Thread attempt 2: Edge physics trying to understand

[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.