How Long Will It Take for the Entire Chain to Slide Off the Table?

[Satvik Pandey]

Yes.
No.
No.
No.

I can find change in mass due to the increase in the length of chain below the table.
$M(x)=\frac { M }{ L } x$
after time interval 'dt' length of chain will be increased by 'dx'
$M(x+dx)=\frac { M }{ L } (x+dx)$
Is this right?
How should I find $v(x)$?

 

[Orodruin]

How should I find $v(x)$?

How fast is the chain moving if in time $dt$ the length below the table increases by $dx$?

Edit: Do note that (as suggested by the form of the solution in post #41) the resulting differential equation is going to be quite nasty.

 

[Satvik Pandey]

How fast is the chain moving if in time $dt$ the length below the table increases by $dx$?

$dx/dt$

 

[Orodruin]

Yes, so how can you express the momentum $p$?

 

[Satvik Pandey]

Yes, so how can you express the momentum $p$?

$p=mv$ or $p=m dx/dt$
But here mass is also changing with time.So is it correct to express momentum like this.

 

[Orodruin]

Yes, the assumption is that the chain below the table is moving with the same velocity and that the part still on top is not moving. So $m$ in this case is the mass of the chain below the table. What do you get if you write this out as a function of $x$? What is the resulting differential equation?

 

[Satvik Pandey]

Yes, the assumption is that the chain below the table is moving with the same velocity and that the part still on top is not moving. So $m$ in this case is the mass of the chain below the table. What do you get if you write this out as a function of $x$? What is the resulting differential equation?

$m(x)=Mx/L$
putting this value I got
$\frac { m }{ l } x\frac { dx }{ dt }$
Is it right?

 

[voko]

Both the mass and the velocity change. What is the derivative of the momentum then?

 

[Satvik Pandey]

Both the mass and the velocity change. What is the derivative of the momentum then?

By product rule
$\frac { d(mv) }{ dt } =m\frac { dv }{ dt } +v\frac { dm }{ dt }$

 

[Orodruin]

So, inserting this and the force into Newton's second law $dp/dt = F$ gives you what?

 

[Satvik Pandey]

I got this $\frac { M }{ L } (x)\frac { dv }{ dt } +v\frac { M }{ L } \frac { dx }{ dt } =\frac { Mgx }{ L }$.
Is it right?

 

[Orodruin]

Yes, now get rid of $v$ in favour of $dx/dt$ as per post #55 and you have a differential equation for $x$. Do you know how to solve such an equation?

 

[Satvik Pandey]

Yes, now get rid of $v$ in favour of $dx/dt$ as per post #55 and you have a differential equation for $x$. Do you know how to solve such an equation?

$x\frac { { d }^{ 2 }x }{ dt^{ 2 } } +\frac { dx }{ dt } \left\{ \frac { dx }{ dt } \right\} =gx$
What is this?
I know to solve linear differential equation with integrating factor method.

 

[Orodruin]

Well, it is a non-linear ordinary differential equation which has a solution that is implicitly given in post #41. Do you have any experience in solving non-linear ODEs? Otherwise this one will be a bit tricky ...

 

[voko]

The good thing is that this one also gets solved by an integrating factor.

 

[haruspex]

By product rule
$\frac { d(mv) }{ dt } =m\frac { dv }{ dt } +v\frac { dm }{ dt }$

Yes, but I'd like to add a word of caution about using this formula in general. In effect, it is saying that a moving object increases in mass (by some magic internal process) by dm while moving at speed v. Nothing really does that. But it also happens to work if dm is mass being added that previously was stationary, i.e. is being instantaneously accelerated by v.

 

[Satvik Pandey]

The good thing is that this one also gets solved by an integrating factor.

How?

 

[Orodruin]

Yes, but I'd like to add a word of caution about using this formula in general. In effect, it is saying that a moving object increases in mass (by some magic internal process) by dm while moving at speed v. Nothing really does that. But it also happens to work if dm is mass being added that previously was stationary, i.e. is being instantaneously accelerated by v.

I would like to add that it also works for any addition of mass (even moving) as long as you take it into account on the force side as well. Heat-like forces in relativity come to mind.

How?

Since this is most likely not the intended solution for your original problem, let me give you a slightly more info than I usually would:

You have $x\ddot x + \dot x^2 = gx$. This can be rewritten
$$
\frac{d}{dt} x\dot x = gx.
$$
Multiply both sides with $x\dot x$ to obtain
$$
x\dot x \frac{d}{dt} x\dot x = gx^2\dot x.
$$
Can you integrate expressions of this type? (note that the LHS is of the form $s(ds/dt)$, where $s=x\dot x$)

 

[Satvik Pandey]

You have $x\ddot x + \dot x^2 = gx$. This can be rewritten
$$
\frac{d}{dt} x\dot x = gx.
$$
Multiply both sides with $x\dot x$ to obtain
$$
x\dot x \frac{d}{dt} x\dot x = gx^2\dot x.
$$
Can you integrate expressions of this type? (note that the LHS is of the form $s(ds/dt)$, where $s=x\dot x$)

I don't know if I am right or not.I did this -

$s\frac { ds }{ dt } =g{ x }^{ 2 }\frac { dx }{ dt }$

$sds=g{ x }^{ 2 }dx$
I can integrate LHS and RHS.But I don't know if I am right or not,till here.

 

[Orodruin]

Looks fine so far.

 

[Satvik Pandey]

Looks fine so far.

On integrating I got
$\frac { { s }^{ 2 } }{ 2 } =g\frac { { x }^{ 3 } }{ 3 }$
should I need to worry about constant or it will be included in constant at end of solution.

On substituting the value of s
${ s }^{ 2 }={ x }^{ 2 }{ \left\{ { \frac { dx }{ dt } } \right\} }^{ 2 }$
${ x }^{ 2 }{ \left\{ { \frac { dx }{ dt } } \right\} ^{ 2 } }\times \frac { 1 }{ 2 } =\frac { { x }^{ 3 } }{ 3 } g$
$\sqrt { \frac { 3 }{ 2gx } } dx=dt$
$\sqrt { \frac { 3 }{ 2g } } \int { \frac { dx }{ \sqrt { x } } } =\int { dt } \\$
$\sqrt { \frac { 3 }{ 2g } } 2\sqrt { x } =t$
Is it correct?

 

[Orodruin]

You should worry about the integration constant. However, you should be able to use your boundary conditions to fix it right away.

Edit: The more straight-forward approach is to simply note that you are integrating both sides from t=0 to an arbitrary time. Thus, the lower bounds of the integrals (after variable substitution) are $s(0)=x(0)\dot x(0)$ and $x(0)$, respectively.

 

[Satvik Pandey]

You should worry about the integration constant. However, you should be able to use your boundary conditions to fix it right away.

Edit: The more straight-forward approach is to simply note that you are integrating both sides from t=0 to an arbitrary time. Thus, the lower bounds of the integrals (after variable substitution) are $s(0)=x(0)\dot x(0)$ and $x(0)$, respectively.

So the constant of 1st integration should be 0.
Should the limits of LHS be from $1/3$ to $1$?(for last integration).

 

[Orodruin]

The constant on the LHS of the first integration is zero due to $\dot x(0)=0$. However, $x(0) = L/3$ (note that the solution earlier in this thread simply called it $x_0$) so the constant on the RHS is non-zero.

 

[Satvik Pandey]

The constant on the LHS of the first integration is zero due to $\dot x(0)=0$. However, $x(0) = L/3$ (note that the solution earlier in this thread simply called it $x_0$) so the constant on the RHS is non-zero.

${ \\ \sqrt { \frac { 3 }{ 2g } } 2\left\{ { \sqrt { x } }_{ 1/3 }^{ 1 } \right\} }=t$
According to question L=1.
So at $t=0$ $x=1/3$ and at $t=t$ $x=1$
Is this correct?

 

[Orodruin]

No, you still have to put the constant in the first integral, which is
$$
\int_{s(0)}^{s(t)} s\, ds = \int_{0}^{x(t)\dot x(t)} s\, ds = \int_{x(0)}^{x(t)} gx^2\, dx.
$$

 

[Satvik Pandey]

Oh!Now I understood that.
$\frac { { s }^{ 2 } }{ 2 } =g\frac { { x }^{ 3 } }{ 3 } +C$
$x(t)=x_{0}$ and $s(t)=0$
$0=g\frac { { x_{0}}^{ 3 } }{ 3 }+C$
or $C=-g\frac { { x_{0}}^{ 3 } }{ 3 }$
or ${ x }^{ 2 }{ \left\{ { \frac { dx }{ dt } } \right\} ^{ 2 } }\times \frac { 1 }{ 2 } =\frac { { x }^{ 3 } }{ 3 } g-g\frac { { x_{ 0 } }^{ 3 } }{ 3 }$

or $t=\sqrt { \frac { 3 }{ 2g } } \int _{ { x }_{ 0 } }^{ x }{ \frac { xdx }{ \sqrt { { x }^{ 3 }-{ x }_{ 0 }^{ 3 } } } }$
How to solve this integral?

 

[voko]

That integral has no representation in elementary functions, you can only compute its numeric value for a given $x_0$..

 

[Orodruin]

It does have a solution in terms of the hypergeometric function ${}_2F_1$, but that is of course not an elementary function and you are probably better off just doing the numerical integration. The end result is, as already mentioned, $t \simeq 0.44\ \rm s$.

 

[Satvik Pandey]

It does have a solution in terms of the hypergeometric function ${}_2F_1$, but that is of course not an elementary function and you are probably better off just doing the numerical integration. The end result is, as already mentioned, $t \simeq 0.44\ \rm s$.

What is numeric integration?

 

[Orodruin]

You compute an integral based on an approximative method using a computer. For example, a very basic approach would be the trapezoid method. The reason to use a computer is that it can use very small step sizes without getting bored or making computational errors due to sleeplessness. However, you would have to use a numerical method suitable for the problem. In this case the trapezoid rule would not be very applicable at the lower integration boundary since the integrand diverges there.

 

[Satvik Pandey]

You compute an integral based on an approximative method using a computer. For example, a very basic approach would be the trapezoid method. The reason to use a computer is that it can use very small step sizes without getting bored or making computational errors due to sleeplessness. However, you would have to use a numerical method suitable for the problem. In this case the trapezoid rule would not be very applicable at the lower integration boundary since the integrand diverges there.

I used the wolfram alpha and also got t=0.44.

 

[Orodruin]

Yes, there is a lot of software that will do the numerical integration for you and chose an appropriate method of integration. I did it with Mathematica, which is a Wolfram product as well. But for simpler things such as this, Wolfram alpha is a good resource.

 

[Satvik Pandey]

Yes, there is a lot of software that will do the numerical integration for you and chose an appropriate method of integration. I did it with Mathematica, which is a Wolfram product as well. But for simpler things such as this, Wolfram alpha is a good resource.

Is there any easy way present to use this change of momentum and force technique to find the answer?

 

[voko]

There is no easy way because this problem is complex. Note that the "change of momentum" approach itself is not necessarily "more correct" than the energy approach; more correctly, one would need to model the transition of the chain links from the motion "on the table" (even if this motion is absence of motion) to the motion "beneath the table" without things like infinite acceleration of links.

 

[Satvik Pandey]

There is no easy way because this problem is complex. Note that the "change of momentum" approach itself is not necessarily "more correct" than the energy approach; more correctly, one would need to model the transition of the chain links from the motion "on the table" (even if this motion is absence of motion) to the motion "beneath the table" without things like infinite acceleration of links.

Thank you voko.:)

 

[Satvik Pandey]

Thank you Orodruin,voko,ehild and haruspex for helping me in solving this problem.:)

 

[haruspex]

There is no easy way because this problem is complex. Note that the "change of momentum" approach itself is not necessarily "more correct" than the energy approach;

I disagree. If they give different answers then one or both are flawed. I don't see any defect in the momentum argument, but I can see that the work conservation assumption is almost surely wrong.

 

[Satvik Pandey]

I disagree. If they give different answers then one or both are flawed. I don't see any defect in the momentum argument, but I can see that the work conservation assumption is almost surely wrong.

I too have the same doubt.
The answer from energy conservation method is 0.55 and answer from momentum technique is 0.44.Why they are't equal?

 

[voko]

I disagree. If they give different answers then one or both are flawed. I don't see any defect in the momentum argument, but I can see that the work conservation assumption is almost surely wrong.

The defect in the momentum approach is, for example, the assumption that the infinitesimal chain element acquires a finite velocity in infinitesimal time, which means infinite acceleration. The assumption that such an infinitesimal element is possible in a chain may itself be considered a flaw.

 

[ehild]

I too have the same doubt.
The answer from energy conservation method is 0.55 and answer from momentum technique is 0.44.Why they are't equal?

Remember, you got 0.55 s from energy conservation assuming that the whole chain moves with the same speed.
With the assumption, that only the vertical part moves, the result is 0.44 s.

ehild

 

[haruspex]

The defect in the momentum approach is, for example, the assumption that the infinitesimal chain element acquires a finite velocity in infinitesimal time, which means infinite acceleration. The assumption that such an infinitesimal element is possible in a chain may itself be considered a flaw.

I don't believe it makes that assumption.

 

[haruspex]

Remember, you got 0.55 s from energy conservation assuming that the whole chain moves with the same speed.
With the assumption, that only the vertical part moves, the result is 0.44 s.

ehild

Sure, but applying work conservation to the bunched chain case produces a rather different equation from that based on momentum conservation.

 

[Orodruin]

Sure, but applying work conservation to the bunched chain case produces a rather different equation from that based on momentum conservation.

I got 0.40 s for the energy conserving approach for the bunched chain. Although I did it quite hastily so there may be errors, but this is similar to what I would expect (loss of work should result in the chain moving slower).

 

[ehild]

Satvick, see that video.

www.youtube.com/watch?v=tj5vm4nnZRo.

It is a simpler problem, assuming outstretched chain, and solved both with F=ma and conservation of energy.
Or read that https://www.google.com/url?sa=t&rct...Sj9th1TF7CuifoQ&bvm=bv.76802529,d.ZWU&cad=rjt

And try to play with a chain. You will see that the outstrecthed one needs more time to slide down then the folded one, shown in the picture.

 

[voko]

I don't believe it makes that assumption.

How do you interpret the term $(dm) v$?

Another interesting assumption in this approach is that the only acting force is that of gravity; specifically, no tension is assumed at the top of the chain column.

 

[Satvik Pandey]

Satvick, see that video.

www.youtube.com/watch?v=tj5vm4nnZRo.

I saw that video.It was quite useful..

Or read that https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB0QFjAA&url=https://wiki.brown.edu/confluence/download/attachments/2752884/Chain.pdf?version=2&modificationDate=1385496323000&ei=U9UuVKLxJcj3OvKVgPgD&usg=AFQjCNHEve_SSVfEbe6RhlWLafvz4l7uvw&sig2=6hZD4Z6Sj9th1TF7CuifoQ&bvm=bv.76802529,d.ZWU&cad=rjt

I understood it, till page(1) but I don't know how to solve that differential equation.
Thank you for helping! :)

 

[ehild]

Do you mean the differential equation $\frac{d^2x(t)}{dt^2}=\frac{g}{L}x(t)$?
It is a very simple second order linear de with constant coefficients, a homogeneous one. You will learn about them soon.
Try the solutions in the form x=e^kt. You get a quadratic equation for k, with two roots, k₁ and k₂. The general solution of the de is the linear combination of the two solutions, belonging to k₁ and k_2. X=c1 e^{k₁ t}+c2 e^k₂t.

ehild

 

[haruspex]

How do you interpret the term $(dm) v$?

ok, that was a bit glib. But it does not assume a nonzero mass undergoes an infinite acceleration. It is only infinite in the limit as dt tends to zero, and dm is also zero in that limit.
The key assumption is that the motion is homogeneous, for the part that is moving. If you want a work conseving version then that assumption must be dropped. This makes the problem far more complex, opening up the possibility of longitudinal oscilations. Thus it may be that even if work is completely conserved, the time taken is greater than the 'simple' calculation suggests.

 

[ehild]

How do you interpret the term $(dm) v$?
.

You can interpret it as an inelastic collision between the moving part and a link at the edge of the table still in rest. That link "sticks" to the moving part and they move together, with a bit less speed. Energy is not conserved.

ehild