Plank with block of mass on top executing vertical SHM

In summary, the plank and block system is undergoing vertical SHM with the equation y = sinωt + √3cosωt. The maximum chance for the mass to break off occurs at the extreme position, and the normal reaction between the block and plank is 0 when this happens. The forces acting on the block are represented by the equation mg - N = mω2A, and to become detached, the block only needs to accelerate downwards more slowly than the plank. The plank follows SHM, so its acceleration is known, and the maximum downward acceleration of the block can be determined by considering when the forces acting on it will fall short.
  • #1
andyrk
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5

Homework Statement


A plank on which a mass of mass m is put, is executing vertical SHM according to the equation y = sinωt + √3cosωt. At what time does the mass break of the plank and what is the value for the mass to break off ?

Homework Equations


The modified equation of motion looks like: y = 2sin(ωt + π/3)
Maximum chance of break off is at extreme position. So mg - N = mω2A

The Attempt at a Solution

.[/B]
My doubt is that why does the maximum chance for the mass to break off occur at an extreme position of the plank and that too at the positive extreme (A) and not the negative extreme (-A)? Can it be explained quantitatively rather than qualitatively?
 
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  • #2
What has to happen for the block to leave the plank? What forces are involved?

I did not check the modified equation of motion, so you should do that if you have not already.

What is the N in mg - N = mω2A ?

Who says the block leaves at the maximum position?
 
  • #3
DEvens said:
What has to happen for the block to leave the plank? What forces are involved?

I did not check the modified equation of motion, so you should do that if you have not already.

What is the N in mg - N = mω2A ?

Who says the block leaves at the maximum position?
The question asks the time when the block breaks off and the minimum ω needed to do so.
N is the normal reaction acting on the block because of the contact between block and plank.
 
  • #4
andyrk said:
why does the maximum chance for the mass to break off occur at an extreme position of the plank
Put that information aside for the moment. At what point in the cycle do you think there's the greatest chance of the mass leaving the plank?
 
  • #5
At the topmost point of course. That's what my logic and intuition says. But I am not able to find a suitable quantitative explanation for this.
 
  • #6
andyrk said:
At the topmost point of course. That's what my logic and intuition says. But I am not able to find a suitable quantitative explanation for this.
Think about forces. What does "stays on the plank" mean in terms of forces?
 
  • #7
Well, the block stays on the plank but its not glued to it or anything like that. So, it can break off if such a situation occurs, can't it?
 
  • #8
andyrk said:
Well, the block stays on the plank but its not glued to it or anything like that. So, it can break off if such a situation occurs, can't it?
Exactly, it's not glued on. What are the forces on it? What will be the relationship between those forces when it separates from the plank?
 
  • #9
When it breaks off then the normal reaction between the block and the plank is 0. The forces acting on the block are represented by the following equation
mg - N = mω2A.

But my question is that why is the net acceleration of the block mω2A? Isn't that supposed to be towards the mean position and if it is then how would the block ever break off if its acceleration is towards the mean position, i.e vertically down at the topmost point and vertically upwards at the bottom-most point?
 
  • #10
andyrk said:
When it breaks off then the normal reaction between the block and the plank is 0. The forces acting on the block are represented by the following equation
mg - N = mω2A.
In general, it is a function of time.
But my question is that why is the net acceleration of the block mω2A? Isn't that supposed to be towards the mean position and if it is then how would the block ever break off if its acceleration is towards the mean position, i.e vertically down at the topmost point and vertically upwards at the bottom-most point?
Because to become detached, the block only needs to accelerate downwards more slowly than the plank.
 
  • #11
haruspex said:
Because to become detached, the block only needs to accelerate downwards more slowly than the plank.
But both the plank and the block have an acceleration of ω2A downwards at the topmost position. Then how can one be slower than the other?
 
  • #12
andyrk said:
But both the plank and the block have an acceleration of ω2A downwards at the topmost position. Then how can one be slower than the other?
As long as the block has that acceleration downwards it will stay with the plank. But how is it guaranteed to have that acceleration down? Its acceleration is a consequence of the forces on it, not of a wish to stay in SHM.
 
  • #13
So
haruspex said:
As long as the block has that acceleration downwards it will stay with the plank. But how is it guaranteed to have that acceleration down? Its acceleration is a consequence of the forces on it, not of a wish to stay in SHM.
So the forces acting on the block would be its own weight and the normal reaction then. But how does that determine whether the acceleration of the block would be lesser than the plank or not?
 
  • #14
andyrk said:
So

So the forces acting on the block would be its own weight and the normal reaction then. But how does that determine whether the acceleration of the block would be lesser than the plank or not?
The plank will follow SHM, so you know the acceleration of that. When will the forces acting on the block fall short of what is required? Consider what the maximum downward acceleration of the block can be.
 
  • #15
The plank follows SHM, and so at all instants when the plank is moving up, it has acceleration towards the mean position and since the block is sitting on top of it, it too has an acceleration towards the mean position even though it is going up. This case is a bit confusing. Because I compare it with the case of an elevator which is accelerating upwards and a man is standing inside it on a weighing scale. As the elevator accelerates upwards, the weighing scale shows the weight of the man more than what it actually is. This is because the normal reaction faced by the man is more than his weight downward (and hence contributing to an acceleration equal to that of the elevator). But this happens when the elevator (in this case a plank) is accelerating upwards. But the plank is accelerating downwards even though it is going up and so the normal reaction experienced by the block is less than its weight downwards and so it has no reason to break off even at the top most point.

What I am trying to come at is that even though the whole system (plank + block) are going up, their net acceleration is down. But I am not able to figure out the case when the accelerations of the two are both downwards but unequal.
 
  • #16
andyrk said:
The plank follows SHM, and so at all instants when the plank is moving up, it has acceleration towards the mean position and since the block is sitting on top of it, it too has an acceleration towards the mean position even though it is going up. This case is a bit confusing. Because I compare it with the case of an elevator which is accelerating upwards and a man is standing inside it on a weighing scale. As the elevator accelerates upwards, the weighing scale shows the weight of the man more than what it actually is. This is because the normal reaction faced by the man is more than his weight downward (and hence contributing to an acceleration equal to that of the elevator). But this happens when the elevator (in this case a plank) is accelerating upwards. But the plank is accelerating downwards even though it is going up and so the normal reaction experienced by the block is less than its weight downwards and so it has no reason to break off even at the top most point.

What I am trying to come at is that even though the whole system (plank + block) are going up, their net acceleration is down. But I am not able to figure out the case when the accelerations of the two are both downwards but unequal.
Answer this: what is the maximum downward acceleration the block can ever have (given that it is not glued on).
 
  • #17
haruspex said:
Answer this: what is the maximum downward acceleration the block can ever have (given that it is not glued on).
2A + mg - N?
 
  • #18
andyrk said:
2A + mg - N?
No. You're overthinking it. There are only two forces acting on the block. The downward force is constant. The upward force has a minimum.
 
  • #19
mg - N?
 
  • #20
andyrk said:
mg - N?
Ok, but the maximum value of that is?
 
  • #21
When the mass breaks off, i.e N = 0. So mg - N = mg. That's the maximum value.
 
  • #22
andyrk said:
When the mass breaks off, i.e N = 0. So mg - N = mg. That's the maximum value.
Right. So the maximum downward acceleration is ?
 
  • #23
haruspex said:
Right. So the maximum downward acceleration is ?
I just told you, mg?
 
  • #24
andyrk said:
I just told you, mg?
That's a force.
 
  • #25
Oh yeah. So it would be g.
 
  • #26
andyrk said:
Oh yeah. So it would be g.
Right. And the maximum downward acceleration that is required (at a certain point in the cycle) to stay in contact with the plank is?
 
  • #27
haruspex said:
Right. And the maximum downward acceleration that is required (at a certain point in the cycle) to stay in contact with the plank is?
You mean the minimum, right? It is ω2A.
 
  • #28
andyrk said:
You mean the minimum, right? It is ω2A.
That's what I was after, but is the maximum that is ever required. So if the required downward acceleration to stay in contact exceeds the maximum downward acceleration available, what equation do you have?
 
  • #29
haruspex said:
That's what I was after, but is the maximum that is ever required. So if the required downward acceleration to stay in contact exceeds the maximum downward acceleration available, what equation do you have?
You mean the downward acceleration of the block exceed the maximum downward acceleration of the plank? Does that mean the block presses even harder to the plank so any possibilities of breaking off the plank are ruled out?
 
  • #30
andyrk said:
You mean the downward acceleration of the block exceed the maximum downward acceleration of the plank? Does that mean the block presses even harder to the plank so any possibilities of breaking off the plank are ruled out?
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?
 
  • #31
Right! So if g < ω2A, then the block leaves contact. But this is a bit confusing. We came to the conclusion that the downward acceleration available to the block is g after we put N = 0 in the equation: mg - N = mω2A. That means, we have already assumed that the block leaves contact before even proving it. And even if we do prove it before hand (I don't know how), then the equation simply reduces down to mg = mω2A and so: g = ω2A. So how do we know if the block leaves contact or not?

And btw, mg - N = mω2A equation holds for the block at the topmost extreme point because the whole system is having an acceleration of ω2A towards the mean position.
 
  • #32
andyrk said:
Right! So if g < ω2A, then the block leaves contact. But this is a bit confusing. We came to the conclusion that the downward acceleration available to the block is g after we put N = 0 in the equation: mg - N = mω2A. That means, we have already assumed that the block leaves contact before even proving it. And even if we do prove it before hand (I don't know how), then the equation simply reduces down to mg = mω2A and so: g = ω2A. So how do we know if the block leaves contact or not?

And btw, mg - N = mω2A equation holds for the block at the topmost extreme point because the whole system is having an acceleration of ω2A towards the mean position.
We showed that the maximum downward acceleration available to the block is g. We showed that at a certain point in the cycle the downward acceleration needed to be ω2A if the block is to remain in contact. Therefore, for the block to remain in contact through the cycle g >= ω2A. Where's the confusion?
 
  • #33
We showed that g is the maximum acceleration of the block by putting N = 0 in mg - N right? But then how do we know that N = 0 in the first place?
 
  • #34
andyrk said:
We showed that g is the maximum acceleration of the block by putting N = 0 in mg - N right? But then how do we know that N = 0 in the first place?
No, we showed it was the maximum downward acceleration by recognising that N >= 0.
 
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  • #35
Finally. I get it! Thanks a lot for the help :D
 

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