Beads on a thread - What stops the acceleration?

[Alettix]

Hello!
I would like to ask for your help with understanding a few things connected to the following problem:

1. Homework Statement
There is an (infinitly) long thread, on which small beads can move without friction. The beads with mass m are lined up on the thread with a constant distance d between them. Whe start to push one bead with the constant force F and continue to do so all the time. The velocity will eventually reach a final, constant value. What is this value of the collisions between the beads are completely inelastic?

Homework Equations

F * Δt = Δp
v * Δt = Δs

3. The solution
When the first bead is accelerated, it will collide into the next one. The beads stuck together and collide in the third one, then the fourth one and so on. If the final velocity is v, the number of new beads attached to the moving "chain" in a time Δt is: n = s/d = v*Δt/d
Therefore the change in mass of the moving "chain" is: Δm = m*n = m*v*Δt/d
Because the velocity is now constant, the change in momentum is only due to the change in mass:
F*Δt = Δp = v*Δm = v²*m*Δt/d
Thus: v = (F*d/m)^(1/2)

Now, there is nothing wrong with the solution. The thing I don't understant is exactly why a constant final velocity is reached. I know that when the total mass approaches infinity, the acceleration will become zero (a=F/m), but is there anything else which causes the acceleration to stop? I feel like the condiction of infinit number of beads is not directly included in the solution. I do therefore wonder: Is there any outer force which counteracts the pushing force F?

I have also thught about an energy aspect. In this case, the work done by F on a distance d should equal the kinetic energy of a new bead, shouldn't it? However, this yields: v = (2F*d/m)^(1/2) , which is one factor √2 to much compared to the answer obtained in the solution above. Why?

Thank you very much!

 

[BvU]

I know that when the total mass approaches infinity, the acceleration will become zero (a=F/m),

Nope. The beads that are moving already are not accelerated further. every ##\Delta p## involves only one extra bead !

Last question: energy is lost because the collisions are inelastic.

 

[Khashishi]

What an interesting and totally unnatural problem to solve!
Remember that work is force times distance. The distance between collisions is fixed, so the average work between each collision is fixed. If you assume that the velocity does asymptote to a value (averaged over the collision period), then the average work is the average kinetic energy of a bead.

Edit: Oops, this is wrong.

 

[Alettix]

Nope. The beads that are moving already are not accelerated further. every ##\Delta p## involves only one extra bead !

How can we be sure that the movig beads do not accelerate futher?

Last question: energy is lost because the collisions are inelastic.

Of course! Thank you so much! This was a very stupid question from me.

 

[Alettix]

What an interesting and totally unnatural problem to solve!
Remember that work is force times distance. The distance between collisions is fixed, so the average work between each collision is fixed. If you assume that the velocity does asymptote to a value (averaged over the collision period), then the average work is the average kinetic energy of a bead.

Thank you so much for your help Sir! The only thing I wonder: Why can we assume that the velocity asymptot to a value?

 

[haruspex]

What an interesting and totally unnatural problem to solve!
Remember that work is force times distance. The distance between collisions is fixed, so the average work between each collision is fixed. If you assume that the velocity does asymptote to a value (averaged over the collision period), then the average work is the average kinetic energy of a bead.

As BvU pointed out, collisions are inelastic here. You cannot assume work conservation.

 

[haruspex]

Why can we assume that the velocity asymptot to a value?

You don't need to assume it. It will drop out of the equations.

 

[Alettix]

You don't need to assume it. It will drop out of the equations.

From which equation Sir? This is the thing I can't really see.

 

[haruspex]

From which equation Sir? This is the thing I can't really see.

We could treat this as a smooth process, but it might be clearer if we treat it as discrete.
Suppose we have just picked up the Nth bead and they're all moving at speed v_N. What will the speed be after picking up the next bead?

 

[J Hann]

Hello!
I would like to ask for your help with understanding a few things connected to the following problem:

1. Homework Statement
There is an (infinitly) long thread, on which small beads can move without friction. The beads with mass m are lined up on the thread with a constant distance d between them. Whe start to push one bead with the constant force F and continue to do so all the time. The velocity will eventually reach a final, constant value. What is this value of the collisions between the beads are completely inelastic?

Homework Equations

F * Δt = Δp
v * Δt = Δs

3. The solution
When the first bead is accelerated, it will collide into the next one. The beads stuck together and collide in the third one, then the fourth one and so on. If the final velocity is v, the number of new beads attached to the moving "chain" in a time Δt is: n = s/d = v*Δt/d
Therefore the change in mass of the moving "chain" is: Δm = m*n = m*v*Δt/d
Because the velocity is now constant, the change in momentum is only due to the change in mass:
F*Δt = Δp = v*Δm = v²*m*Δt/d
Thus: v = (F*d/m)^(1/2)

I also have a problem with the given solution.
Even though the collisions are inelastic momentum is still conserved.
Isn't then M V1 = (m + M) V2
then V2 = M V1 / (m + M)

Now, there is nothing wrong with the solution. The thing I don't understant is exactly why a constant final velocity is reached. I know that when the total mass approaches infinity, the acceleration will become zero (a=F/m), but is there anything else which causes the acceleration to stop? I feel like the condiction of infinit number of beads is not directly included in the solution. I do therefore wonder: Is there any outer force which counteracts the pushing force F?

I have also thught about an energy aspect. In this case, the work done by F on a distance d should equal the kinetic energy of a new bead, shouldn't it? However, this yields: v = (2F*d/m)^(1/2) , which is one factor √2 to much compared to the answer obtained in the solution above. Why?

Thank you very much!

Hello!
I would like to ask for your help with understanding a few things connected to the following problem:

1. Homework Statement
There is an (infinitly) long thread, on which small beads can move without friction. The beads with mass m are lined up on the thread with a constant distance d between them. Whe start to push one bead with the constant force F and continue to do so all the time. The velocity will eventually reach a final, constant value. What is this value of the collisions between the beads are completely inelastic?

Homework Equations

F * Δt = Δp
v * Δt = Δs

3. The solution
When the first bead is accelerated, it will collide into the next one. The beads stuck together and collide in the third one, then the fourth one and so on. If the final velocity is v, the number of new beads attached to the moving "chain" in a time Δt is: n = s/d = v*Δt/d
Therefore the change in mass of the moving "chain" is: Δm = m*n = m*v*Δt/d
Because the velocity is now constant, the change in momentum is only due to the change in mass:
F*Δt = Δp = v*Δm = v²*m*Δt/d
Thus: v = (F*d/m)^(1/2)

Now, there is nothing wrong with the solution. The thing I don't understant is exactly why a constant final velocity is reached. I know that when the total mass approaches infinity, the acceleration will become zero (a=F/m), but is there anything else which causes the acceleration to stop? I feel like the condiction of infinit number of beads is not directly included in the solution. I do therefore wonder: Is there any outer force which counteracts the pushing force F?

I have also thught about an energy aspect. In this case, the work done by F on a distance d should equal the kinetic energy of a new bead, shouldn't it? However, this yields: v = (2F*d/m)^(1/2) , which is one factor √2 to much compared to the answer obtained in the solution above. Why?

Thank you very much!

 

[Alettix]

We could treat this as a smooth process, but it might be clearer if we treat it as discrete.
Suppose we have just picked up the Nth bead and they're all moving at speed v_N. What will the speed be after picking up the next bead?

Well, if we suppose that a work W = F*d*k is done between the beads, where k is the efficiency (0<k<1), energy conservation gives:
Fdk+v_N²Nm/2 = v_N+1²(N+1)m/2
which yields:
v_N+1 = ((2Fdk/(N+1)m)+(v_N²N/m))^(1/2)

 

[haruspex]

Well, if we suppose that a work W = F*d*k is done between the beads, where k is the efficiency (0<k<1), energy conservation gives:
Fdk+v_N²Nm/2 = v_N+1²(N+1)m/2
which yields:
v_N+1 = ((2Fdk/(N+1)m)+(v_N²N/m))^(1/2)

You are told the collisions are completely inelastic. What is conserved?

 

[J Hann]

F dt = M dV + V dM
Consider a glob of putty thrown at a "fixed" wall and sticking to the wall..
Then the impulse on the wall is F dt = V dM because the wall does not move.
Then you get the term V^2 = F d / M where M = L m / d since dM / dt = m V / d.
So the accumulated beads behave as a fixed wall which is being moved by a force F.
Also, this holds at all times after t = 0, and the impulse as each bead is added continually corresponds to the force F.
So one could accumulate an unlimited mass of beads with the application of force F at speed V.

 

[Alettix]

You are told the collisions are completely inelastic. What is conserved?

Momentum is conserved! Which means that: ## V_N *Nm = V_{N+1} *(N+1)m##
so ## V_{N+1} = \frac{V_N *N}{N+1}##
What should be done next? :)

 

[Alettix]

F dt = M dV + V dM
Consider a glob of putty thrown at a "fixed" wall and sticking to the wall..
Then the impulse on the wall is F dt = V dM because the wall does not move.
Then you get the term V^2 = F d / M where M = L m / d since dM / dt = m V / d.
So the accumulated beads behave as a fixed wall which is being moved by a force F.
Also, this holds at all times after t = 0, and the impulse as each bead is added continually corresponds to the force F.
So one could accumulate an unlimited mass of beads with the application of force F at speed V.

Thank you for your reply! I like the example with the wall, but exactly how do I relate it to the bead problem? I can't see this clearly because the are not still after a collision, but go on moving.
I do also have some problems following your mathematics, would you please mind explaining it a bit more? :)
I am also slightly unsure about F dt = M dV + V dM , because my textbook says d(mv)/(dt) ≠ dm/dt + dv/dt and that this can be motivated with "symmetry" (although this is not done).

 

[J Hann]

The accumulated string of beads (wall) is moving at constant speed so the problem takes dV / dt = 0.
The beads evidently are taken to be so close together so as to simulate a constant accumulation of mass.
The applied force to the wall then just equals the impulse of the beads striking the wall so the net force on the wall is zero.
So the accumulated mass of beads moves at constant speed from t = zero to any indefinite time.
I'm not familiar with the example to which your textbook refers in which dP/dt would not equal M dV/dt + V dM/dt.

 

[Alettix]

The accumulated string of beads (wall) is moving at constant speed so the problem takes dV / dt = 0.

I know it is assumed in the problem, but how does one motivate that the speed will become constant?

The beads evidently are taken to be so close together so as to simulate a constant accumulation of mass.
The applied force to the wall then just equals the impulse of the beads striking the wall so the net force on the wall is zero.

What physical consequences are there of a constant accumulation of mass? (will mass not accumulation constantly even if the speed is not constant?) In there is an impuls on the wall, how can the force be zero?

 

[haruspex]

Momentum is conserved! Which means that: ## V_N *Nm = V_{N+1} *(N+1)m##
so ## V_{N+1} = \frac{V_N *N}{N+1}##
What should be done next? :)

No, you've forgotten about the force. The velocity will increase before it strikes the next bead.

Now, there is a much easier way, but I went this route because you asked how we know it will converge to a finite speed.
If you assume a steady state is reached, you can work in the reference frame of the moving mass. The force F applied one side is to be balanced by the steady stream of impacts from beads on the other side. You can just write down the solution.

 

[haruspex]

I'm not familiar with the example to which your textbook refers in which dP/dt would not equal M dV/dt + V dM/dt.

That equation is popular but not really valid, and many rail against it.
Logically, the dm/dt term is saying that mass is magically created or destroyed. In reality, it is coming from or going to some other system, so may bring with it or take with it some momentum. The equation can work, but only if the added mass had no initial momentum (or departing mass takes none with it).

 

[Alettix]

No, you've forgotten about the force. The velocity will increase before it strikes the next bead.

so should the equation be:
## (V_N + a* \Delta t) Nm = V_{N+1} *(N+1)m ##
? where ##a = F/m## and ##\Delta t = \sqrt 2d/a ## thus:
## (V_N + \sqrt \frac{2Fd}{m}) Nm = V_{N+1} *(N+1)m ## ?

Now, there is a much easier way, but I went this route because you asked how we know it will converge to a finite speed.
If you assume a steady state is reached, you can work in the reference frame of the moving mass. The force F applied one side is to be balanced by the steady stream of impacts from beads on the other side. You can just write down the solution.

If it is assumed that a steady state is reached (but I do still not see exactly how we can assume this), then the impuls of the new bead must be balanced by the force pushing from the other side. In the frame of reference of the moving mass, the mass is moving towards it with velocity v. Thus:
## F * \Delta t = \Delta p = mv ##
The time is the time between two collisons, thus ## v = \sqrt \frac{Fd}{m} ## , which is the correct answer.

 

[haruspex]

a=F/m

How much accumulated mass is there at this time?

 

[Alettix]

How much accumulated mass is there at this time?

The mass is N*m before the collision. So the equation becomes:
##(v_N+\sqrt \frac{2Fd}{Nm})Nm=v_{N+1}∗(N+1)m##

which simplyfies to:
##v_{N+1} - v_N = \frac{\sqrt \frac{2FdN}{m} - v_{N+1}}{N} ##

when N → ∞ , does this expression approach 0?

 

[haruspex]

##(v_N+\sqrt \frac{2Fd}{Nm})##

That's still wrong.
If an object traveling at speed u undergoes an acceleration a for a distance s, what is the new speed?

It will simplify the algebra if you substitute k for Fd/m and w_n for v_n².

 

[Alettix]

That's still wrong.
If an object traveling at speed u undergoes an acceleration a for a distance s, what is the new speed?

It will simplify the algebra if you substitute k for Fd/m and w_n for v_n².

Energy conservation yields ## v^2 = w + k ##. Because energy is conserved during the acceleration, isn't it?
Or should I use dynamics to calculate this? Like obtaining t from ##\frac{at^2}{2} + v_Nt = d## and putting it into Δv =a*t?

 

[haruspex]

Energy conservation yields ## v^2 = w + k ##. Because energy is conserved during the acceleration, isn't it?
Or should I use dynamics to calculate this? Like obtaining t from ##\frac{at^2}{2} + v_Nt = d## and putting it into Δv =a*t?

All methods shoild lead to the same answer. Most direct is to use the SUVAT equation that involves initial and final speeds, acceleration and distance. This is the same as the energy equation, but with mass canceled out.

 

[Alettix]

All methods shoild lead to the same answer. Most direct is to use the SUVAT equation that involves initial and final speeds, acceleration and distance. This is the same as the energy equation, but with mass canceled out.

Okay, using SUVAT, the speed just before the collision is:
##v_x = \sqrt (v_N^2 +2da) ##
By momentum conservation we therefore have:
## Nm *\sqrt (v_N +2da) = (N+1)m* v_{N+1}##
--> ## v_{N+1} = \frac{N}{N+1} *\sqrt (v_N^2 +2da) =\frac{N}{N+1} *\sqrt (v_N^2 +2d \frac{F}{m}) ##
Is this right? How do I proceed?
I can see that for N→∞
N/N+1 → 1, but what happens with the expression under the rootsign?

 

[haruspex]

Okay, using SUVAT, the speed just before the collision is:
##v_x = \sqrt (v_N^2 +2da) ##
By momentum conservation we therefore have:
## Nm *\sqrt (v_N +2da) = (N+1)m* v_{N+1}##
--> ## v_{N+1} = \frac{N}{N+1} *\sqrt (v_N^2 +2da) =\frac{N}{N+1} *\sqrt (v_N^2 +2d \frac{F}{m}) ##
Is this right? How do I proceed?
I can see that for N→∞
N/N+1 → 1, but what happens with the expression under the rootsign?

You've gone back to taking a to be F/m.
When you have fixed that, consider whether a change of variable would get rid of the surd.

 

[Alettix]

You've gone back to taking a to be F/m.
When you have fixed that, consider whether a change of variable would get rid of the surd.

Oh, I am so sorry! It should of course be a=F/mN, or using your k = Fd/m and v_N² =w:
##v_{N+1} = \frac{N}{N+1} * \sqrt{w + 2k/N}##
Now there might be a miss in my knowledge of english, but what is "surd"? What should I get rid of?

 

[haruspex]

Oh, I am so sorry! It should of course be a=F/mN, or using your k = Fd/m and v_N² =w:
##v_{N+1} = \frac{N}{N+1} * \sqrt{w + 2k/N}##
Now there might be a miss in my knowledge of english, but what is "surd"? What should I get rid of?

A surd is a fractional power, in this case the square root.
The w=v_N² step is good, but generalise it: w_N=v_N²

 

[Alettix]

A surd is a fractional power, in this case the square root.
The w=v_N² step is good, but generalise it: w_N=v_N²

Okay, so the expression is ##v_{N+1} = \frac{N}{N+1} * \sqrt{w_N + 2k/N}##
I could get rid of the root-sign by squaring both sides, but that would yield several variables with the power of two which probably is not desired.
The other way is to rewrite the expression under the root-sign as a square, but I can't see how.

 

[haruspex]

Okay, so the expression is ##v_{N+1} = \frac{N}{N+1} * \sqrt{w_N + 2k/N}##
I could get rid of the root-sign by squaring both sides, but that would yield several variables with the power of two which probably is not desired.
The other way is to rewrite the expression under the root-sign as a square, but I can't see how.

Squaring both sides is exactly right. Turn all the v terms into corresponding w terms.

 

[Alettix]

Squaring both sides is exactly right. Turn all the v terms into corresponding w terms.

Oh, yes! So now we have:
## w_{N+1} = (\frac{N}{N+1})^2 * (w_N + \frac{2k}{N}) ##
And when N → ∞ we got ##\frac{N}{N+1} → 1## and ## 2k/N → 0 ##
thus: ## w_{N+1} → w_N ##
is this right? :)

 

[haruspex]

Oh, yes! So now we have:
## w_{N+1} = (\frac{N}{N+1})^2 * (w_N + \frac{2k}{N}) ##
And when N → ∞ we got ##\frac{N}{N+1} → 1## and ## 2k/N → 0 ##
thus: ## w_{N+1} → w_N ##
is this right? :)

Yes, but that doesn't show w_N converges. Consider a_n+1=(1+1/n)a_n. That does not converge. (Indeed, solution is a_n=n.)
Multiply through the equation by (N+1)². This should suggest another change of variable.

 

[Alettix]

Yes, but that doesn't show w_N converges. Consider a_n+1=(1+1/n)a_n. That does not converge. (Indeed, solution is a_n=n.)

But if ## w_{N+1} = w_N##, it means that the velocity does not change when more beads are added, thus it is constant, isn't it? I also graphed your function, and for big n-s, ##a_{n+1}## always approached the constant ##a_n##, how comes this does not count as a convergance?

Multiply through the equation by (N+1)². This should suggest another change of variable.

If we say: ##m_N=w_N*N^2##, we now have:
##m_{N+1} = m_N + 2kN##
but in this case, ##m_{N+1}## does not approach ##m_N## when ##N→∞##
Have I done the wrong change of varibale?

 

[haruspex]

But if ## w_{N+1} = w_N##, it means that the velocity does not change when more beads are added, thus it is constant, isn't it?

You have not proved that equality. You only showed the ratio tends to 1. That is not the same thing.

I also graphed your function, and for big n-s, ##a_{n+1}## always approached the constant ##a_n##, how comes this does not count as a convergance?

a_n is not a constant. As I posted, the solution to my a_n recurrence relation is a_n=n. The sequence {n} does not converge to a constant.

If we say: ##m_N=w_N*N^2##, we now have:
##m_{N+1} = m_N + 2kN##

That's the right step. Can you see how to rewrite that as m_N= sum of a series? You can easily sum that series and find the exact algebraic form of v_N.