Recent content by Charles Link

Item of interest is that I was logged off of PF earlier the other day, so that when I viewed this thread, I got an AI summary of the discussion. The AI seemed to do a better job of summarizing the discussion and pointing out a couple of the key points than I could have. From all appearances...

It works for the case where there are no torques, but the expression for the torques gets much more complicated with any accelerations at the reference point. That's where @haruspex kept his expression simple with conservation of angular momentum, (by keeping his reference point fixed and...

One comment to the above is by attaching the observation point to the rod, it is now in an accelerated frame. If you keep the reference point in a stationary frame, you then have correction terms to ## J_B ## that make it the same as ## J_{cm}=I_{cm} \omega ##. In any case, yes, you do get...

It may be worth mentioning that as odd as angular momentum can be there is the case where we get the same results from all of the various reference points in a stationary frame when the body is rotating freely about the center of mass with no linear translational velocity. In that case angular...

To add to post 64, it may be worth mentioning the center of percussion. If you have a door swing on a hinge or are swinging a baseball bat, there will be a reference point on the door or bat called the center of percussion about which the angular momentum is zero. Thereby if an object...

It might be worth taking an extra look at what we have observed in this problem from working it from a couple of different reference points. The angular momentum appears to be kind of a strange bird. It is different when observed from different reference points in the same inertial frame of...

They solved for ## u ## using ## \cos{\theta}=1/3 ##. (Their result does agree with what we both got). I anticipate that's why there is no ## \theta ## dependence. They apparently plugged in that value whenever ## \theta ## occurred. Meanwhile there are no torques about P. (The angular...

an update to my post 46 that I just included an "Edit": I recomputed ## u ##, and I now agree with @haruspex post 45 result for ## u=2v/(32-27 \sin^2{\theta} ) ##. Looks like we are now in complete agreement on our results, using both the point of attachment as the reference and the center...

In post 31 @haruspex seems to offer a solution to what occurs after the impulse response. The discussion though focused mostly on the impulse response and @haruspex discovered a very important oversight in his calculations which showed how the impulse response of the string affected his ##...

@kuruman Somewhat good, but you forgot to include ## v_{cm} ## to the velocity of the left end of the rod for the first part. Once you make this correction, I think your numbers will agree with what @haruspex and I previously computed for the zero attachment case. See posts 43 and 45. We get...

The rotation rate will increase, but in any case, I think you successfully located what was causing the big dilemma, and I found it also somewhat independently afterward when I studied Becker's derivation and looked carefully at his ## r_i \times m_i \dot{r}_i ## terms. See my post 51.(I...

@haruspex Please see my post 51. Together I think we successfully solved it=at least the first part. What it does after the initial impulsive response, I still have yet to figure out, but perhaps your solution in post 31 is how the second part works.

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 looking to be a very subtle one. I think I may have located the source of the difficulty, but I need to look it over carefully. We have angular momentum ## J=\sum...

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 have worked it how you did, had I worked it using the reference point you did. Consider balancing a yardstick on your finger, and accelerating it upward. It...

But the rod moves at the point of attachment. For the case of ## \theta=0 ## degrees, we are also in agreement, because the string keeps the point of attachment from moving. When ## \theta ## is some arbitrary angle, our results differ, and I believe the problem is the torque at the...