take the point of the sphere contacting the plane. It's velocity relative to the plane is significant. A force is created at that point as to convert rotational momentum to translational momentum (Force at the distance creates torque opposing rotational momentum).. I don really know how energy would be lost in this case, or any other.
Assuming that the plane is horizontal and that the sphere has been initially in rotation (but above the plane, with no velocity). After it falls, it will hit the plane.
Take the point of the sphere contacting the plane. It's velocity relative to the plane is significant. A force is created at that point as to convert rotational momentum to translational momentum (Force at the distance creates torque opposing rotational momentum).. I don really know how energy would be lost in this case, or any other.
Author
Ok I understand that part,
but what if the point has no linear velocity, and has some angular velocity,
that is the sphere is dropped onto a surface while it is spinning perpendicular to the surface. eg a spinning top.
How would I calculate the friction force, or torque that would work against the spinning?
Another example would be 2 spheres held together by a (rigid) rod.
if they were spinning around the centre of mass and hit the ground the friction force would be in opposition to the torque.
(Though that raises another question, if I'm conserving energy, it would actually apply torque to the ground, but if the ground has infinite mass, then energy would not be conserved?)
[Edited by - johnnyBravo on August 6, 2007 8:02:16 PM]
but what if the point has no linear velocity, and has some angular velocity,
that is the sphere is dropped onto a surface while it is spinning perpendicular to the surface. eg a spinning top.
How would I calculate the friction force, or torque that would work against the spinning?
Another example would be 2 spheres held together by a (rigid) rod.
if they were spinning around the centre of mass and hit the ground the friction force would be in opposition to the torque.
(Though that raises another question, if I'm conserving energy, it would actually apply torque to the ground, but if the ground has infinite mass, then energy would not be conserved?)
[Edited by - johnnyBravo on August 6, 2007 8:02:16 PM]
Well, if a sphere is spinning perpendicular to the plane, then what is the contact point's velocity relative to the plane? Zero, it has no velocity. Therefore, no friction force.
In the case of two spheres connected by a rod you'll have to consider two contact points. Assuming you're spinning about the center of mass (which, with equal mass spheres, should be the center of the rod) then you'll have to points of contact moving in opposite directions on the plane. Thus they will have opposite friction forces. The torque created from these forces, however, will be in the same direction. Remember, torque = r x F where r is the radius from the center of torque to the point at which the force is being applied. Using the right hand rule and assuming you spun the dumbell in a clockwise manner, both torques should be facing down (or into the plane). The two friction forces added together cancel each other out, so in the end we get a logical answer: The spinning dumbell will not being moving with any linear velocity, but its angular velocity will decrease as a result of the torque being applied to it.
Does that help? It's kind of hard to explain some of this stuff with just text. If you need clarification, just say so and I'll try my best.
In the case of two spheres connected by a rod you'll have to consider two contact points. Assuming you're spinning about the center of mass (which, with equal mass spheres, should be the center of the rod) then you'll have to points of contact moving in opposite directions on the plane. Thus they will have opposite friction forces. The torque created from these forces, however, will be in the same direction. Remember, torque = r x F where r is the radius from the center of torque to the point at which the force is being applied. Using the right hand rule and assuming you spun the dumbell in a clockwise manner, both torques should be facing down (or into the plane). The two friction forces added together cancel each other out, so in the end we get a logical answer: The spinning dumbell will not being moving with any linear velocity, but its angular velocity will decrease as a result of the torque being applied to it.
Does that help? It's kind of hard to explain some of this stuff with just text. If you need clarification, just say so and I'll try my best.
Author
Quote:Well, if a sphere is spinning perpendicular to the plane, then what is the contact point's velocity relative to the plane? Zero, it has no velocity. Therefore, no friction force.
I see, as in the real world multiple points of the sphere would be touching as it deformed against the ground.
Quote:
both torques
I'd assume you would treat the dumbell as a single object once they're joined, so you would treat it as one torque, but I do see what you are saying with those calculations.
I guess I have been kind of obsessing about conservation of KE,
I've been thinking about a scenario in my head, say you had two cubes being pushed together on their flat surfaces, and they are both spinning on the 'normal' axis but in the opposite direction. Such that the friction would cancel each others angular rotation, with no linear movement (due to friction forces cancelling each other out).
This would be in a rigid body scenario, from this I cannot see how ke (= Iw^2) could be conserved, because the angular rotations would be less for both after the collision.
(this is just something I'm thinking about ..for now, as I'm only dealing with spheres right now)
....
Now just back onto my last problem, I can set a coefficient of restitution for the change in linear velocity along the normal axis for how much energy I want it to lose, but with the tangential motion, how do you think I could use my coefficient to set how much energy to lose? I can't really fathom on where I would put it.
thanks alot.
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