I Thread attempt 2: Edge physics trying to understand

[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.