I Thread attempt 2: Edge physics trying to understand

[paradisePhysicist]

My first thread was closed for being too confusing to read. I will try to make it more clear.

Also some more parameters: Rod and solid has 0 restitution, solid is anchored and does not move, it is instantly deleted upon collision.

 

[Nugatory]

You will have to give us some details about your physics simulator and what it's assuming about the collision between the rod and the object before we can be sure what's going on.

However, I can suggest some things that you may not have considered:
- If the collision with the object is elastic, then the total kinetic energy of the system will be conserved.
- If the mass of the object is large compared with that of the rod (implied by saying that the object does not move) then all the kinetic energy will stay with the rod.
- If the rod is rotating after the collision then some of the kinetic energy will go into that rotation, reducing the kinetic energy available for the rod's forward motion.

 

[jbriggs444]

As I understand it, you have a hypothetical rigid rod. The left end is brought to rest with a brief impulse. After this event is complete, the object which was the source of that impulse becomes irrelevant and is removed from the simulation.

One might approach this problem by writing down several equations. One equation would involve linear momentum and the unknown impulse. One would involve angular momentum and the unknown impulse. A third would relate the final rotation rate with the final linear velocity based on the known velocity of the left endpoint on the rod.

If the rod is not modeled as being completely rigid then the problem changes. One would need to consider the duration of the impulse and the speed of [transverse] waves in the rod.

 

[A.T.]

solid is anchored and does not move, it is instantly deleted upon collision.

If it disappears instantly, it has no effect at all. Otherwise, it depends how long it is in the way.

 

[jrmichler]

Keep in mind that simulation is not reality. Simulation is a series of equations. Reality is mass, velocity, stiffness, modulus of elasticity, air resistance, yield strength, an infinite number of natural frequencies, gravity, and many other variables.

Your simulation does not includes most of the factors of a real world collision, therefore it will not respond like a real world collision. A question to ask is: "In what ways does my simulation differ from reality?".

All simulations differ from reality. The goal is make any particular simulation good enough to answer a specific problem.

 

[paradisePhysicist]

- If the collision with the object is elastic, then the total kinetic energy of the system will be conserved.

Elasticity is zero since restitution of both objects is zero.

- If the mass of the object is large compared with that of the rod (implied by saying that the object does not move) then all the kinetic energy will stay with the rod.

Not sure how that works. Wouldn't a lighter object allow the rod to keep more of its kinetic energy?

- If the rod is rotating after the collision then some of the kinetic energy will go into that rotation, reducing the kinetic energy available for the rod's forward motion.

Yes. But the strange thing about rotational energy is, if you apply a force to a side of any object, the object will move 100% of the velocity of the force, even if a rotation is added.

So if you have a rectangle, and its mass is 1, and you put a force of 1 to the center of its mass, it will not rotate but have a velocity of 1. But if you put the force to its edge, it will still have a velocity of one but a bonus angular velocity as well.

It is important to note though that this is a completely different scenario than the OP though. The OP is talking about hitting a wall, whereas what I mention here, is applying an arbitrary thrust force.

As I understand it, you have a hypothetical rigid rod. The left end is brought to rest with a brief impulse. After this event is complete, the object which was the source of that impulse becomes irrelevant and is removed from the simulation.

Yes.

One might approach this problem by writing down several equations. One equation would involve linear momentum and the unknown impulse. One would involve angular momentum and the unknown impulse. A third would relate the final rotation rate with the final linear velocity based on the known velocity of the left endpoint on the rod.

What I have so far (in my head, the sims used in the OP are not mine) is an equation that applies a force to a point (the angle of force is the normal of the wall collision, though I'm not sure this is actually the correct way to do it.) And it tries different forces until it approaches the correct force that will push the left end of the rod outside of the collision next frame. That is my method, its not very CPU efficient but seems realistic to me. (Actually a more realistic method would probably be first going backwards in time to the first point of contact, then doing something.)

I don't know what method Box2d or algodoo uses, it probably uses calculus or something.

If the rod is not modeled as being completely rigid then the problem changes. One would need to consider the duration of the impulse and the speed of [transverse] waves in the rod.

Rod is completely rigid. Is the speed of transverse waves of rigid bodies the speed of light or slower? You are talking about the speed at which atoms are connected to each other right?

If it disappears instantly, it has no effect at all. Otherwise, it depends how long it is in the way.

Good point, but, in the simulation I have tested random frame rates and update rates, and it sticks at 75% no matter what.

Keep in mind that simulation is not reality. Simulation is a series of equations. Reality is mass, velocity, stiffness, modulus of elasticity, air resistance, yield strength, an infinite number of natural frequencies, gravity, and many other variables.

Your simulation does not includes most of the factors of a real world collision, therefore it will not respond like a real world collision. A question to ask is: "In what ways does my simulation differ from reality?".

All simulations differ from reality. The goal is make any particular simulation good enough to answer a specific problem.

Correct. My goal isn't to go for hyper realisim, but the purpose of using these sims is to find a simple equation that I can create a simplistic physics system by. My original thesis was that physics is fundamentally about mass and inertia, and that if I have a very long rod, and a tiny edge tip is hit, that the rod will continue with roughly 99% of its velocity.

But putting it in the sims confuses me and no longer does it seem like a simple case, due to the 75% problem.

The sims I have tried are Box2d, Algodoo, and a 3rd party one which allows different solvers.

 

[jbriggs444]

Rod is completely rigid. Is the speed of transverse waves of rigid bodies the speed of light or slower? You are talking about the speed at which atoms are connected to each other right?

Completely rigid = infinite speed of sound. Not limited by speed of light. This is a Newtonian simulation. Speed of light does not enter in. This is a simulation. Atoms do not enter in either.

 

[Nugatory]

Not sure how that works. Wouldn't a lighter object allow the rod to keep more of its kinetic energy?

No, it's the other way around. If the object is lighter than the force on it from the collision (equal and opposite to the force on the rod, by Newton's third law) will accelerate it more so that it gains more kinetic energy; conversely when the object is heavier it accelerates less and more of the kinetic energy stays with the rod.

You've specified that the object does not move at all. It's kinetic energy is $mv^2/2$, and if it isn't moving $v=0$ and its kinetic energy is zero before and after. So none of the original kinetic energy carried by the rod went into the object, and elastic collisions conserve kinetic energy so it didn't just disappear; it all had to stay with the rod.

Yes. But the strange thing about rotational energy is, if you apply a force to a side of any object, the object will move 100% of the velocity of the force, even if a rotation is added.
So if you have a rectangle, and its mass is 1, and you put a force of 1 to the center of its mass, it will not rotate but have a velocity of 1. But if you put the force to its edge, it will still have a velocity of one but a bonus angular velocity as well.

When I apply a one Newton force to the center of mass of an object weighing one kilogram for one second, the object will indeed end up moving at one meter per second and the point where I am applying the force will move 1/2 meter. The kinetic energy of the object will be 1/2 Joule ($E_k=mv^2/2=Fd$).

However, if I apply the same one Newton force to the edge of the object causing it to rotate, I will find that the point where I'm applying the force moves more than 1/2 meter during the second that I'm applying the force. That is, the $d$ in $W=Fd$ is greater, so the force is doing more work. That's the source of the "bonus" energy that you're thinking of: it takes more energy to start an object moving and turning than just moving.

My original thesis was that physics is fundamentally about mass and inertia, and that if I have a very long rod, and a tiny edge tip is hit, that the rod will continue with roughly 99% of its velocity.

It will depend on the details of the collision. Energy is conserved and none of it goes into the object, so it all stays with the rod. The sum of the rotational and linear kinetic energy after the collision will be equal to the initial kinetic energy before the collision. All the collision is doing is turning some of the linear kinetic energy into rotational kinetic energy; by varying that amount you ill be able to vary the final speed.

 

[jbriggs444]

Collisions between rigid objects involve infinite forces applied over infinitesimal intervals and infinitesimal displacements. Lots of room for indeterminate results. Lots of room for erroneous reasoning.

Just because both objects are rigid, that does not mean that mechanical energy is conserved. The collision is specified as inelastic. Mechanical energy is not conserved.

However, it is possible to pose an exactly equivalent problem using an elastic collision. Replace the inelastic block with an elastic rubber ball. Tune the mass of the ball so that the rod tip is brought to rest by an elastic collision with the ball.

 

[paradisePhysicist]

No, it's the other way around. If the object is lighter than the force on it from the collision (equal and opposite to the force on the rod, by Newton's third law) will accelerate it more so that it gains more kinetic energy; conversely when the object is heavier it accelerates less and more of the kinetic energy stays with the rod.

You've specified that the object does not move at all. It's kinetic energy is $mv^2/2$, and if it isn't moving $v=0$ and its kinetic energy is zero before and after. So none of the original kinetic energy carried by the rod went into the object, and elastic collisions conserve kinetic energy so it didn't just disappear; it all had to stay with the rod.

Oh ok I see.

When I apply a one Newton force to the center of mass of an object weighing one kilogram for one second, the object will indeed end up moving at one meter per second and the point where I am applying the force will move 1/2 meter. The kinetic energy of the object will be 1/2 Joule ($E_k=mv^2/2=Fd$).

However, if I apply the same one Newton force to the edge of the object causing it to rotate, I will find that the point where I'm applying the force moves more than 1/2 meter during the second that I'm applying the force. That is, the $d$ in $W=Fd$ is greater, so the force is doing more work. That's the source of the "bonus" energy that you're thinking of: it takes more energy to start an object moving and turning than just moving.

Yes but the funny thing is that the center of the object (if center of mass is in middle) will go 1 meter per second in both scenarios (at least according to a physics guy I talked to and Box2d.)

It will depend on the details of the collision. Energy is conserved and none of it goes into the object, so it all stays with the rod. The sum of the rotational and linear kinetic energy after the collision will be equal to the initial kinetic energy before the collision. All the collision is doing is turning some of the linear kinetic energy into rotational kinetic energy; by varying that amount you ill be able to vary the final speed.

I'm not sure how to go about solving this, but it seems like it has potential.

Collisions between rigid objects involve infinite forces applied over infinitesimal intervals and infinitesimal displacements. Lots of room for indeterminate results. Lots of room for erroneous reasoning.

Just because both objects are rigid, that does not mean that mechanical energy is conserved. The collision is specified as inelastic. Mechanical energy is not conserved.

However, it is possible to pose an exactly equivalent problem using an elastic collision. Replace the inelastic block with an elastic rubber ball. Tune the mass of the ball so that the rod tip is brought to rest by an elastic collision with the ball.

So basically, there is no "correct" answer and it cannot be solved? And the only realistic physics we can make is ball physics?

 

[jbriggs444]

So basically, there is no "correct" answer and it cannot be solved? And the only realistic physics we can make is ball physics?

One can have rigidity with predictability. Usually what is required is an elasticity parameter. What you can do is to look at the limiting behavior as rigidity is increased without bound while elasticity (i.e. coefficient of restitution) is held constant.

Just because the algebraic result takes an indeterminate form does not mean that the limit cannot be evaluated.

 

[Nugatory]

Yes but the funny thing is that the center of the object (if center of mass is in middle) will go 1 meter per second in both scenarios (at least according to a physics guy I talked to and Box2d.)

Yes, but think about that means. The center of mass of the box moving at one meter per second, and the point on the rod where the force is applied is rotating forward relative to the center of mass - so that point is moving at more than one meter per second so covers more than 1/2 meter in the second that the force is applied. More distance means more work being done, and the “extra” work is what provides the energy needed for rotation.

 

[Nugatory]

I'm not sure how to go about solving this, but it seems like it has potential.

The kinetic energy from the motion of the center of mass of the rod is $mv^2/2$. The kinetic energy from the rotation of the rod is $I\omega^2/2$ where $\omega$ is the rotation rate (in radians per second) and $I$ is a constant that depends on the mass and shape of the rotating object; for an ideal rod of length $L$ the formula is $I=mL^2/12$. (Don’t take my word for this! google for “moment of inertia” and “rotational kinetic energy” and look for the meaning behind the formulas!).
Conservation of energy and a bit of algebra will let you see what combinations of rotation and speed are possible if the collision transfers various amount of energy.

So basically, there is no "correct" answer and it cannot be solved? And the only realistic physics we can make is ball physics?

It’s actually the other way around. The ideal ball physics is unrealistic, but the math involved is straightforward enough that it doesn’t obscure the underlying physical principles (conservation laws, Newton’s laws, ...) so that’s what we use in teaching examples.

 

[hutchphd]

.

I must be missing something here because this all seems inordinately obscure. Force from object brings end of stick to momentary stop (restitution=0). Angular momentum about that point conserved in process. Write down angular momentum in terms of velocity. Answer falls out...what am I missing?

 

[paradisePhysicist]

One can have rigidity with predictability. Usually what is required is an elasticity parameter. What you can do is to look at the limiting behavior as rigidity is increased without bound while elasticity (i.e. coefficient of restitution) is held constant.

Just because the algebraic result takes an indeterminate form does not mean that the limit cannot be evaluated.

I don't know what you mean by limiting behavior, but I get the vibe I must make a soft body sim then slowly make it more and more rigid. Seems kinda complicated.

Yes, but think about that means. The center of mass of the box moving at one meter per second, and the point on the rod where the force is applied is rotating forward relative to the center of mass - so that point is moving at more than one meter per second so covers more than 1/2 meter in the second that the force is applied. More distance means more work being done, and the “extra” work is what provides the energy needed for rotation.

I think what helps me understand the most is when I realize that when one edge is moving forward, the other is moving backward relative to the COM, thus both are actually cancelling each other out. I just thought of this btw.

The kinetic energy from the motion of the center of mass of the rod is $mv^2/2$. The kinetic energy from the rotation of the rod is $I\omega^2/2$ where $\omega$ is the rotation rate (in radians per second) and $I$ is a constant that depends on the mass and shape of the rotating object; for an ideal rod of length $L$ the formula is $I=mL^2/12$. (Don’t take my word for this! google for “moment of inertia” and “rotational kinetic energy” and look for the meaning behind the formulas!).
Conservation of energy and a bit of algebra will let you see what combinations of rotation and speed are possible if the collision transfers various amount of energy.
It’s actually the other way around. The ideal ball physics is unrealistic, but the math involved is straightforward enough that it doesn’t obscure the underlying physical principles (conservation laws, Newton’s laws, ...) so that’s what we use in teaching examples.

I want the most realistic, but simple. Also, the using Energy thing (rather than force and mass) seems like an interesting approach, however the problem is that it has multiple solutions.

.

I must be missing something here because this all seems inordinately obscure. Force from object brings end of stick to momentary stop (restitution=0). Angular momentum about that point conserved in process. Write down angular momentum in terms of velocity. Answer falls out...what am I missing?

Therein lay the problem...there are multiple solutions. It sounds too easy. This is not a case of a door locked to a hinge, but sliding going on the corner of solid block. If you just project the center of the moving rod forward, and get the angle for torque, it doesn't seem like it'd be the correct solution, though perhaps I'm misunderstanding.

 

[hutchphd]

Therein lay the problem...there are multiple solutions. It sounds too easy. This is not a case of a door locked to a hinge, but sliding going on the corner of solid block. If you just project the center of the moving rod forward, and get the angle for torque, it doesn't seem like it'd be the correct solution, though perhaps I'm misunderstanding.

The original statement of he problem seems unambiguous to me: big block stops the end of stick with no bounce (restitution=0). Block disappears (no more forces on stick)
The unique solution as I described follows immediately and matches simulations.
(The result cannot depend upon mass or length because there is nothing in the problem to set a scale,. The stick is rigid and the wall infinitely massive.)
I don't understand all the ancillary analysis. Hope this helps!

 

[Nugatory]

Therein lay the problem...there are multiple solutions. It sounds too easy.

When the rod strikes the object, the speed of that end of the rod will change. For any given change in speed, there is only one solution and it really is as easy as @hutchphd says to calculate it from there.
The ambiguity that leads to multiple solutions is that it is not completely clear that you intended that the end of the rod would come to a complete stop as opposed to being slowed but continuing to move forward in the instant after the collision.

, but I get the vibe I must make a soft body sim then slowly make it more and more rigid. Seems kinda complicated.

Much much much better is to solve this problem analytically, using what you know about conservation of energy and angular momentum.

 

[Dale]

So if you have a rectangle, and its mass is 1, and you put a force of 1 to the center of its mass, it will not rotate but have a velocity of 1. But if you put the force to its edge, it will still have a velocity of one but a bonus angular velocity

That is correct, assuming the force is applied for 1 unit of time in both cases. Btw, it is best to include your units when describing dimensionful quantities

My original thesis was that physics is fundamentally about mass and inertia

You cannot determine that through simulations.

Yes but the funny thing is that the center of the object (if center of mass is in middle) will go 1 meter per second in both scenarios

Yes, that is Newton’s 2nd law

 

[paradisePhysicist]

The original statement of he problem seems unambiguous to me: big block stops the end of stick with no bounce (restitution=0). Block disappears (no more forces on stick)
The unique solution as I described follows immediately and matches simulations.
(The result cannot depend upon mass or length because there is nothing in the problem to set a scale,. The stick is rigid and the wall infinitely massive.)
I don't understand all the ancillary analysis. Hope this helps!

When the rod strikes the object, the speed of that end of the rod will change. For any given change in speed, there is only one solution and it really is as easy as @hutchphd says to calculate it from there.
The ambiguity that leads to multiple solutions is that it is not completely clear that you intended that the end of the rod would come to a complete stop as opposed to being slowed but continuing to move forward in the instant after the collision.
Much much much better is to solve this problem analytically, using what you know about conservation of energy and angular momentum.

Hmm so you are saying to simply, push the contact point downwards vertically until it is outside the solid?
I tried that it didn't work, probably not understanding though.

I think the problem is that the rod rotates so at some times its outside of the collision. Also that its an edge with a tip.

Can you show me a picture of what hutch means?

You cannot determine that through simulations.

Why not?

 

[hutchphd]

The original statement of he problem seems unambiguous to me: big block stops the end of stick with no bounce (restitution=0). Block disappears (no more forces on stick)
The unique solution as I described follows immediately and matches simulations.
(The result cannot depend upon mass or length because there is nothing in the problem to set a scale,. The stick is rigid and the wall infinitely massive.)
I don't understand all the ancillary analysis. Hope this helps!

I am trying to understand what is unclear. The force is applied "percussively" to one end of a finite length stick. Angular momentum about this point of contact is always conserved.
1)If restitution = zero the end of the stick does not rebound and angular momentum conservation alone allows simple solution.
2) If restitution is 1 then additionally energy conservation allows complete solution and characterizes the "bounce" of the stick

In between these extremes it is more complicated...as it is in any collision

 

[Dale]

Why not?

Because there is no way to verify that the simulation is an accurate representation of reality. That can only be done by experiment.

 

[paradisePhysicist]

I am trying to understand what is unclear. The force is applied "percussively" to one end of a finite length stick. Angular momentum about this point of contact is always conserved.
1)If restitution = zero the end of the stick does not rebound and angular momentum conservation alone allows simple solution.
2) If restitution is 1 then additionally energy conservation allows complete solution and characterizes the "bounce" of the stick

In between these extremes it is more complicated...as it is in any collision

Can I see a picture please? lol.

Because there is no way to verify that the simulation is an accurate representation of reality. That can only be done by experiment.

I see your point. Unfortunately, I don't have access to a gravity free vacuum in which to test my theory.

 

[jbriggs444]

Collisions are normally understood to be brief. The tip of the rod is not "pushed" downward. It is "bumped" downward. Once the bump is over, it is over. There is no further force and, often, no further contact.

You have several basic possibilities depending on the exact make-up of the rod and block.

1. The tip of the [rigid] rod bounces off of the [elastic] block. But you do not want this. You want zero restitution. No bounce.

2. The tip of the [rigid] rod stops at the [plastic] block. As if it hit a soft pad and came to rest. No bounce, no restitution. You seem to want this.

3. The tip of the [rigid] rod stops at the [gummy] block and sticks. The rod pivots around the point of attachment. You do not seem to want this.

4. The tip of the [plastic] rod stops at the [rigid] block. But the rest of the rod continues on its way, blissfully unaware that a collision has taken place. The rod flexes more and more and the point of contact is under a continuing force. You want zero restitution. So the rod stays bent.

5. The tip of the [flexible] rod stops at the [rigid] block. The rest of the rod continues on its way, blissfully unaware that a collision has taken place. The rod flexes more and more and the point of contact is under a continuing force. Then it unbends and rebounds. It may or may not continue to vibrate back and forth after the collision is over.

Let us suppose that you choose 2. Let r be the length of the rod.

You have a rigid rod with known velocity and momentum. Multiply the momentum by r/2 you have the the associated angular momentum about the point of contact. The zero restitution assumption means that the rod tip comes to a stop. So you know that the motion immediately following the collision can be [momentarily] seen as a pure rotation about the point of contact. You also know that the interaction involved zero torque about the point of contact.

As @hutchphd points out, conservation of angular momentum allows you to read off the resulting rotation rate (in rad/sec) almost immediately. Divide angular momentum by the moment of inertia of a uniform rod about its endpoint.

Multiply that rotation rate by r/2 and you have the resulting velocity of the center of mass. Done.

 

[paradisePhysicist]

Collisions are normally understood to be brief. The tip of the rod is not "pushed" downward. It is "bumped" downward. Once the bump is over, it is over. There is no further force and, often, no further contact.

You have several basic possibilities depending on the exact make-up of the rod and block.

1. The tip of the [rigid] rod bounces off of the [elastic] block. But you do not want this. You want zero restitution. No bounce.

Yes, so far so good.

2. The tip of the [rigid] rod stops at the [plastic] block. As if it hit a soft pad and came to rest. No bounce, no restitution. You seem to want this.

Nope. I want it to be rigid. When it is fast, it bounces. But imagine a long heavy rod hitting something slowly. There would be almost no bounce, a nearly 0 restitution.

3. The tip of the [rigid] rod stops at the [gummy] block and sticks. The rod pivots around the point of attachment. You do not seem to want this.

That is correct, I do not want gummy sticks.

4. The tip of the [plastic] rod stops at the [rigid] block. But the rest of the rod continues on its way, blissfully unaware that a collision has taken place. The rod flexes more and more and the point of contact is under a continuing force. You want zero restitution. So the rod stays bent.

Nope I just want no bounce, but I want a rigid rod.

5. The tip of the [flexible] rod stops at the [rigid] block. The rest of the rod continues on its way, blissfully unaware that a collision has taken place. The rod flexes more and more and the point of contact is under a continuing force. Then it unbends and rebounds. It may or may not continue to vibrate back and forth after the collision is over.

Nope. No flex, just rigid.

Let us suppose that you choose 2. Let r be the length of the rod.

You have a rigid rod with known velocity and momentum. Multiply the momentum by r/2 you have the the associated angular momentum about the point of contact. The zero restitution assumption means that the rod tip comes to a stop. So you know that the motion immediately following the collision can be [momentarily] seen as a pure rotation about the point of contact. You also know that the interaction involved zero torque about the point of contact.

As @hutchphd points out, conservation of angular momentum allows you to read off the resulting rotation rate (in rad/sec) almost immediately. Divide angular momentum by the moment of inertia of a uniform rod about its endpoint.

Multiply that rotation rate by r/2 and you have the resulting velocity of the center of mass. Done.

Im not sure how to get the rotation rate. My guess is to get the sum of torques of both ends, but I only know how to apply torques to the center of mass, not an arbitrary point. You said use r/2 but that doesn't seem like it would apply to an arbitrary point, since r is defined as the length of the rod.

 

[jbriggs444]

Nope I just want no bounce, but I want a rigid rod.

Then you want case #2 in spite of your protestations to the contrary.

 

[paradisePhysicist]

Then you want case #2 in spite of your protestations to the contrary.

Maybe a little cushion so it does not bounce at all, but not so much that it is visually apparent.

 

[jbriggs444]

Maybe a little cushion so it does not bounce at all, but not so much that it is visually apparent.

Right. But no cushion is needed on the rod. The cushion can be on the block.

 

[hutchphd]

This problem can in principle be solved by tracking the motion of each atom in the rod and doing the calculation using interatomic forces. If that is your method of choice, give it a go. Might take a while.
If you wish an exact macroscopic answer, intelligent use of momentum conservation will provide it in minutes, as previously and repeatedly described. Your choice.

 

[paradisePhysicist]

Ok I have decided will go with the point method, even though not 100% accurate. Need to search for links on teetertotters and such in terms of rotational balance.

Once I have achieved my own simulation, I will finally find out if box2d and algodoo are truly realistic.

 

[hutchphd]

Ok I have decided will go with the point method, even though not 100% accurate. Need to search for links on teetertotters and such in terms of rotational balance.

Once I have achieved my own simulation, I will finally find out if box2d and algodoo are truly realistic.

You don't need a simulation. The analytic answers are exact for the problem you described. As I understand it box2d and agodoo agree which the exact analysis. If you write a bad simulation, it will give you a different result.

 

[paradisePhysicist]

You don't need a simulation. The analytic answers are exact for the problem you described. As I understand it box2d and agodoo agree which the exact analysis. If you write a bad simulation, it will give you a different result.

It's just I'm still not understanding where it says how to calculate the actual rotational acceleration.

He says "
Let r be the length of the rod.

You have a rigid rod with known velocity and momentum. Multiply the momentum by r/2 you have the the associated angular momentum about the point of contact "

That doesn't make sense. For instance, if the middle of the rod is hit, there ought not to be any angular momentum at all, yet the equation r/2 will generate angular momentum anyway. The equation seems missing certain variables, like the relative distance to the COM of the point, etc.

In the meantime, his equation may be good enough for an infinitesimally small tip point maybe? Which should theoretically be good enough to see if the 2d physics sims are truly accurate.

I guess I should make a separate thread because, I'm kind of looking for an equation that can handle tip sizes larger than zero, which is what I meant about looking for links about teetertotter equations.

 

[jbriggs444]

That doesn't make sense. For instance, if the middle of the rod is hit

The scenario I thought we were discussing is a hit on the end of the rod.

If you hit the rod elsewhere then the calculation of angular momentum about the point of impact will multiply linear momentum by the offset of the center of mass from the point of impact.

And when you go on to divide angular momentum by moment of inertia to get rotation rate, you'll need moment of inertia about the point of impact.

In the special case of an impact at the center of the rod, the rod comes to a halt, obviously.

 

[paradisePhysicist]

The scenario I thought we were discussing is a hit on the end of the rod.

If you hit the rod elsewhere then the calculation of angular momentum about the point of impact will multiply linear momentum by the offset of the center of mass from the point of impact.

Sorry for being a bit vague. I kind of wanted to use the equation to conduct further edge tests that were nonzero values.

 

[paradisePhysicist]

The scenario I thought we were discussing is a hit on the end of the rod.

If you hit the rod elsewhere then the calculation of angular momentum about the point of impact will multiply linear momentum by the offset of the center of mass from the point of impact.

And when you go on to divide angular momentum by moment of inertia to get rotation rate, you'll need moment of inertia about the point of impact.

In the special case of an impact at the center of the rod, the rod comes to a halt, obviously.

Hmm. Sounded easy at first, but I must ask: When you say I need moment of inertia around point of impact...does that mean the standard rod equation i=m/12*length*length no longer applies? Or do I proceed with the standard COM equations to get angular rate?

 

[jbriggs444]

Hmm. Sounded easy at first, but I must ask: When you say I need moment of inertia around point of impact...does that mean the standard rod equation i=m/12*length*length no longer applies? Or do I proceed with the standard COM equations to get angular rate?

It is easy.

First, realize that the moment of inertia of a rod about its end is m/3 * length * length.

So treat your rod rotating about an axis somewhere in the middle as two rods, one to the left and one to the right.

i = m/3 * left * left * left/length + m/3 * right * right * right/length

Obviously, "left" is the length of the left piece and "right" is the length of the right piece. [You get two factors of "left" for the length of the left piece and one factor of "left" to reflect the fraction of mass on the left piece. Similar for the right piece]

Now, for extra credit, use this approach to derive the moment of inertia of a uniform thin rod rotating about its center.

Alternately, the parallel axis theorem can be used. If the moment of inertia of the rod rotating about its center of mass is m/12 * length * length then the moment of inertia about a point at offset r from the center of mass is an additional m * r * r.

For extra credit, use this approach to derive the moment of inertia of a uniform thin rod rotating about its end.

 

[hutchphd]

That doesn't make sense.

If you write down the solution as described, you will see that the r and m cancel out and

v_final=(3/4)v_initial​

One of the joys of doing physics is that many times results obtain that are wonderfully surprising. If you know the answer beforehand, why bother with the math?? It always good to double-check such results, but when bunch of competent practitioners tell you your intuition is quite incorrect you should listen.

If you would prefer to demand that it just can't be so, good luck to you, maybe you can be President of the U.S. someday

 

[paradisePhysicist]

If you write down the solution as described, you will see that the r and m cancel out and

v_final=(3/4)v_initial​

One of the joys of doing physics is that many times results obtain that are wonderfully surprising. If you know the answer beforehand, why bother with the math?? It always good to double-check such results, but when bunch of competent practitioners tell you your intuition is quite incorrect you should listen.

If you would prefer to demand that it just can't be so, good luck to you, maybe you can be President of the U.S. someday

No, I meant it doesn't make sense for dynamic points. But yes your equation is good for a contact point with zero thickness on the edge.

 

[paradisePhysicist]

It is easy.

First, realize that the moment of inertia of a rod about its end is m/3 * length * length.

So treat your rod rotating about an axis somewhere in the middle as two rods, one to the left and one to the right.

i = m/3 * left * left * left/length + m/3 * right * right * right/length

Obviously, "left" is the length of the left piece and "right" is the length of the right piece. [You get two factors of "left" for the length of the left piece and one factor of "left" to reflect the fraction of mass on the left piece. Similar for the right piece]

Now, for extra credit, use this approach to derive the moment of inertia of a uniform thin rod rotating about its center.

Brilliant. Cant believe I didnt think of this myself.

Although wouldn't the mass also have to be changed too? Like this: i = left_mass/3 * left_length*left_length + right_mass/3 * right_length*right_length? Or is that wrong?

 

[jbriggs444]

Although wouldn't the mass also have to be changed too? Like this: i = left_mass/3 * left_length*left_length + right_mass/3 * left_length*left_length? Or is that wrong?

Yes, that's what the factor of left_length/length is for on the left hand term and right_length/length on the right hand term. [If you'd looked quickly before I edited the corrections in, you might have spotted that goof]

 

[paradisePhysicist]

Yes, that's what the factor of left_length/length is for on the left hand term and right_length/length on the right hand term.

I made a typo before I could edit.

 

[paradisePhysicist]

Yep, it checks out. Because its actually m/3/2/2 so its the same as m/12.

 

[hutchphd]

My apologies for missing that you were also worried about other possible points of contact. I got exasperated!

 

[paradisePhysicist]

My apologies for missing that you were also worried about other possible points of contact. I got exasperated!

Lol no worries. I had a headache and not focused at the time, so I wasn't thinking that clearly either.

 

[hmmm27]

dangit I finally had something potentially useful to post, and the answer's been given.

Which I will anyways, because I've had an immense amount of fun playing/exercising with the stick-figure physics presented - including almost successfully doing some integrals on the energy that needs to dumped to stop one side of the rod, cold :

I'd suggest that an unobtanium collision be given handwavium properties: either...

- reflection (which is the one which requires the "block" be moved), or

- emission - where all the KE involved in the collision is spontaneously emitted as photons - which also has the benefit of being instantaneous (within the constraints of spacetime), which is the one the OP was looking for.