Law of conservation of energy wrong?

[vibgyor]

i revived this closed thread yesterday becoz whe i chanced upon tthis i thought i could take the thread starter's place and post my doubts. hope the experimental set up is clear ad iwant you to tell me if the exp is perfomed ehat would the velocity of links could be and how they all add up so that LoC holds true. i have expressed my doubts in the previous post .

 

[Doc Al]

No, your scenario is not clearly defined. Is the chained stretched out on the table (as in the OP's problem) or bunched up at the edge? That makes a big difference! (In the first case all parts of the chain will move at the same speed--at least until the chain hits the floor. But not so for the second case--here only the hanging part is moving. Mechanical energy is not conserved in that second case--some of it is converted to thermal energy as each chain link is jerked from rest to the speed of the falling chain.)

now for the eqn mgh =1/2 m v^2 to hold good , every link has to hit the ground at exactly 28 .28 metres per second ( g = 10 m/s^2) and remain the same except the last 40 metres of chain...

As has been pointed out several times already, this calculation treats each link as if it were in free fall, which is not true regardless of which scenario you are talking about.

Here's what you should do if you would like to continue this discussion: Define your scenario clearly and present your calculations. Your initial attempt was shown to be incorrect, so it's up to you to try again.

 

[vibgyor]

i described the scenario as a bunched up chain or lace at the edge in my opening post.

"
Mechanical energy is not conserved in that second case--some of it is converted to thermal energy as each chain link is jerked from rest to the speed of the falling . ""

i thought about that too and in that case I am not sure how much of mechanical
energy gets converted to thermal energy.

here, the force of gravity acting on 40 metres of chain or lace that is vertically running between the floor and the plane as to be balanced at some point and at this point there will be no further acceleration.rgt.? now gravity acting on (g =10m/^2) 40 metres of chain balances out when x metres chain hit the floor per sec. (and v = xm/s). and how to arrive at the value of x which is the velocity(can i call it terminal..?) of the chain .

 

[Dale]

Are you certain that there even is a "terminal" or "steady-state" velocity? You certainly have not shown that yet. And the one calculation that you have shown is clearly wrong since the equation doesn't apply for your situation.

Since this is your problem you can certainly make your chain into a "perfectly flexible rope" and "neglect friction and heat". However, even doing all of that will not make the problem very easy to solve. You will have to do some calculus, perhaps even some differential equations, to handle the KE lost due to the collision with the earth.

If I get bored sometime I may try to work this problem just for kicks, but don't hold your breath.

 

[Dale]

I had to babysit tonite, so I got bored. So here is my problem statement:

A length L of perfectly flexible rope is coiled at the edge of a cliff of height H (H<L). One end of the rope begins to fall off the edge under gravity, with the rest of the rope smoothly uncoiling as each coil is pulled down in turn. Determine the speed of the rope as it falls from the time that the leading end of the rope hits the ground to the time that the trailing end of the rope leaves the cliff. Neglect friction.

First, I parameterized the system with a parameter y which represents the amount of rope that has already hit the ground. The parameter y ranges from 0 to L-H. I then wrote the conservation of energy expression in terms of y:

$$g H L \rho =\frac{1}{2} H \rho v(y)^2+g H \left(-\frac{H}{2}+L-y\right) \rho +\frac{1}{2} \rho \int_0^y v(\psi )^2 \, d\psi$$

Where $$\rho$$ is the linear density of the rope, and v is the velocity of the rope as a function of y. The left hand side is the initial potential energy before the rope begins to fall. The first term on the right hand side is the kinetic energy in the falling rope, the second term is the gravitational potential energy, the third term is the (plastic) collision energy as the rope hits the ground.

Setting y=0 and solving for v(0) we obtain:
$$v(0)=\sqrt{g H}$$

Differentiating both sides wrt y we obtain the following differential equation:
$$0=\frac{1}{2} \rho v(y)^2+H \rho v'(y) v(y)-g H \rho$$

Solving this differential equation with the above initial condition we obtain:
$$v(y)=\sqrt{2 g H-e^{-\frac{y}{H}} g H}$$

Thus there is no constant velocity although the velocity exponentially approaches:
$$v(\infty )=\sqrt{2} \sqrt{g H}=\sqrt{2} v(0)$$

Bottom line, there is no problem with conservation of energy here, only a problem with misapplying a formula and making several assumptions about the results without doing the math. Conservation of energy has been proved in general, as mentioned by arildno and others, but it is not terribly difficult to come up with a complicated scenario with so many variables and details that even the best "problem solver" might work it incorrectly. Whenever you find that some complicated scenario seems to violate some known physical law you can be sure that "the devil is in the details". There is always some ambiguity in the problem set up or some error in the work or assumptions. If you really want to demonstrate a violation of a physical law look for a very simple and clear example where the violation cannot possibly be due to any other factor.

 

[vector3]

I can't believe I read this whole thread...

I'm not leaving however, without posting something...

I once heard that if you put a scale in an elevator and stood on it while the elevator was free falling the scale would read zero. I'm suggesting that the weight of the chain above and the Plane A equals the weight of chain in free fall below the plane regardless of how much chain is above or below plane A. In other words, the chain does not weight anything above the plane or below the plane... until it hits the ground...?

Personally I believe links enter the gravitational field with the current velocity that the first link is at. I assume the chain will not break if the links are quickly accelerated and that there is a slight tension between the links in order to pull a new link into the field.

Any given link would then have it's normal sum of potential energy plus it's kinetic energy. I guess I don't see the significance that plane A has on the problem? Except to make it more clear that that mgh + 1/2mv^2 is the total energy.

The problem starts with no gravitational field. If no g then then no mGh. Also no initial 1/2 m v^2.

 

[stewartcs]

I can't believe I read this whole thread either.

I can't believe this thread hasn't been moved https://www.physicsforums.com/forumdisplay.php?f=5" yet either.

 

[sadhu]

No, that's logically impossible.

Part of the chain is moving at a finite speed and the other part is at rest. However you try to arrange it, when the next link starts moving it has to accelerate "instantaneously" from rest, to match the speed of the moving part.

It is impossible for moving part to accelerate at 9.8m/s, because part of its KE will used to accelerate the stationary part to the same speed. It will only accelerate at 9.8m/s when all the chain has left the table.

This has nothing to do with imagining a "real" chain with finite sized links - the same applies to an idealized flexible string.

what struck me was the word "heap of links"

i think that the chain was kept on something like a table

in respect to it the statement i have quoted is correct only if we assume that no internal frictional forces are acting within the chain.

the very first link will pull the next and it willdo the same to next ..

hence at any instant the force is conveyed to all links and they start moving
but assume that somehow only some links start moving ..
then there must be some force preventing others to move ..
the force must also act on links just starting to move and must do some negtive work..

when this negative work is added to net K.E
i am sure it equalise the net potential energy

 

[sadhu]

I had to babysit tonite, so I got bored. So here is my problem statement:

A length L of perfectly flexible rope is coiled at the edge of a cliff of height H (H<L). One end of the rope begins to fall off the edge under gravity, with the rest of the rope smoothly uncoiling as each coil is pulled down in turn. Determine the speed of the rope as it falls from the time that the leading end of the rope hits the ground to the time that the trailing end of the rope leaves the cliff. Neglect friction.

First, I parameterized the system with a parameter y which represents the amount of rope that has already hit the ground. The parameter y ranges from 0 to L-H. I then wrote the conservation of energy expression in terms of y:

$$g H L \rho =\frac{1}{2} H \rho v(y)^2+g H \left(-\frac{H}{2}+L-y\right) \rho +\frac{1}{2} \rho \int_0^y v(\psi )^2 \, d\psi$$

Where $$\rho$$ is the linear density of the rope, and v is the velocity of the rope as a function of y. The left hand side is the initial potential energy before the rope begins to fall. The first term on the right hand side is the kinetic energy in the falling rope, the second term is the gravitational potential energy, the third term is the (plastic) collision energy as the rope hits the ground.

Setting y=0 and solving for v(0) we obtain:
$$v(0)=\sqrt{g H}$$

Differentiating both sides wrt y we obtain the following differential equation:
$$0=\frac{1}{2} \rho v(y)^2+H \rho v'(y) v(y)-g H \rho$$

Solving this differential equation with the above initial condition we obtain:
$$v(y)=\sqrt{2 g H-e^{-\frac{y}{H}} g H}$$

Thus there is no constant velocity although the velocity exponentially approaches:
$$v(\infty )=\sqrt{2} \sqrt{g H}=\sqrt{2} v(0)$$

Bottom line, there is no problem with conservation of energy here, only a problem with misapplying a formula and making several assumptions about the results without doing the math. Conservation of energy has been proved in general, as mentioned by arildno and others, but it is not terribly difficult to come up with a complicated scenario with so many variables and details that even the best "problem solver" might work it incorrectly. Whenever you find that some complicated scenario seems to violate some known physical law you can be sure that "the devil is in the details". There is always some ambiguity in the problem set up or some error in the work or assumptions. If you really want to demonstrate a violation of a physical law look for a very simple and clear example where the violation cannot possibly be due to any other factor.

well i can't understand your assumption.

to neglect friction and to uncoil the pile one by one

what i only figured out that you prooved that velocity cannot remain same , by taking law of con. of energy as true

 

[Dale]

well i can't understand your assumption.

to neglect friction and to uncoil the pile one by one

Sorry about that. I didn't know how to express the "smoothly uncoiling" assumption well. What I wanted to describe was that I was assuming that no energy was lost in uncoiling the rope, and that the velocity of all of the rope at the top of the cliff was zero (i.e. all the KE was in the falling rope).

 

[sadhu]

well yesterday i thought to derive a mathematical formulation to include frictional forces
wthin the chain so as to complete the proof

what i thought was this...

if we only consider friction forces then the force must reach up to that link where net friction due to the links before it equalise the net force due to gravity,but then i thought what when suppose the orientation of pile disturbes and friciton decreases
then also (according yo my assumption)then the force must reach up to that link where net friction due to the links before it equalise the net force due to gravity,but how come all links come to same velocity from rest instantaniously
more over whatever the velocity the rope was having in one orientation that will remain same for all orie. that's impossible

that factor which i left was "bends"

imagine a rope coiled in circle,now you began to uncoil it from outside,let you apply a constant force and let the rope do reduce it radius due to friction,

at any instant the force acting on the coiled rope depend upon the angle of force with respect to that uncoiling end,the tangetial stress ,etc
then only we can suppose that one end of rope can move while other do not..

this is just one case but imagine a rope spread randomly thus the above factors will act randomly and hence the accelration of rope will also be random ,but less then g.

if you consider a straight rope it has no bends hence the rope will move only if net friction is more that gravity at that instant and acceration keep on increasing
but whole rope move at once...

logically the terminal velocity is never reached ..if i am correct

well don't expect me to come up with a mathematical model to proove it

 

[Doc Al]

modeling a falling chain

Sorry about that. I didn't know how to express the "smoothly uncoiling" assumption well. What I wanted to describe was that I was assuming that no energy was lost in uncoiling the rope, and that the velocity of all of the rope at the top of the cliff was zero (i.e. all the KE was in the falling rope).

There are two simplified ways to model this problem of a chain falling over an edge:
(1) Have the chain layed out in a line (not coiled) so that the entire chain has one speed.
(2) Have the chain coiled so that only the "falling" section of chain has a velocity and the coiled portion has speed zero.

Using model #1, mechanical energy is conserved as the chain falls. (Ignoring the collision of chain with the ground.)

But with model #2, as each segment of chain goes over the edge it is jerked from a speed of zero to the speed of the falling segment. I don't see anyway of doing this without losing some mechanical energy to thermal energy--a point I made in post #37. (Each segment undergoes an inelastic collision.) So you cannot assume conservation of mechanical energy.

 

[Dale]

But with model #2, as each segment of chain goes over the edge it is jerked from a speed of zero to the speed of the falling segment. I don't see anyway of doing this without losing some mechanical energy to thermal energy--a point I made in post #37. (Each segment undergoes an inelastic collision.) So you cannot assume conservation of mechanical energy.

Of course you cannot physically uncoil a rope without losing some mechanical energy to thermal energy. Just as you cannot physically push a box up an incline without losing some mechanical energy to thermal energy, or a roller coaster cannot go without losing some thermal energy. We solve these kinds of problems all the time anyway. Just because it is physically impossible doesn't mean that you cannot get some useful insight from the solution.

 

[Himanshu]

But with model #2, as each segment of chain goes over the edge it is jerked from a speed of zero to the speed of the falling segment. I don't see anyway of doing this without losing some mechanical energy to thermal energy--a point I made in post #37. (Each segment undergoes an inelastic collision.) So you cannot assume conservation of mechanical energy.

Can't we assume the collision to be elastic?

 

[Doc Al]

the uncoiling rope

Of course you cannot physically uncoil a rope without losing some mechanical energy to thermal energy. Just as you cannot physically push a box up an incline without losing some mechanical energy to thermal energy, or a roller coaster cannot go without losing some thermal energy. We solve these kinds of problems all the time anyway. Just because it is physically impossible doesn't mean that you cannot get some useful insight from the solution.

It's a different kind of loss than simply loss due to friction. (I'm perfect OK with assuming a frictionless rope! ) And I think you'll get more useful--and more interesting--insight by trying to figure out the energy loss, rather than just by assuming energy conservation.

Can't we assume the collision to be elastic?

Well, no. Especially since the original theme of this thread was to question whether mechanical energy was conserved.

Let's crank it out. Let the amount of rope hanging over the edge be y; let the mass per unit length be $\lambda$. Applying Newton's 2nd law (in momentum form, since the hanging mass changes):

$$F = dp/dt = d(mv)/dt = (dm/dt)v + m(dv/dt)$$

$$\lambda y g = \lambda \dot{y}^2 + \lambda y\ddot{y}$$

The solution to this is:
$$y = \frac{gt^2}{6}$$

Thus:
$$\dot{y} = \frac{gt}{3}$$

After a length L of rope has slid off the table, the KE will be:
$$(1/2)mv^2 = \frac{1}{3}\lambda g L^2$$

But the decrease in PE is:
$$mg(L/2) = \frac{1}{2}\lambda g L^2$$

Note that the decrease in PE is greater than the resulting KE; the difference is the mechanical energy "lost". (Unless I made an error somewhere. )

 

[Himanshu]

Well I can even put it like this.

Your explanation of modeling a falling chain in post#47 is completely logical if I replace the "discrete" chain with a "continous" rope. There has some amount of work expended on the rope in increasing its speed from zero to the speed of the falling section.
If this explanation is logical for "continous" rope it should be applicable to a "discrete" chain Therefore there cannot an elastic collision between the links. Am I right.

In your analytical solution of the previous post I am stuck at the fifth equation. How did you equate the two sides.