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

[Satvik Pandey]

Homework Statement

There is a chain of uniform density on a table with negligible friction. The length of the entire chain is 1 m. Initially, one-third of the chain is hanging over the edge of the table. How long will it take the chain (in seconds) to slide off the table?

Homework Equations

The Attempt at a Solution

I tried to use conservation of energy.
Initially position of CM of hanging part of chain wrt to table is 1/3.
And mass of part of chain hanging down the table is 2m/3.
So ${ E }_{ pi }=\frac { 2mg }{ 9 }$
If the chain falls by distance '$x$'
then position of CM of hanging part of chain wrt to table will be $(3x+2)/6$
And mass of part of chain hanging down the table will be $(3x+2)m/3$
So ${ E }_{ px }=\frac { { (2+3x) }^{ 2 } }{ 18 } mg$
The potential energy is converted into kinetic energy.

So $\frac { { (2+3x) }^{ 2 } }{ 18 } mg-\frac { 2mg }{ 9 } =\frac { 1 }{ 2 } \times \frac { (2+3x)m }{ 3 } { v }^{ 2 }$

on simplifying I got
$\frac { { (2+3x) }^{ 2 } }{ 18 } mg-\frac { 2mg }{ 9 } =\frac { 1 }{ 2 } \times \frac { (2+3x)m }{ 3 } { v }^{ 2 }$

or $\frac { 1 }{ { v }^{ 2 } } =\frac { (3x+2) }{ x(3x+4)g }$

or $\int { dt } =\int { \sqrt { \frac { (3x+2) }{ x(3x+4)g } } dx }$
Is this correct till here?

 

[Orodruin]

then position of CM of hanging part of chain wrt to table will be (3x+2)/6

This would bean the CoM of the hanging part would initially be at 1/3 m, i.e., at the end of the hanging part.

And mass of part of chain hanging down the table will be (3x+2)m/3

The same thing here, initially the chain would have 2/3 of its mass hanging down the side.

 

[Satvik Pandey]

This would bean the CoM of the hanging part would initially be at 1/3 m, i.e., at the end of the hanging part.
The same thing here, initially the chain would have 2/3 of its mass hanging down the side.

I understood that.Is my last equation in #post1 wrong(I think so),if yes then could you please explain why is it wrong.
How should I proceed to solve this question.

 

[Satvik Pandey]

I understood that.Is my last equation in #post1 wrong(I think so),if yes then could you please explain why is it wrong.

EDIT: (I don't think so):(

 

[voko]

I would do this differently. Work out the problem generically, then plug in the details. For example. Let $x$ denote the vertical coordinate of the chain's free end. The potential energy of the chain is $U(x)$. The mass of the moving part of the chain is $m(x)$. Conservation of energy then gives $$U(x_0) - U(x) = {m(x) v^2 \over 2}.$$ Which yields $$ t = \int\limits_{x_0}^x dx \sqrt{m(x) \over 2\left[U(x_0) - U(x)\right]}.$$ Now express $U(x)$ and $m(x)$ correctly and you have the problem solved.

 

[ehild]

I would do this differently. Work out the problem generically, then plug in the details. For example. Let $x$ denote the vertical coordinate of the chain's free end. The potential energy of the chain is $U(x)$. The mass of the moving part of the chain is $m(x)$.

I agree, but I imagine that the whole chain moves, so m does not depend on x. KE=1/2 mv². The potential energy depends on the hanging mass, which is mx/L. Initially L/3 is the hanging length.

ehild

 

[Satvik Pandey]

I would do this differently. Work out the problem generically, then plug in the details. For example. Let $x$ denote the vertical coordinate of the chain's free end. The potential energy of the chain is $U(x)$. The mass of the moving part of the chain is $m(x)$. Conservation of energy then gives $$U(x_0) - U(x) = {m(x) v^2 \over 2}.$$ Which yields $$ t = \int\limits_{x_0}^x dx \sqrt{m(x) \over 2\left[U(x_0) - U(x)\right]}.$$ Now express $U(x)$ and $m(x)$ correctly and you have the problem solved.

As the chains free end is at distance $x$ so the CoM of chain will be at a distance $\frac { x }{ 2 }$.
As the mass is distributed uniformly so mass of chain is $\frac { M }{ L } x$

So $M(x)=\frac { M }{ L } x$

and $U(x)=\frac { M{ x }^{ 2 } g}{ 2L }$

 

[voko]

I imagine that the whole chain moves

Unfortunately, it is not clear from the initial description whether the entire chain moves, or a part of it. Satvik must clarify this.

 

[voko]

and $U(x)=\frac { M{ x }^{ 2 } g}{ 2L }$

How come the potential energy increases as $x$ increases?

Please see my previous message, too.

 

[ehild]

Unfortunately, it is not clear from the initial description whether the entire chain moves, or a part of it. Satvik must clarify this.

Anyway, the length of the chain is constant and all its parts must reach the edge somehow. It can be said that the whole thing is piled up at the edge of the table and has zero volume but it is far from being physical. You are right, it must have been specified by the problem writer.
I used to give the same problem to my pupils, but the chain was a snake, and I added the picture.

ehild

 

[ehild]

How come the potential energy increases as $x$ increases?

Please see my previous message, too.

The potential energy is negative. Add a minus sign.

 

[Satvik Pandey]

The potential energy is negative. Add a minus sign.

Oh! Sorry as chain comes down its potential decreases so there should be minus sign.

As $\triangle U=\frac { 1 }{ 2 } m{ v }^{ 2 }$

So $\frac { 2mg }{ 9 } -\frac { m{ x }^{ 2 }g }{ 2 } =\frac { m{ v }^{ 2 } }{ 2 }$

or $\frac { 2g }{ 9 } -\frac { { x }^{ 2 }g }{ 2 } =\frac { { v }^{ 2 } }{ 2 }$

 

[Satvik Pandey]

Unfortunately, it is not clear from the initial description whether the entire chain moves, or a part of it. Satvik must clarify this.

Sorry.:s

 

[Satvik Pandey]

Anyway, the length of the chain is constant and all its parts must reach the edge somehow. It can be said that the whole thing is piled up at the edge of the table and has zero volume but it is far from being physical. You are right, it must have been specified by the problem writer.
I used to give the same problem to my pupils, but the chain was a snake, and I added the picture.

ehild

Snake instead of chain!.
Nice substitution.:)

 

[haruspex]

If the chain is heaped up at the edge of the table initially, or if it is laid out straight, but you are going to ignore the horizontal velocity of the descending part, then you cannot use conservation of energy. (This problem is come up so many times on this forum, in different guises.). Use momentum.
If you want to take it as initially straight and incorporate the horizontal movement in descent, the problem becomes very difficult. You need to consider the shape of the descending arc in order to get the PE right. See 'chain fountain'.

 

[voko]

So $\frac { 2mg }{ 9 } -\frac { m{ x }^{ 2 }g }{ 2 } =\frac { m{ v }^{ 2 } }{ 2 }$

Where does 2/9 come from?

Sorry.:s

This is hardly a clarification.

 

[voko]

If the chain is heaped up at the edge of the table initially [...] then you cannot use conservation of energy.

Is this what you intended to say? Frankly, I do not see why (I assume the volume of the chain is negligible).

 

[Satvik Pandey]

Where does 2/9 come from?

Initially $2/3$ of chain is hanging from the table sa $-\frac { m{ x }^{ 2 }g }{ 2 }$ becomes $2mg/9$
also L(length of the chain is 1m).

 

[voko]

Initially $2/3$ of chain is hanging from the table

Really? Then why does the problem say "one-third of the chain is hanging over the edge of the table"?

 

[Satvik Pandey]

Is this what you intended to say? Frankly, I do not see why (I assume the volume of the chain is negligible).

I saw this question online on a website.I think this is poorly stated problem.However ehild's explanation in #post6 sound good to me.

 

[Satvik Pandey]

Really? Then why does the problem say "one-third of the chain is hanging over the edge of the table"?

It should be $\frac { mg }{ 18 } -\frac { m{ x }^{ 2 }g }{ 2 } =\frac { m{ v }^{ 2 } }{ 2 }$

or $\frac { mg-9m{ x }^{ 2 }g }{ 18 } =\frac { m{ v }^{ 2 } }{ 2 }$

or $t=\int _{ 1/3 }^{ 1 }{ \sqrt { \frac { 9 }{ g(1-9{ x }^{ 2 }) } } dx }$

 

[ehild]

For the integral, substitute 3x=cosh(u)

ehild

 

[voko]

Still incorrect. The left-hand side is always negative.

 

[Satvik Pandey]

Yes there were sign mistakes
As ${ K }_{ i }+{ U }_{ i }={ K }_{ f }+{ U }_{ f }$

so $0-\frac { mg }{ 18 } =\frac { m{ v }^{ 2 } }{ 2 } +\frac { -m{ { x }^{ 2 } }g }{ 2 }$

or $\frac { m{ v }^{ 2 } }{ 2 } =\frac { m{ x }^{ 2 }g }{ 2 } -\frac { mg }{ 18 }$

or $\frac { m{ v }^{ 2 } }{ 2 } =\frac { 9{ mx }^{ 2 }g-mg }{ 18 }$

or $9{ v }^{ 2 }=(9{ x }^{ 2 }-1)g$

or $\frac { dt }{ dx } =\sqrt { \frac { 9 }{ (9{ x }^{ 2 }-1)g } }$

or $dt=\sqrt { \frac { 9 }{ (9{ x }^{ 2 }-1)g } } dx$

or $\int _{ 0 }^{ t }{ dt } =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 9 }{ (9{ x }^{ 2 }-1)g } } dx }$

or $t=\frac { 3 }{ \sqrt { g } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 1 }{ (9{ x }^{ 2 }-1) } } } dx$

or $t=\frac { 3 }{ \sqrt { g } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 1 }{ 9\left( { x }^{ 2 }-\frac { 1 }{ 9 } \right) } } } dx$

or $t=\frac { 1 }{ \sqrt { g } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 1 }{ \left( { x }^{ 2 }-\frac { 1 }{ 9 } \right) } } dx }$

or $t=\frac { 1 }{ \sqrt { g } } \log { \left| x+\sqrt { { x }^{ 2 }-\frac { 1 }{ 9 } } \right| }$

or $t=\frac { 1 }{ \sqrt { g } } \log { \left| \frac { 3x+\sqrt { 9{ x }^{ 2 }-1 } }{ 3 } \right| }$
where limits are - lower limit is $1/3$ and upper limit is $1$

Is it right?

 

[voko]

The radical magically disappeared from the integral. Not good.

 

[Satvik Pandey]

The radical magically disappeared from the integral. Not good.

Sorry,that was a typo.I have edited it. Is it right now?

 

[voko]

I do not think you computed the integral correctly. Show your steps.

 

[Satvik Pandey]

I do not think you computed the integral correctly. Show your steps.

I just used this formula $\int { \frac { dx }{ \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } } =log\left| x+\sqrt { { x }^{ 2 }-{ a }^{ 2 } } \right| +C$.

Here in this case $a=1/3$

 

[haruspex]

Is this what you intended to say? Frankly, I do not see why (I assume the volume of the chain is negligible).

One should certainly not assume conservation of energy without some justification. If you think about how the process would operate in detail (for the bunched chain), you run into some interesting issues. Once the chain is moving, each link has to accelerate instantly from rest to a nonzero speed as it leaves the table. That sounds like a non-conservative impulse to me. If you try to make the process smoother, you see KE going into other modes, such vibrations (longitudinal or lateral) in the chain.
Much safer is to use conservation of momentum. And voila, you get a different answer.
For the case of the chain starting stretched out across the table, clearly some of the energy goes into horizontal motion, which persists as the chain descends.

 

[voko]

I just used this formula $\int { \frac { dx }{ \sqrt { { x }^{ 2 }-{ a }^{ 2 } } } } =log\left| x+\sqrt { { x }^{ 2 }-{ a }^{ 2 } } \right| +C$.

Here in this case $a=1/3$

Hmm, indeed. I am not sure why that seemed incorrect to me initially.

 

[Satvik Pandey]

Hmm, indeed. I am not sure why that seemed incorrect to me initially.

Are limits which I have mentioned in #post24 correct?

 

[voko]

One should certainly not assume conservation of energy without some justification. If you think about how the process would operate in detail (for the bunched chain), you run into some interesting issues.

Clearly the simple picture of the chain just falling down (or moving horizontally and then vertically) is much too simplified. However, it seems to me that any more complex model will just make this sort of problem unsuitable for beginners. So I guess one has to ignore the discrepancies with the reality.

Your point about a different answer when using momentum is well-taken. I hope Satvik will try that.

 

[voko]

Are limits which I have mentioned in #post24 correct?

Yes, they seem correct to me.

 

[Satvik Pandey]

Yes, they seem correct to me.

So $t=\frac { 1 }{ \sqrt { g } } { \log { \left| x+\sqrt { { x }^{ 2 }-\frac { 1 }{ 9 } } \right| } }_{ 1/3 }^{ 1 }$

$=\frac { 1 }{ \sqrt { g } } \left\{ \log { \left| 1+\sqrt { 1-\frac { 1 }{ 9 } } \right| } -\log { \left| \frac { 1 }{ 3 } +\sqrt { \frac { 1 }{ 9 } -\frac { 1 }{ 9 } } \right| } \right\}$

$\frac { 1 }{ \sqrt { g } } \left\{ \log { \left| 1+\sqrt { \frac { 8 }{ 9 } } \right| } -\log { \left| \frac { 1 }{ 3 } +0 \right| } \right\}$

$\frac { 1 }{ \sqrt { g } } \left\{ \log { \left| \frac { 3+2\sqrt { 2 } }{ 3 } \right| } -\log { \left| \frac { 1 }{ 3 } \right| } \right\}$

$\frac { 1 }{ \sqrt { g } } \left\{ \log { \left| 1.94 \right| } -\log { \left| 0.33 \right| } \right\}$

$\frac { 1 }{ \sqrt { g } } \left\{ 0.28-(-0.48) \right\}$

$t=\frac { 1 }{ \sqrt { 9.8 } } \left\{ 0.76 \right\} =0.243sec$
Is this correct?
Is't it too small?

 

[Orodruin]

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:

##\frac { m{ v }^{ 2 } }{ 2 } =\frac { m{ x }^{ 2 }g }{ 2 } -\frac { mg }{ 18 }##

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.