Recent content by k4ff3

Doc-Al: Sorry, I did not know that. However, it seems like the you overlook the fact that the rod is inelastic. The same thing goes for the answers in the thread you linked to. And even if the rod was elastic, I do not buy your answer that the information/displacement will propagate as a...

You have a straight, light, hard, inelastic rod at hand. It's pretty long, long enough so that you can hold it out of your window and into your friends window at the other side of the street. The way you and your friend communicate is through morse code. You move the rod a distance x so that...

Corrected. Thanks. Still stuck though. Despite the article. I know how to use the chain rule when I have variables. But it's when I'm dealing purely with operators I struggle. I.e \frac{ \partial}{\partial u} How can you apply a chain rule on this guy?

Nice to be back here at PF and to physics after a year off in the software industry. Now it's time to catch up again :) I feel like I never really get the grasp of manipulating operators. In QM there's a lot of trixing and mixing going on, and I really would like to learn to do the magic...

The absolute value of the velocity is the same, but in the opposite direction.

When the masses are comparable, the axis of rotation exerts a force on the MGR under the collision, so this is not similar to 1-D linear elastic collision, as you state. But when the mass of the MGR is so big that the mass of the "person" bouncing off the MGR again is negligible, then the axis...

Doc Al's explanation of the cases is correct. Anyway, since I'm so good at asking questions, what is "a highly contrived scenario"? I've tried to look it up, but I can't make sense out of it in my native language. An online dictionary says: "Contrived = Obviously planned or calculated; not...

It's Okay - you can keep it ;) If this is a typo, and that you really mean PE => KE = (...) where arrow means "is being transformed to", I agree. Because PE = KE does not follow from the energy conservation law. Yes, I do :) Yes, I did :) That's true! No energy is therefore lost to the...

I have corrected the rotational PE typo, I meant of course rotational KE. What does it mean that the point of application of the force do not move? Why does it move when you apply the same force on the COM? By leaking I meant that the body was constantly losing PE, i.e PE was transformed...

Thank you for being so patient. In the rolling case (r): How can the gravitational PE be transformed into rotational KE? In the sliding case (s): How can the gravitational PE "go into whatever is applying the force f"? How does nature know that she should transform exactly the same amount of...

I must be stupid.. I really don't get it. Sorry

.. so the PE=mgh does not apply for rotating bodies, because PE is being transformed into KE (so that PE<mgh)??

These two figures illustrate the rolling case and the sliding case respectively: Rollin' --- Slidin' [PLAIN]http://img213.imageshack.us/img213/8353/frce.png Let me break down my questions: Which one wins the race? If it's a dead race, as some here claim, then what about the KE...

So we have concluded that the rolling cylinder and the sliding cylinder will have the same translational motion. If they where to race down a slope, it would be a dead one. What about kinetic energy? It arises from both translation and rotation, meaning that the rolling cylinder will have more...

I really don't think that's called math. Anyway, consider this: You have a rod laying down on a frictionless surface. If you hit the rod with a force near the edges two things will happen: a) it will get translational motion in the direction of the force, b) it will get rotational motion due...