Tension in the string holding a falling chain

[Nathanael]

Homework Statement

A uniform chain of length L and [constant] mass per unit length λ is suspended at one A end by an inextensible light string. The other end of the chain B is held at rest at level of end A of the chain. [See image.] Now if the end B of the chain is released under gravity, then find the tension in the string when the end B has fallen a distance y

Homework Equations

F=dp/dt

The Attempt at a Solution

I was thinking the tension would be equal to the weight of the chain $gλ(\frac{L}{2}+y)$ plus the change in momentum per time of the chain that is transitioning from the right to left side (in the image).

The momentum per time would be equal to the speed times the mass per time, and the mass per time would be the speed times the linear density, so I get:

$T=gλ(\frac{L}{2}+y)+λv^2$

v can be written in terms of y as $v=\sqrt{2gy}$ so I get:

$T=gλ(\frac{L}{2}+3y)$

But this is apparently incorrect

 

[mfb]

If the right end went down by y, how long is the left side?

 

[Nathanael]

If the right end went down by y, how long is the left side?

L/2 + y

 

[mfb]

If the right side is fully down (y=L) then the length of the left side is L/2+L > L? I see a problem with that assumption.

 

[Nathanael]

Oh, wow... hahaha

The length is (L+y)/2 thanks

 

[Nathanael]

$T=\frac{gλ}{2}(L+5y)$

But it still says this is the wrong answer

 

[mfb]

You have to change it at two points.

 

[Nathanael]

Yes I just realized this The amount of mass transitioning per second would not be λv, it would be λv/2

Thanks mfb :) and sorry for being careless

 

[Vibhor]

The speed of the bottommost tip of the left string is $\sqrt{2g(\frac{y}{2})}$ .

But what is the speed of the bottommost tip of the right string ? Is it $\sqrt{2gy}$ ?

Thanks

 

[mfb]

No, they should differ by a factor 2.

This thread is 2 years old.

 

[Vibhor]

This thread is 2 years old.

I know . I hope this should not be a problem .

I came across the same question and somehow worked and got the correct answer . But thinking more about the setup , wasn't sure about some things .

No, they should differ by a factor 2.

You mean , the speed of the bottommost tip of the right string is $2\sqrt{2g(\frac{y}{2})}$ ? Could you explain why they should differ by a factor 2.

Shouldn't the speed of the topmost point on right string and that of the bottommost point on right string be same ? Isn't speed of the topmost point $\sqrt{2gy}$ ?

 

[mfb]

The length of the right side gets shorter, so if we track "the position of the currently lowest point" (identical for left and right side) and "the position of the upper end of the right side", their speeds differ (by the factor of 2 mentioned before).

 

[Vibhor]

The thing I am not understanding is that , despite the right string being in free fall , how can the two ends move with different speeds ?

 

[mfb]

The right string part gets shorter. This is only possible if its ends have different speed.

 

[Vibhor]

I agree , but isn't right part of chain in free fall which means every part of the chain has same acceleration 'g' ?

@TSny , @haruspex

 

[TSny]

but isn't right part of chain in free fall which means every part of the chain has same acceleration 'g' ?

Yes.

It's important to distinguish between the downward velocity of a point of the right side of the chain and the downward velocity of the lowest point of the string. Specifically, consider a red dot painted on one of the links on the right side. Also, consider a blue horizontal line that moves down so that it is always level with the lowest point of the chain. Note that the red dot will catch up to the blue line.

 

[Vibhor]

Thank you TSny .

Could you please reply to post#13 and post#15 .

 

[Vibhor]

If the two end points are moving with different speeds then clearly they cannot be having same acceleration , but then correct answer was obtained by treating the lowermost point to be in free fall .

I am still not clear how the two ends are moving with different speeds despite having same acceleration 'g' .

Please explain this inconsistency .

 

[TSny]

I think you are right that all points on the right portion of the chain have the same free-fall acceleration, g. Once a link on the right side catches up to the blue line, it is abruptly brought to rest.

 

[haruspex]

how the two ends are moving with different speeds despite having same acceleration 'g' .

It depends what you mean by the speed of the bottom of the right hand section of chain.
An individual link approaching the bottom moves at the same speed as the top of the section, gt, or √(2gy). The locus of "the bottom of the chain" descends at half that speed. Consequently the transference from right to left is also at the half speed.
It's a bit like rolling contact of a wheel. The locus of the point of contact moves with the same velocity as the centre of the wheel, but the part of the wheel instantaneously at the point of contact is stationary.

 

[Vibhor]

I think you are right that all points on the right portion of the chain have the same free-fall acceleration, g. Once a link on the right side catches up to the blue line, it is abruptly brought to rest.

OK . But how would you explain that speed of the differential element being added to left string is half the speed of the top element of the right string considering the fact that bottom of right string is brought to rest ?

 

[TSny]

I would say that that the speed of the differential element (red dot) just before it is brought to rest is equal to the speed of the top element of the right segment at that instant. But you also need to consider how much mass is brought to rest per unit time if you are trying to get the force.

 

[Vibhor]

I would say that that the speed of the differential element (red dot) just before it is brought to rest is equal to the speed of the top element of the right segment at that instant. But you also need to consider how much mass is brought to rest per unit time if you are trying to get the force.

I think the bottom tip of right string does not come to rest .Instead it's speed is reduced to half that of the top of right string 'v'.OTOH , the speed of the bottom of the left string is increased from 0 to 'v/2' .

 

[TSny]

The links in the left segment are at rest. When a link in the right segment becomes part of the left segment, it comes to rest.

 

[Vibhor]

Consequently the transference from right to left is also at the half speed.

OK . But then how is the left string in free fall at the same ?

I am not able to understand the Free fall acceleration of left part of string .

 

[Vibhor]

The links in the left segment are at rest. When a link in the right segment becomes part of the left segment, it comes to rest.

But the speed of the differential mass being added to left string is √[2g(y/2)] ??

 

[haruspex]

But then how is the left string in free fall at the same ?

It isn't.
Suppose at some time the top of the right section is descending at speed v. In time dt, the top of the right section descends vdt. The length of the right section reduces by vdt/2. The length of the left section grows by vdt/2. A length vdt/2 is transferred from the right to the left; in doing so it goes from descending at speed v to being stationary. Its change in momentum is (λvdt/2)v=λv²dt/2.

 

[TSny]

But the speed of the differential mass being added to left string is √[2g(y/2)] ??

No, the speed of the red dot just before it comes to rest is the same as the speed of the top end of the right segment. This speed is the speed obtained by free falling a distance y.

 

[Vibhor]

It isn't.
Suppose at some time the top of the right section is descending at speed v. In time dt, the top of the right section descends vdt. The length of the right section reduces by vdt/2. The length of the left section grows by vdt/2. A length vdt/2 is transferred from the right to the left; in doing so it goes from descending at speed v to being stationary. Its change in momentum is (λvdt/2)v=λv²dt/2.

So that is how thrust force on left part is coming out to be λgy ?

But why do we get the same result assuming end point of left string to be in free fall moving a distance y/2 ?

 

[Vibhor]

No, the speed of the red dot just before it comes to rest is the same as the speed of the top end of the right segment. This speed is the speed obtained by free falling a distance y.

OK.

Which part of the string is in free fall ?Is left part in free fall ? Are you suggesting that only the right part is in free fall ? Why do we get correct answer by assuming left part to be in free fall ?

 

[TSny]

So that is how thrust force on left part is coming out to be λgy ?

yes

 

[TSny]

Which part of the string is in free fall ?Is left part in free fall ?

No parts of the left side are in free fall. All parts of the left side are at rest. The left side is just getting longer as more links are added to the left side.

Are you suggesting that only the right part is in free fall ?

Yes

Why do we get correct answer by assuming left part to be in free fall ?

Can you show how you get the answer assuming the left side is in free fall?

 

[Vibhor]

Can you show how you get the answer assuming the left side is in free fall?

Thrust force is λv² . If we assume tip of left string to be in free fall ,it falls by a distance y/2 . Hence the speed is √(2g(y/2)). This is how I got the thrust force λgy .

 

[TSny]

Thrust force is λv² . If we assume tip of left string to be in free fall ,it falls by a distance y/2 . Hence the speed is √(2g(y/2)). This is how I got the thrust force λgy .

I'm afraid I don't follow your argument.

haruspex outlined how to get the force in #27.

The rate, dm/dt, at which mass on the right is coming to rest is the rate at which mass on the right is reaching the (moving) blue line. This is λ times the velocity of the right side relative to the blue line: dm/dt = λ(v_{red dot} - v_{blue line}) =λ(v - v/2) = λv/2, where v is the velocity of free fall of the right side: v = √(2gy). The magnitude of the force required to bring elements of the right side to rest is equal to the magnitude of the rate of change of momentum of these elements:
F = (dm/dt)⋅v = (λv/2)⋅v = λv²/2 = λgy.

 

[Vibhor]

The rate, dm/dt, at which mass on the right is coming to rest is the rate at which mass on the right is reaching the (moving) blue line. This is λ times the velocity of the right side relative to the blue line: dm/dt = λ(v_{red dot} - v_{blue line}) =λ(v - v/2) = λv/2, where v is the velocity of free fall of the right side: v = √(2gy). The magnitude of the force required to bring elements of the right side to rest is equal to the magnitude of the rate of change of momentum of these elements:
F = (dm/dt)⋅v = (λv/2)⋅v = λv²/2 = λgy.

Thanks .This clears up the confusion.