Charles Link's latest activity

The rotation rate will increase, but in any case, I think you successfully located what was causing the big dilemma, and I found it...

@haruspex Please see my post 51. Together I think we successfully solved it=at least the first part. What it does after the initial...

and a follow-on: I'm taking a very close look at Becker's discussion and derivation of the angular momentum law, and this one is...

This one seems to be open to interpretation if the point on the body that sits there accelerates, and it does in this case. I would...

But the rod moves at the point of attachment. For the case of ## \theta=0 ## degrees, we are also in agreement, because the string...

Looks very good. We now agree at 90 degrees, but in a way, that is expected. We agreed for ## \int T \, dt =0 ## for 90 degrees, and...

Ha! Found my error. Missed two terms… Let the post-impact angular velocity be ##\omega## anticlockwise and the speed of the left end...

and a follow-on: It just occurred to me with your angular momentum calculation, that when using the reference point the way you did...

and a follow-on: We both agree that ## \int T \, dt =0 ## for ## \theta=90 ## degrees. You have a result (post 31) that ## u=v...

Would you be happy to post your working, in a PM maybe?

@haruspex Your solution in post 31 looks to be a very good one. Thank you. :) I've looked over my solution carefully=I will need to...

No, I don’t fix that point. I wrote: I then expressed the motions of the rod and bullet in terms of these: That is, in the lab frame...

This is super. We now have something to compare to, and I think it shows some very interesting features. I'm not going to try to...

and a follow-on: again I'm not so sure about your point of reference. The center of mass is stated in Becker to be one of the...

I computed ## \int T \, dt ##, and with what I computed, the velocity of the point of attachment is moving perpendicular to the string...