Equation of motion of a chain with moving support

[Jenny Physics]

Homework Statement

Derive the equation of motion of a chain whose supporting point moves vertically with motion Y(t).
Assume displacement of the chain in the vertical y direction is much smaller than the chain length.
Thus displacement of the chain happens only in the x direction.

Relevant Equations

u is the transverse displacement, g is gravity, L is the chain length and $\rho$ is the mass density per unit length of the chain. $\hat{x},\hat{y}$ are the unit vectors along the x and y directions.

In the figure assume the "ceiling" moves with motion $Y(t)$, i.e. it is a point support.
Applying Newton's law in the vertical direction
$T(y).\hat{y}=\rho y[g+\frac{d^{2}Y}{dt^{2}}]$
If $\theta$ is the angle between $T$ and $\hat{y}$ that means $|T|\cos\theta=\rho y[g+\frac{d^{2}Y}{dt^{2}}]$
In the horizontal direction
$[T(y)-T(y-\Delta y)].\hat{x}=\rho\Delta y\frac{\partial^{2}u}{\partial t^{2}}$
Taking the limit $\Delta y\rightarrow 0$
$\frac{\partial}{\partial y}[T(y).\hat{x}]\equiv \frac{\partial}{\partial y}[|T(y)|\sin\theta]\equiv \frac{\partial}{\partial y}[\rho y[g+\frac{d^{2}Y}{dt^{2}}]\tan\theta]=\rho y\frac{\partial^{2}u}{\partial t^{2}}$
since $\tan\theta=\frac{\partial u}{\partial y}$ the equation of motion is (assuming $\rho=constant$)
$(g+\frac{d^{2}Y}{dt^{2}})\frac{\partial}{\partial y}[ y\frac{\partial u}{\partial y}]=y\frac{\partial^{2}u}{\partial t^{2}}$

Is this right? In particular should I include the acceleration of the supporting point moving vertically ($d^{2}Y/dt^{2}$)?
Usually the motion of the supporting point is used when we set the boundary condition at the top of the string, but here I am including it in the equation itself, so I suspect it is wrong?

 

[haruspex]

In the diagram, the top appears attached to a ceiling. Is that supposed to be moving as Y(t)? If it moves down fast it will contact other parts of the chain. I feel it should be a point support.

Is this a typo: "Thus displacement of the chain happens only in the x direction"? Should it say in the y direction?

Your first equation seems wrong. Shouldn’t you be applying it just to an element of the chain (so it's the difference in the tensions) and not assume it has the same acceleration as the support?

 

[Jenny Physics]

In the diagram, the top appears attached to a ceiling. Is that supposed to be moving as Y(t)? If it moves down fast it will contact other parts of the chain. I feel it should be a point support.

Is this a typo: "Thus displacement of the chain happens only in the x direction"? Should it say in the y direction?

Your first equation seems wrong. Shouldn’t you be applying it just to an element of the chain (so it's the difference in the tensions) and not assume it has the same acceleration as the support?

Ceiling: Yes it should be a point support, the drawing is a bit misleading.
Displacement meaning as far as the transverse displacement of the chain. The chain as a whole does move vertically but its elements relative displacement is negligible vertically (thats how I understand it).
Thinking about the first equation is wrong...

 

[haruspex]

The chain as a whole does move vertically but its elements relative displacement is negligible vertically (thats how I understand it).

Yes, but more importantly, the fact that the vertical displacements are small means the horizontal displacements that result are much smaller and should be ignored. If you consider the slope of an element length a as dθ then the vertical end-to-end displacement is a sin(dθ) ≈ adθ, whereas the horizontal shrinkage is a(1-cos(dθ))≈ a(dθ)²/2.

This means you can identify a physical element of the chain as being from x to x+dx. What is its mass?
And y=y(x,t).

 

[Jenny Physics]

Yes, but more importantly, the fact that the vertical displacements are small means the horizontal displacements that result are much smaller and should be ignored. If you consider the slope of an element length a as dθ then the vertical end-to-end displacement is a sin(dθ) ≈ adθ, whereas the horizontal shrinkage is a(1-cos(dθ))≈ a(dθ)²/2.

This means you can identify a physical element of the chain as being from x to x+dx. What is its mass?
And y=y(x,t).

I don't understand the $a(1-\cos d\theta)$ shouldn't it be $a\cos d\theta\approx a[1-(d\theta)^{2}/2]$?
As far as the vertical equation, I believe you are saying because both elements at x and $x+dx$ experiment the same vertical acceleration of the moving support it should be
$T(y).\hat{y}=\rho yg$ (why would the acceleration cancel?)

 

[haruspex]

I don't understand the $a(1-\cos d\theta)$ shouldn't it be $a\cos d\theta\approx a[1-(d\theta)^{2}/2]$?
As far as the vertical equation, I believe you are saying because both elements at x and $x+dx$ experiment the same vertical acceleration of the moving support it should be
$T(y).\hat{y}=\rho yg$ (why would the acceleration cancel?)

Yes, $a\cos d\theta\approx a[1-(d\theta)^{2}/2]$. If a horizontal section length a is tilted at angle dθ then its horizontal extent reduces to $a\cos (d\theta)$, and so reduces by $a-a\cos (d\theta)=a(1-\cos (d\theta))\approx a(1-(1-(d\theta)^{2}/2)))=a(d\theta)^{2}/2$.
That makes it a second order small quantity compared with the change in vertical extent, which is why it can be ignored.

From x to x+dx is one element, length dx. You can write an expression for its mass.
It has a slope $\frac{\partial y}{\partial x}$.
The tension varies along the chain, also as a function of x and t. What is the net vertical force on the element and its resulting vertical acceleration?
You might also need to consider the torque on it, so as to relate the rate of change of slope of the element to $\frac{\partial ^2y}{\partial x^2}$. I haven't tried this myself yet, so I am not sure it is needed.

 

[Jenny Physics]

Yes, $a\cos d\theta\approx a[1-(d\theta)^{2}/2]$. If a horizontal section length a is tilted at angle dθ then its horizontal extent reduces to $a\cos (d\theta)$, and so reduces by $a-a\cos (d\theta)=a(1-\cos (d\theta))\approx a(1-(1-(d\theta)^{2}/2)))=a(d\theta)^{2}/2$.
That makes it a second order small quantity compared with the change in vertical extent, which is why it can be ignored.

Since at rest the chain elements all lie vertically,
shouldnt it be instead if a vertical section length $a$ is tilted at angle $d\theta$ with the vertical then its vertical extent reduces to $a\cos(d\theta)$ and so its vertical size reduces by $a-a\cos (d\theta)=a(1-\cos (d\theta))\approx a(1-(1-(d\theta)^{2}/2)))=a(d\theta)^{2}/2$ while its horizontal extent increases to $a\sin d\theta\approx a d\theta$. In other words it is the same logic as what you say but with the horizontal, vertical directions swapped: the vertical motion is second order compared to the horizontal motion.

The vertical equation of motion is then basically the equilibrium between the gravity, acceleration of the moving support and tension which leads to the original equation I wrote above (basically the tension has to match the gravity of the portion of the chain below point $y$ which is $\rho y g$ + plus the acceleration of the moving support)

$T(y).\hat{y}=\rho y[g+\frac{d^{2}Y}{dt^{2}}]$

 

[haruspex]

Since at rest the chain elements all lie vertically,
shouldnt it be instead if a vertical section length $a$ is tilted at angle $d\theta$ with the vertical then its vertical extent reduces to $a\cos(d\theta)$ and so its vertical size reduces by $a-a\cos (d\theta)=a(1-\cos (d\theta))\approx a(1-(1-(d\theta)^{2}/2)))=a(d\theta)^{2}/2$ while its horizontal extent increases to $a\sin d\theta\approx a d\theta$. In other words it is the same logic as what you say but with the horizontal, vertical directions swapped: the vertical motion is second order compared to the horizontal motion.

The vertical equation of motion is then basically the equilibrium between the gravity, acceleration of the moving support and tension which leads to the original equation I wrote above (basically the tension has to match the gravity of the portion of the chain below point $y$ which is $\rho y g$ + plus the acceleration of the moving support)

$T(y).\hat{y}=\rho y[g+\frac{d^{2}Y}{dt^{2}}]$

Ah, I see I have quite misinterpreted the set up. Partly I was misled by the diagram.
But I'm not sure what the diagram should look like. Since the motion is driven by displacement in the y direction, it is nonsense to say the displacement is only in the x direction. Whether the lateral displacement exceeds the vertical displacement depends on the magnitude of $\ddot Y$. If it is always less than g then there will be no lateral movement.
Perhaps it means we can ignore the local variation in vertical displacement, i.e. treat it as almost Y everywhere.
Which way it will displace laterally at any point is a matter of chance. In practice, it seems likely to produce chaotic motion.

Maybe I am still missing something, but it seems like an awfully ill-defined question to me. @PeroK, @BvU, @kuruman, @TSny, @Steve4Physics: can any of you suggest how we should read this?

 

[wrobel]

I would employ the Hamilton least action principle in the moving frame with potential of inertial forces

 

[haruspex]

I would employ the Hamilton least action principle in the moving frame with potential of inertial forces

But how will that cope with my observation that there will be no lateral oscillation unless the applied acceleration exceeds a threshold?

 

[wrobel]

By the Hamilton principle I have got the following
$$u_{tt}=g(t)(xu_x)_x$$
it is accomplished under assumption that $|u|,|u_x|$ are small and the vertical velocities are small relative the moving frame;
y=u(t,x) is the shape of the chain relative the moving frame

UPD: I directed y horizontally ; x -- vertically down and the origin is at the pending point

 

[Jenny Physics]

By the Hamilton principle I have got the following
$$u_{tt}=g(t)(xu_x)_x$$
it is accomplished under assumption that $|u|,|u_x|$ are small and the vertical velocities are small relative the moving frame;
y=u(t,x) is the shape of the chain relative the moving frame

UPD: I directed y horizontally ; x -- vertically down and the origin is at the pending point

So $g(t)=g-\ddot{Y}$? (thinking of non inertial force). Or is it $g(t)=g+\ddot{Y}$?

 

[Jenny Physics]

Ah, I see I have quite misinterpreted the set up. Partly I was misled by the diagram.
But I'm not sure what the diagram should look like. Since the motion is driven by displacement in the y direction, it is nonsense to say the displacement is only in the x direction. Whether the lateral displacement exceeds the vertical displacement depends on the magnitude of $\ddot Y$. If it is always less than g then there will be no lateral movement.
Perhaps it means we can ignore the local variation in vertical displacement, i.e. treat it as almost Y everywhere.
Which way it will displace laterally at any point is a matter of chance. In practice, it seems likely to produce chaotic motion.

Maybe I am still missing something, but it seems like an awfully ill-defined question to me. @PeroK, @BvU, @kuruman, @TSny, @Steve4Physics: can any of you suggest how we should read this?

Yes it is ill-defined. I believe the intention of small vertical displacements means $Y(t)$ is small as well (as in a small oscillation of the supporting point say).

 

[haruspex]

By the Hamilton principle I have got the following
$$u_{tt}=g(t)(xu_x)_x$$
it is accomplished under assumption that $|u|,|u_x|$ are small and the vertical velocities are small relative the moving frame;
y=u(t,x) is the shape of the chain relative the moving frame

UPD: I directed y horizontally ; x -- vertically down and the origin is at the pending point

Does that handle the applied acceleration (positive up) being always greater than -g? I.e. does it produce u=0 as the only solution?

 

[Jenny Physics]

By the Hamilton principle I have got the following
$$u_{tt}=g(t)(xu_x)_x$$UPD: I directed y horizontally ; x -- vertically down and the origin is at the pending point

Typically when we have a wave equation like $$u_{tt}=[g-\ddot{Y}](xu_x)_x$$ we solve it by imposing a boundary condition at the support that would be in this case $u(L,t)=Y(t)$. So $Y(t)$ appears in the equation (as the double time derivative) and in the boundary condition?

 

[wrobel]

In my formulas the pending point is $x=y=0$ and we solve the initial value problem $u\mid_{t=0}=\hat u(x),\quad u_t\mid_{t=0}= v(x),\quad u(t,0)=0$.
To get rid of dependence on t in g(t) one should change the time $t=\psi(\tau)$

Does that handle the applied acceleration (positive up) being always greater than -g? I.e. does it produce u=0 as the only solution?

Yes it handles the applied acceleration.
The equation is obtained in the assumption that the chain does not lose tension. The solution of IVP is unique and depends on initial conditions
For correctness of IVP it must be $g(t)>0$

 

[haruspex]

The equation is obtained in the assumption that the chain does not lose tension.

If all of the chain is always under tension, how is it ever anything other than straight and vertical?

 

[wrobel]

If all of the chain is always under tension, how is it ever anything other than straight and vertical?

If the initial conditions are not identical zero the chain can oscillate little bit about its vertical equilibrium.
The key word here is " little bit" The pendulum can also oscillate keeping a tension of the string. I do not see a problem here.

 

[wrobel]

I added $u(t,0)=0$ to post #16

 

[haruspex]

If the initial conditions are not identical zero the chain can oscillate little bit about its vertical equilibrium.
The key word here is " little bit" The pendulum can also oscillate keeping a tension of the string. I do not see a problem here.

Consider the bottommost link. If it is never in free fall, why would it move sideways? Now move up one link: same argument.
To get sideways movement, we need a segment accelerating faster downwards than the link below it. It's a chain, not an elastic string.

 

[wrobel]

Consider the bottommost link. If it is never in free fall, why would it move sideways? Now move up one link: same argument.

It is not an argument. It is some strange speculation. I am sorry but I rather believe to the Hamilton principle.

 

[haruspex]

It is not an argument. It is some strange speculation. I am sorry but I rather believe to the Hamilton principle.

Since I do not know how you obtained your equation in post #11 I am unable to challenge it.
But I think I now see why we differ on this. I have assumed hanging vertically as the initial condition, where you are assuming some small perturbation.
Fair enough.

Reviewing this, my difficulty with the question as given was that it did not specify to assume i) some small initial deviation from straight, ii) that the tension is never zero.