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

[voko]

Do you understand that $\log$ in your formulae means "natural logarithm"?

 

[Satvik Pandey]

Do you understand that $\log$ in your formulae means "natural logarithm"?

No I took common logarithm.
Should the answer be $2.303 (0.243)=0.566$?

 

[voko]

Yes, I get a number close to this.

However, as discussed above, you may want to use a non-energetic method to solve this.

 

[Satvik Pandey]

While your current line of thought will lead you to the correct expression in the end, let me just mention an alternative approach. You have:

where $v = \dot x$. You could differentiate this relation with respect to $t$ and end up with a differential equation that is fairly straight forward to solve instead of having to deal with the rather messy integration which will require substituition, looking in tables, or recognition.

Do you want me to differentiate $v = \dot x$ wrt t.
$a=\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } }$

 

[Orodruin]

Do you want me to differentiate $v = \dot x$ wrt t.
$a=\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } }$

The suggestion was to differentiate the part of your post that I quoted. You should obtain a second order linear homogeneous ODE for x.

 

[haruspex]

Yes, I get a number close to this.

However, as discussed above, you may want to use a non-energetic method to solve this.

The bunched chain assumption with the momentum approach gives me $t = \int_{x=x_0}\left[\frac{2g}{3}\left(x-\frac{{x_0}^3}{x^2}\right)\right]^{-\frac 12}.dx$, where x is the hanging length at time t.

 

[voko]

The bunched chain assumption with the momentum approach gives me $t = \int_{x=x_0}\left[\frac{2g}{3}\left(x-\frac{{x_0}^3}{x^2}\right)\right]^{-\frac 12}.dx$, where x is the hanging length at time t.

I also get this. Evaluated with the initial conditions given, this results in 0.44 seconds.

 

[Satvik Pandey]

I also get this. Evaluated with the initial conditions given, this results in 0.44 seconds.

In problem the chain was lying straight on the table.Suppose if the chain was heaped on the table such that only the part of chain which is hanging from the table travels with velocity "v",then I think that I should not use equation in #post5 to find the time.(as haruspex suggested)
How can I use conservation of momentum here?

 

[haruspex]

In problem the chain was lying straight on the table.Suppose if the chain was heaped on the table such that only the part of chain which is hanging from the table travels with velocity "v",then I think that I should not use equation in #post5 to find the time.(as haruspex suggested)
How can I use conservation of momentum here?

Suppose there's length x hanging at time t. In time dt, what force acts to accelerate the system? What momentum does it impart? What is the change in momentum of the chain?

You say that in the actual problem, the chain starts straight and taut on the table. In that cas, the technical difficulty is that the chain wil not fall straight down but describe an arc. Search net for 'chain fountain effect'. To make the problem work, it needs to specify some kind of smooth guide to deflect the chain down as it leaves the table. Some formulations have the chain falling through a hole in the table, though I'm not sure that completely resolves it.

 

[Satvik Pandey]

Suppose there's length x hanging at time t. In time dt, what force acts to accelerate the system? What momentum does it impart? What is the change in momentum of the chain?

The force which acts on the chain is $Mxg/L$
As $\triangle p/ \triangle t=F$

So $\triangle p=\frac { Mxg }{ L } dt\quad$

 

[voko]

And what is $\Delta p$?

 

[Satvik Pandey]

And what is $\Delta p$?

Change in momentum.

 

[voko]

Change in momentum.

Very nice. Now you just need to express it via the known variables.

 

[Satvik Pandey]

Very nice. Now you just need to express it via the known variables.

Δp=mv-mu(final momentum-initial momentum)
I am confused.In time interval 'dt' the length of chain hanging below will be increased by 'dl'.Hence it's mass of chain below the table is also increased.Should I assume this change in mass negligible?
Also in 'dt' velocity will also increase by some amount.Should I assume this change in velocity negligible?
Should I use conservation of energy to find $v(x)$?

 

[voko]

Δp=mv-mu(final momentum-initial momentum)

Yes.

I am confused.In time interval 'dt' the length of chain hanging below will be increased by 'dl'.Hence it's mass of chain below the table is also increased.Should I assume this change in mass negligible?

No.

Also in 'dt' velocity will also increase by some amount.Should I assume this change in velocity negligible?

No.

Should I use conservation of energy to find $v(x)$?

No.

 

[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.