Plank with block of mass on top executing vertical SHM

[Jediknight]

so this posts a month old, I am probably good to propose some answers, a(max)=ω^2(A)=g at the minimum frequency to make block leave plank, so just algebra from there right

 

[haruspex]

so this posts a month old, I am probably good to propose some answers, a(max)=ω^2(A)=g at the minimum frequency to make block leave plank, so just algebra from there right

Yes.

 

[andyrk]

While pondering on this problem once again, I got some more doubts with it. Why is mg - N = mω²x and not: mg - N = -mω²x, since acceleration is
-ω²x and not ω²x?

 

[haruspex]

While pondering on this problem once again, I got some more doubts with it. Why is mg - N = mω²x and not: mg - N = -mω²x, since acceleration is
-ω²x and not ω²x?

It might depend on the conventions adopted for the positive direction and what 'g' means. If we take up as positive and g as a signed value (i.e. its value will be negative) then the equation is mg+N=-mω²x. Other conventions can lead to other forms. Since N will be set to zero, it may be that the equation was simply wrong in a way that didn't matter.

 

[andyrk]

It might depend on the conventions adopted for the positive direction and what 'g' means. If we take up as positive and g as a signed value (i.e. its value will be negative) then the equation is mg+N=-mω²x. Other conventions can lead to other forms. Since N will be set to zero, it may be that the equation was simply wrong in a way that didn't matter.

The convention is that up is positive and down is negative.

 

[andyrk]

No, the other way around. When the maximum downward acceleration available to the block (g) is less than the downward acceleration required to stay in contact (which can be up to $\omega^2A$) the block will lose contact with the plank, no?

How do we know that the downward acceleration of the plank would be greater than the block? We just saw that g = $\omega^2A$. And also, the downward acceleration of the block at the topmost point is $\omega^2A$. So shouldn't the two be same?

 

[andyrk]

Anybody there?

 

[haruspex]

How do we know that the downward acceleration of the plank would be greater than the block? We just saw that g = $\omega^2A$. And also, the downward acceleration of the block at the topmost point is $\omega^2A$. So shouldn't the two be same?

We don't know the plank's downward acceleration is greater than the block's. I said if it is ever greater.
g = $\omega^2A$ Is the limiting condition, i.e. when it is in imminent danger of leaving the plank at some point in the oscillation. It is not a general fact.
I don't know what you mean about "the two" being the same. What two?

Seems to me you have a basic problem with handling conditionals, the difference between a general statement and a statement that only applies when certain conditions are met. I don't think I can make it any clearer. If you still don't get it, I can only recommend that you read and reread my posts.

 

[andyrk]

I don't know what you mean about "the two" being the same. What two?

By two I mean the downward acceleration of the block and the plank at the topmost point. If they are the same, then why would the block leave contact? Now you are saying if they are same. But what I am trying to say that they have to be same, there's no if associated with it, it just has to be the same.

 

[haruspex]

By two I mean the downward acceleration of the block and the plank at the topmost point. If they are the same, then why would the block leave contact? Now you are saying if they are same. But what I am trying to say that they have to be same, there's no if associated with it, it just has to be the same.

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.

 

[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 acceleration of the plank at the top is ω²A, right? And the acceleration of the block is g, right? And we also proved that g = ω²A (by putting N = 0 in the force equation of the block). So, we just showed that the accelerations are same at the topmost point and not different. But wait. Had they been the same, then the block wouldn't have left contact and hence N wouldn't have been 0. So what am I doing wrong here?

 

[haruspex]

The acceleration of the plank at the top is ω²A, right? And the acceleration of the block is g, right? And we also proved that g = ω²A (by putting N = 0 in the force equation of the block). So, we just showed that the accelerations are same at the topmost point and not different. But wait. Had they been the same, then the block wouldn't have left contact and hence N wouldn't have been 0. So what am I doing wrong here?

We could only write N=0 in that equation on the basis that the block was almost leaving the plank but not quite. If the block leaves the plank then N will still be zero, but the equation we plugged it into is no longe valid. The equation assumes they remain in contact.

 

[andyrk]

I am unable to make any sense of what you are saying. Are you then saying that the downward acceleration of the block is less than g and that of the plank is g? If block leaves contact at the top and its acceleration is g, then does that mean that the acceleration of the plank is greater than g, i.e. greater than ω²A? But how is this possible??

 

[haruspex]

I am unable to make any sense of what you are saying. Are you then saying that the downward acceleration of the block is less than g and that of the plank is g? If block leaves contact at the top and its acceleration is g, then does that mean that the acceleration of the plank is greater than g, i.e. greater than ω²A? But how is this possible??

At the top position, the downward acceleration of the plank will always be ω²A.
The downward acceleration of the block at that position will be min{ω²A, g}. If ω²A <= g they stay in contact. If ω²A > g they will lose contact.

 

[andyrk]

The downward acceleration of the block at that position will be min{ω2A, g}

But we just found out that ω²A = g by putting N = 0 in the equation mg - N = mω²A. So isn't ω²A = g at the topmost position? Which means that min{ω2A, g} = g or ω²A since they are one and the same thing? So that means that they should not leave contact but that is not the case as it is clearly mentioned that the block leaves contact. This is what is confusing me.

 

[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.