Plank with block of mass on top executing vertical SHM

[haruspex]

But we just found out that ω²A = g by putting N = 0 in the equation mg - N = mω²A.

As I wrote, that equation is only valid by assuming they stay in contact.

 

[andyrk]

As I wrote, that equation is only valid by assuming they stay in contact.

How can they stay in contact if N = 0?

 

[haruspex]

How can they stay in contact if N = 0?

That's the boundary case. Whether you consider that to be contact or not is somewhat a matter of taste. I would have said it could be that that N=0 and yet there is no separation between them, but it really doesn't matter. In the same way, you can argue that no two real physical quantities are ever exactly equal. But they can be near enough equal that you can't measure the difference, so who cares?

 

[SammyS]

That's the boundary case. Whether you consider that to be contact or not is somewhat a matter of taste. I would have said it could be that that N=0 and yet there is no separation between them, but it really doesn't matter. In the same way, you can argue that no two real physical quantities are ever exactly equal. But they can be near enough equal that you can't measure the difference, so who cares?

... and in this limiting case, N = 0 for only an instant, when the plank is at it's very highest position. (whatever an instant is physically)

 

[andyrk]

That's the boundary case. Whether you consider that to be contact or not is somewhat a matter of taste. I would have said it could be that that N=0 and yet there is no separation between them, but it really doesn't matter. In the same way, you can argue that no two real physical quantities are ever exactly equal. But they can be near enough equal that you can't measure the difference, so who cares?

What I am feeling is that if g = ω²A and acceleration of plank > ω²A then that means ω²A of plank is different from that of the block.

 

[haruspex]

What I am feeling is that if g = ω²A and acceleration of plank > ω²A then that means ω²A of plank is different from that of the block.

The acceleration of the plank cannot exceed ω²A.

 

[andyrk]

The acceleration of the plank cannot exceed ω²A.

And if acceleration of block = ω²A, then I don't see how could they get separated? But we found out that acceleration of the block ω²A by assuming they get separated, which is just very ironical. How could two object which are accelerating at the same rate ever get separated?

 

[haruspex]

But we found out that acceleration of the block ω²A by assuming they get separated,

No, by assuming that it was right on the boundary between separating and not separating.

 

[andyrk]

No, by assuming that it was right on the boundary between separating and not separating.

So you mean to say that acceleration of the block is ω²A when it is just about to separate but hasn't yet separated? Also called as "Just about to break off"? But then the acceleration of the block is ω²x at all times where x is the distance from mean position. Why then, is the block not on the verge of breaking off at all these times?

 

[haruspex]

So you mean to say that acceleration of the block is ω²A when it is just about to separate but hasn't yet separated? Also called as "Just about to break off"?

No, I didn't say that. The minimum normal force, N_min, is at the top of the movement, agreed? When N_min>0 and the plank is at the top of its movement the block will have acceleration ω²A, and it will be less than g. Suppose we gradually increase A somehow. As ω²A gets closer to g, N_min will decrease. At the point where ω²A equals g, N_min will be zero and the block is on the verge of losing contact. The slightest further increase in A will cause the block to lose contact.

 

[andyrk]

No, I didn't say that. The minimum normal force, N_min, is at the top of the movement, agreed? When N_min>0 and the plank is at the top of its movement the block will have acceleration ω²A, and it will be less than g. Suppose we gradually increase A somehow. As ω²A gets closer to g, N_min will decrease. At the point where ω²A equals g, N_min will be zero and the block is on the verge of losing contact. The slightest further increase in A will cause the block to lose contact.

So when you say "we gradually increase A somehow" you mean increasing the amplitude of the SHM that the whole system (plank+block) is executing? So the break-off point is reached at a certain amplitude? But how can you increase A since it is fixed and if you change A you change the equation of SHM, which I don't think can be done.

 

[haruspex]

So when you say "we gradually increase A somehow" you mean increasing the amplitude of the SHM that the whole system (plank+block) is executing? So the break-off point is reached at a certain amplitude? But how can you increase A since it is fixed?

I don't care how we increase it. Maybe we just run hundreds of different experiments, each with a slightly higher A than the one before.

 

[andyrk]

I don't care how we increase it. Maybe we just run hundreds of different experiments, each with a slightly higher A than the one before.

The equation of SHM is fixed for the problem and so is the amplitude. So how will you go about proving that the block leaves contact/is about to leave contact with the plank even though their accelerations are same at the highest point? This problem has become so complicated.

 

[SammyS]

So when you say "we gradually increase A somehow" you mean increasing the amplitude of the SHM that the whole system (plank+block) is executing? So the break-off point is reached at a certain amplitude? But how can you increase A since it is fixed ...

How about rather than increasing A, we see start with small enough ω so that the block maintains contact with the plank, then similar to what haruspex has suggested, we run successive experiments each time with slightly larger ω until we find the minimum value of ω that results in the block losing contact with the plank. Certainly you can't object to increasing ω in this fashion.

 

[andyrk]

How about rather than increasing A, we see start with small enough ω so that the block maintains contact with the plank, then similar to what haruspex has suggested, we run successive experiments each time with slightly larger ω until we find the minimum value of ω that results in the block losing contact with the plank. Certainly you can't object to increasing ω in this fashion.

Rightly said. So the block would leave contact with the plank at a specific ω. But when ω²A = g, then the block is on the verge of losing contact, right? Any slight increase in ω would make the block to lose contact with the plank. But right now the acceleration of the block equals that of the plank, right? So doesn't that happen at all instants, even when the plank is not at its highest point? Or does it not because N is not 0 at all other instants except for the topmost point?

 

[andyrk]

How about rather than increasing A, we see start with small enough ω so that the block maintains contact with the plank, then similar to what haruspex has suggested, we run successive experiments each time with slightly larger ω until we find the minimum value of ω that results in the block losing contact with the plank. Certainly you can't object to increasing ω in this fashion.

So finally, N = 0 has nothing to do with different accelerations of the block and the plank. They are the same at the topmost point and all points, yet N = 0 because of the algebraic reasoning we provided earlier. Is that correct?

 

[andyrk]

Anybody there?

 

[SammyS]

Anybody there?

Many are here.

That last question has no simple answer because it rather ill-posed.

Many posts ago, when harupex was having a difficult time trying to get you to understand reasoning involved in deciding under what conditions the block would lose contact with the plank, I considered writing a post that would start from scratch, answering the Original Post, and express the acceleration of the block as a function of time and as a function of position; similarly finding the force (as a function of time and then of position) which the plank would need to exert on the block if the block were to maintain contact.

... but then it would seem that harupex would make some headway. A post or two later things would be all messed up again.

In post # 65, what do you mean by "right now" when you say, "But right now the acceleration ..." . Except for that puzzling statement, I might have answered that you had made correct statements.

I'll think about whether it's worthwhile to answer post #66.

 

[andyrk]

In post # 65, what do you mean by "right now" when you say, "But right now the acceleration ..." .

By that I mean when the plank and the block are the extreme position of the SHM, i.e. A distance from the mean position.

 

[SammyS]

By that I mean when the plank and the block are the extreme position of the SHM, i.e. A distance from the mean position.

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.

 

[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.