Plank with block of mass on top executing vertical SHM

[andyrk]

If the block maintains contact with the plank, then they have the same acceleration at all positions.

I could just leave it at that, but ...

Even in the limiting case in limiting in which Aω² = g, they maintain contact at all positions, because there would be no position at which the normal force would need to be negative to keep them in contact.

But you didn't answer my question. I was asking that N = 0 (block on the verge of leaving contact) even though the acceleration of the block and plank is the same and not different. So am I correct on this part or not? And if they are not different then can you explain how?

 

[andyrk]

Why doesn't anyone reply until I post a message like this? Anybody there?

 

[SammyS]

Why doesn't anyone reply until I post a message like this? :/

Wow! 52 minutes and no answer to an ill-formed request. It's not a complete sentence nor a complete question.

What is it precisely that you are trying to ask in Post # 71?

B.T.W. I have other things I am doing in between looking at this website from time to time.

 

[andyrk]

What is it precisely that you are trying to ask in Post # 71?

The post explains it itself. N = 0 at the top, but the acceleration of the plank is not greater than the block. They are equal. And if they are not equal, i.e. the acceleration of the plank is greater than the block can you prove it?

 

[haruspex]

Anybody there?

Not me, I've given up. The problem is not complicated at all, and I'm at a loss to understand why it still baffles you. Take your objection to my considering varying A. It strikes me as so inane as to be deliberately obtuse. Byeee...

 

[SammyS]

The post explains it itself.

I think not. It was incomprehensible.

N = 0 at the top, but the acceleration of the plank is not greater than the block. They are equal.

Yes. Provided that you are referring to the case in which Aω² = g.

And if they are not equal, i.e. the acceleration of the plank is greater than the block can you prove it?

If Aω² > g, then at some point the block loses contact unless it is fastened to the plank, in which case the plank can exert downward (i.e. negative) force on the block.

 

[andyrk]

The value of g is fixed. Give the plank enough amplitude and its acceleration at the top of the cycle will exceed g. The block will have less downward acceleration than the plank and lose contact with it. The normal force cannot be negative.

The fact that you are mentioning something like acceleration of the plank being greater than the block, confuses me. The solution I am looking at doesn't say anything like that. It just simply puts N = 0 in the equation mg - N = mω²A to get the value of ω. But you say that this equation holds only when the block is in contact? And N can't go beneath 0 since that can't happen. N = 0 means contact lost or about to lose contact (but hasn't yet). So are you saying the same thing? It hasn't yet? But will? How will it lose contact if accelerations are not different? And for this you then begin your story of comparing accelerations of the block with that of the plank involving greater than (leaves contact completely) or less than equal to (stays in touch). Which I am simply not able to comprehend! But I just showed you that the accelerations are equal, so why do involve greater than or less than equal to cases, even though they are not required and meaningless?

Basically, you say that N = 0 then acceleration of block and plank is g. Now you say that acceleration of plank is increased somehow (by increasing ω).. But then we would again have to write the equation mg - N = mω²A with new ω and put N = 0 again so as to get the new value of ω at which the block leaves contact at the topmost point.

 

[SammyS]

The value of g is fixed. Give the plank enough amplitude and its acceleration at the top of the cycle will exceed g. The block will have less downward acceleration than the plank and lose contact with it. The normal force cannot be negative.

The fact that you are mentioning something like acceleration of the plank being greater than the block, confuses me. The solution I am looking at doesn't say anything like that. It just simply puts N = 0 in the equation mg - N = mω²A to get the value of ω. But you say that this equation holds only when the block is in contact? And N can't go beneath 0 since that can't happen. N = 0 means contact lost or about to lose contact (but hasn't yet). So are you saying the same thing? It hasn't yet? But will? How will it lose contact if accelerations are not different? And for this you then begin your story of comparing accelerations of the block with that of the plank involving greater than (leaves contact completely) or less than equal to (stays in touch). Which I am simply not able to comprehend! But I just showed you that the accelerations are equal, so why do involve greater than or less than equal to cases, even though they are not required and meaningless?

Basically, you say that N = 0 then acceleration of block and plank is g. Now you say that acceleration of plank is increased somehow (by increasing ω).. But then we would again have to write the equation mg - N = mω²A with new ω and put N = 0 again so as to get the new value of ω at which the block leaves contact at the topmost point.

You quote haruspex but I believe he recently said he no longer participates in this thread.

If you would number your questions, it would be a bit easier (still pretty difficult) to respond directly to each one, or to a group of them. I count at least six questions here.

Also, you mention "The solution I am looking ..." . What solution is that, and what precisely is the problem it is the solution to. - Please be complete.

 

[andyrk]

I got my answer.