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

[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.