Recent content by pluspolaritons

Got it! Thanks!

The selection rules state that: Δl=±1 Δml=0,±1 I can't figure out a case where Δml=0 and Δl=±1, can anyone help by giving an example? Thanks.

Thanks, I think it's clearer now. So, F only acts on A but not B right? This is because F is in direct contact with A but not B, correct? But since B is in direct contact with A, it will feel the force of A on B.

Thanks Andrew. I know the solution to this problem, but I need to know what is maa and mba. Why is the force of A on B equals to mba instead of maa? Then what is maa?

Hi, Please take a look at this figure: Screenshot by Lightshot There is a force that is applied so that block B will not fall under the influence of gravity. I have set up the problem as: F= (ma+mb)a μR=mbg where μ is the coefficient of friction and R is the reaction force...

What I meant is shouldn't the friction force be in the opposite direction of the centripetal force (i.e pointing outward)? I thought that was the definition of friction.

technician - i didn't mean that the coin will fly away from the center, but just simply it will fly away. 256bits/technician/rcgldr - It is possible that the coin will return to circular path if the turntable is big enough. But once the static friction is overcome it takes less force to keep...

Shouldn't the friction force always be in a direction opposite to the force applied? In this case the force is the centripetal force and the friction force should be in a direction opposite to the centripetal force?

OK here's an after thought, please let me know if I'm correct: Maybe the coin does in fact go towards the center. But the point is it is no longer in a circular path, so it will eventually fly off the turntable once the static friction is overcome.

If I put a coin on a turntable at some distance away from the center and start turning the turntable eventually there will be a speed where the coin will fly off from the turntable. If we put this into calculation. We will equate the centripetal force and static friction to find out the...

What about rotational angular momentum, e.g a disc that is rotating. Does the angular momentum depends on where my point of reference is?

@ModusPwnd - I should have written 'wavefunctions' instead of 'functions'. Anyway, I found a pretty good website that helps to explain this a bit: http://quantummechanics.ucsd.edu/ph130a/130_notes/node140.html So from what I understand, In QM, if we want our observable to be real and...

What does it mean when we say that two functions are orthogonal (the physical meaning, not the mathematical one)? I tried to search for the physical meaning and from what I read, it means that the two states are mutually exclusive. Can anyone elaborate more on this? Why do we impose...

Oh right, I was thinking the pivot is at the end of the stick. Ok you're right, the pivot should be at the center of mass. Good catch, I didn't notice I was using 'L' for two different things. Thanks for pointing that out. Thanks again!

Hi, I think I understood the problem now thanks to cepheid. I was assuming that the angular momentum of the particle is zero at all point without using any reference point.