Find Force to Lift Chain of Length L and Mass ρ Up

[Saitama]

Homework Statement

The end of a chain of length L and mass per unit length ρ, which is piled up on a horizontal platform is lifted vertically with a constant velocity u by a variable force F. Find F as a function of height x of the end above platform.
A)ρ(gx+2u^2)
B)ρ(gx+u^2)
C)ρ(2gx+ρu^2)
D)ρ(u^2-gx)

Homework Equations

The Attempt at a Solution

The chain goes up with a constant speed so at any instant the net force should be zero i.e. F=mg. If the height of the end is x, then m=ρx therefore F=ρgx. But there is no such option.

Any help is appreciated. Thanks!

 

[ehild]

Think: The part of the chain still on the ground has zero speed. The force has to accelerate new and new pieces to speed u.

ehild

 

[tms]

The mass of the chain being pulled up by the variable force is also variable.

 

[Saitama]

Think: The part of the chain still on the ground has zero speed. The force has to accelerate new and new pieces to speed u.

ehild

I understand it now but I am unable to form an equation to find F.

 

[tms]

Start from first principles:
\begin{align}<br /> F &amp;= \frac{dp}{dt} \\<br /> &amp;= v\, \frac{dm}{dt} + m\, \frac{dv}{dt}.<br /> \end{align}<br />

 

[Saitama]

Start from first principles:
\begin{align}<br /> F &amp;= \frac{dp}{dt} \\<br /> &amp;= v\, \frac{dm}{dt} + m\, \frac{dv}{dt}.<br /> \end{align}<br />

F=u \cdot \frac{d(ρx)}{dt}+ρx \frac{dv}{dt}
I can't figure out what should replace dv/dt here.

Or use energy. The work done when increasing the length by dx is the difference of the initial and final energies.

I have formulated an expression for the work done but not sure how to proceed further. Here's the expression I have reached:
W=-ρgxdx-\frac{ρu^2dx}{2}

 

[ehild]

The velocity is constant, isn't it? And think: what force you have to use in the first equation.
As for the method concerning change of energy, you have to use dW instead of W and it is the infinitesimal work Fdx when moving the end of the rope up by dx. Fdx=d(PE+KE). And mind the signs. But something is wrong, I do not see what. The two methods give different results.

ehild

 

[Saitama]

The velocity is constant, isn't it?

So the term dv/dt equals to zero.

And think: what force you have to use in the first equation.

Do you mean I need to add an extra term in the LHS? I think the first equation should be this:
F-ρgx=ρu^2
F=ρgx+ρu^2
Ahah, I reached the answer, thanks a lot!

As for the method concerning change of energy, something is wrong, I do not see what. You have to use dW instead of W and it is the infinitesimal work Fdx when lifting dx length.

Yes, you are right, I need to use dW and I guess I made a sign error. The infinitesimal work should be
dW=ρgxdx+\frac{ρu^2dx}{2}
Substituting dW=Fdx doesn't give me the right answer.

 

[Saitama]

ehild, can you please explain why the energy method doesn't work?

 

[ehild]

I do not know. But I am inclined to think that the other method can be wrong. There is force also between the two parts of the chain. I asked the other HH-s, no reply yet.

ehild

 

[D H]

The work-energy theorem is correct. A mis-application of F=dp/dt yields ρ(gx+u^2), which apparently is the supposedly correct answer. It's not. One has to be extremely careful in applying either F=ma or F=dp/dt to variable mass systems. Do it wrong (which is remarkably easy to do for variable mass systems) and you'll get the wrong answer. As a clue that this approach is wrong, I found an old version of the problem here, http://www.iitk.ac.in/phy/oldfiles/phy102N/Problem_Sheet_6.pdf , question #8. From that page, emphasis mine:

One end of an open-link chain of length L and mass ρ per unit length, piled on a platform, is lifted vertically with a constant velocity v by a variable force F. Find F as a function of the height x of the end above the platform. Also find the energy lost during the lifting of the chain.​

So, let's use that answer of F=ρ(gx+u^2) to calculate the work performed. The net force is that force less gravity, or F_net=ρu^2. Integrating this to find the work performed while lifting the chain to a height x at a constant velocity u yields W=ρxu^2, which is twice the change in kinetic energy. Is the difference between this calculated amount of work performed and the change in the kinetic energy the amount of "energy lost during the lifting of the chain"? No. It's the amount by which the author of the problem messed up.

 

[tms]

The velocity is constant, isn't it? And think: what force you have to use in the first equation.

The force in the equation is the net force, of course. It is composed of the upwards force and gravity.

 

[ehild]

The force in the equation is the net force, of course. It is composed of the upwards force and gravity.

Read DH's post. The upward force is not the lifting force alone.

ehild

 

[tms]

Read DH's post. The upward force is not the lifting force alone.

I did read it (after I made my own post). What other force is there?

 

[ehild]

Normal force from the ground.

ehild

 

[tms]

That only acts on the part of the chain still on the ground, not in motion. Also, according to DH, the force calculated from F = dp/dt is too large, so if there is another force it must be downwards.

 

[D H]

Start from first principles:
F = \frac{dp}{dt}

<br /> OK, let&#039;s do that.<br /> <br /> Suppose that at some point in time t the length of the chain being held off the platform is x(t). This length of chain is moving upwards at a constant velocity u with respect to the platform. Given that the mass per unit length of the chain is \rho, this means the momentum of the chain with respect to the platform is p(t)=\rho x(t) u, directed upward. Some very short time \Delta t later, the length of the chain moving upward is x(t)+u\Delta t, making the momentum p(t+\Delta t)=\rho (x(t) + u\Delta t)u. The change in momentum is \Delta p = \rho u^2 \Delta t. Applying F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t yields F_{\text{net}}=\rho u^2. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, F_{\text{tot}} = \rho gx + \rho u^2. That&#039;s answer (B). Done!<br /> <br /> Or maybe we&#039;re not done. It&#039;s always good to do a sanity check.<br /> <br /> Let&#039;s see how much work is done by this force and compare this to the change in kinetic energy. The net force is F_{\text{net}}=\rho u^2, a constant. Calculating the work performed by this constant net force yields W=\int_0^x F\,dl = \rho x u^2. The change in kinetic energy is half this amount. At this point we can do one of two things:<br /> (a) Attribute this discrepancy to energy that is somehow lost.<br /> (b) Figure out where we went wrong.<br /> <br /> The right option is (b), figure out where we went wrong. Energy is not somehow lost. It&#039;s a conserved quantity. Where we went wrong was in attributing all of this <i>F_net</i> to the hoist that is lifting the chain. We weren&#039;t solving the chain-lifted-off-a-platform problem. We were instead solving this problem:<div style="margin-left: 20px"><i>The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate dm/dt = \rho u with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.</i>&#8203;</div>It&#039;s best not to solve problems in universes where magic occurs.

 

[tms]

So, let's use that answer of F=ρ(gx+u^2) to calculate the work performed. The net force is that force less gravity, or F_net=ρu^2. Integrating this to find the work performed while lifting the chain to a height x at a constant velocity u yields W=ρxu^2, which is twice the change in kinetic energy.

It's late, and I'm not thinking all that clearly, but I think the problem is with the integration. \int F\,dx takes care of a force that varies with position acting on each bit of mass, but in the problem each bit of mass moves a different distance, so the varying force acts on each bit over a varying distance.

 

[Saitama]

OK, let's do that.

Suppose that at some point in time t the length of the chain being held off the platform is x(t). This length of chain is moving upwards at a constant velocity u with respect to the platform. Given that the mass per unit length of the chain is \rho, this means the momentum of the chain with respect to the platform is p(t)=\rho x(t) u, directed upward. Some very short time \Delta t later, the length of the chain moving upward is x(t)+u\Delta t, making the momentum p(t+\Delta t)=\rho (x(t) + u\Delta t)u. The change in momentum is \Delta p = \rho u^2 \Delta t. Applying F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t yields F_{\text{net}}=\rho u^2. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, F_{\text{tot}} = \rho gx + \rho u^2. That's answer (B). Done!

Or maybe we're not done. It's always good to do a sanity check.

Let's see how much work is done by this force and compare this to the change in kinetic energy. The net force is F_{\text{net}}=\rho u^2, a constant. Calculating the work performed by this constant net force yields W=\int_0^x F\,dl = \rho x u^2. The change in kinetic energy is half this amount. At this point we can do one of two things:
(a) Attribute this discrepancy to energy that is somehow lost.
(b) Figure out where we went wrong.

The right option is (b), figure out where we went wrong. Energy is not somehow lost. It's a conserved quantity. Where we went wrong was in attributing all of this F_net to the hoist that is lifting the chain. We weren't solving the chain-lifted-off-a-platform problem. We were instead solving this problem:

The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate dm/dt = \rho u with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.​

It's best not to solve problems in universes where magic occurs.

You explained it nicely. Thanks!

But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it?

Can you give links to some good resources where I can learn more about these variable mass systems?

 

[sankalpmittal]

You explained it nicely. Thanks!

But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it?

Can you give links to some good resources where I can learn more about these variable mass systems?

This question did surprise me too. I completed H.C. Verma work and energy and there was 57th question in which we had a chain just touching the ground and it was released from rest. It was asked that out of chain of length L , x length of it strike the floor. No heap formed. We had to calculate force exerted by chain on floor as a function of displacement x. I tried and got the wrong answer until I realized that weight of chain also acts in addition to change in momentum. In this case also , normal reaction acts. By using this logic , I get the same answer as DH. Mechanical energy is not conserved as I suspect, may be some non conservative forces be acting in chain+earth system.

But again, work energy theorem,i.e. W_net=ΔK.E. still applies. Did you try applying it ? If work energy theorem fails, then I am sure that this question is somewhat a magic. Work energy theorem applies to every inertial Earth system, as per H.C. Verma.

 

[sankalpmittal]

OK, let's do that.

Suppose that at some point in time t the length of the chain being held off the platform is x(t). This length of chain is moving upwards at a constant velocity u with respect to the platform. Given that the mass per unit length of the chain is \rho, this means the momentum of the chain with respect to the platform is p(t)=\rho x(t) u, directed upward. Some very short time \Delta t later, the length of the chain moving upward is x(t)+u\Delta t, making the momentum p(t+\Delta t)=\rho (x(t) + u\Delta t)u. The change in momentum is \Delta p = \rho u^2 \Delta t. Applying F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t yields F_{\text{net}}=\rho u^2. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, F_{\text{tot}} = \rho gx + \rho u^2. That's answer (B). Done!

Or maybe we're not done. It's always good to do a sanity check.

Let's see how much work is done by this force and compare this to the change in kinetic energy. The net force is F_{\text{net}}=\rho u^2, a constant. Calculating the work performed by this constant net force yields W=\int_0^x F\,dl = \rho x u^2. The change in kinetic energy is half this amount. At this point we can do one of two things:
(a) Attribute this discrepancy to energy that is somehow lost.
(b) Figure out where we went wrong.

The right option is (b), figure out where we went wrong. Energy is not somehow lost. It's a conserved quantity. Where we went wrong was in attributing all of this F_net to the hoist that is lifting the chain. We weren't solving the chain-lifted-off-a-platform problem. We were instead solving this problem:

The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate dm/dt = \rho u with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.​

It's best not to solve problems in universes where magic occurs.

Wait , I may be stupid , but let's see. W_net=ΔKE

F_net = ρu²-ρgx

Now as x=u²/2g

F_net = ρu²-ρu²/2
F net =ρu²/2

Integrating net F with respect to dx from 0 to x,

W=ρu²x/2

So W=ρu²x/2 which is equal to ΔKE. As length x is in air , its incorrect to consider normal reaction. I think this is where you blundered.

Edit: As per me the correct answer is D. Then only this question is justified by work energy theorem.

 

[D H]

But do you mean that the question given is wrong?

Yep. In a multiple choice type question, the question itself is wrong when the correct answer is not one of listed answers. That's the case here.

This is not the first time this has happened. Textbooks can be erroneous. Multiple choice: The problem apparently is particularly bad
(a) In far too many online automated physics homework problem systems.
(b) In far too many Indian physics texts.
(c) Both of the above are true.

Apparently India has rather weak enforcement of its copyright laws, which leads Indian textbook authors to copy problems from one another. It's much easier to copy some other author's already-worked problems rather than to create new ones. That's fine (other than copyright issues) if the problems are well written and the solutions are correct. It's not so fine if the question is poorly written, if the correct answer isn't present in the multiple choice list, or if the solution in the answer guide / worked example is incorrect. We've chased down a number of problems with Indian physics texts at this site, occasionally seeing the exact same bad question replicated in dozens of different texts.

If I was asked to calculate the energy lost, how can I find it?

The correct answer is that energy is not lost, at least not ideally. Work is being done against gravity in this problem Gravity is a conservative force. There are no losses in working against gravity.

In reality, there will be some energy loss due to entropy. (Note: the energy isn't really "lost". It's just converted to unusable energy: Heat.) A cable or chain may heat up a bit due to non-conservative interactions as the cable/chain grows taut, links shift, etc. We can ignore all that messiness by assuming an ideal chain. Even a terribly constructed, non-ideal chain will not heat up to the extent implied by the wrong answer to this problem.

Can you give links to some good resources where I can learn more about these variable mass systems?

The best thing to do in an introductory physics class is to try to pose the question without worrying about variable mass. It's messy, and easy to mess up. IMO, it's not a subject introductory physics classes should delve around in much (other than perhaps rockets as an exemplar). This particular problem can easily be answered without looking at variable mass. Simply look at the chain of fixed length L as a whole.

 

[TSny]

I’ll add my 2 cents to the confusion! For any system $F_{net}^{ext} = M a_{cm}$ where M is the total mass of the system.

Let the system be the entire chain alone. The net external force is $F_{net}^{ext} = F_{lift} – Mg + F_N$ where $F_{lift}$ is the lifting force and $F_N$ is the normal force acting on the part of the chain still resting on the platform.

When amount $x$ of chain is off the table, the center of mass of the section that is off the table is at $x/2$ above the platform and has mass $\rho x$. So, relative to the platform, the location of the center of mass of the whole chain is

$x_{cm} = \frac{(\rho x)(x/2)}{M} = \frac{\rho}{2M}x^2$.

Taking the second time derivative of this yields $a_{cm} = \frac{\rho}{M}(\dot{x}^2 + x\ddot{x}) = \frac{\rho}{M}u^2$ since $\ddot{x} = 0$

So, we have $F_{lift} – Mg + F_N = \rho u^2$ or

$F_{lift}= \rho u^2 + Mg - F_N$

If we take the normal force to equal the weight of the part of the chain resting on the table, then $F_N = (M-\rho x)g$ and we get

$F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)$

 

[ehild]

I’ll add my 2 cents to the confusion!
$F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)$

It is very convincing 2 cents :) But what is the problem with the energy then?


ehild

 

[tms]

It is very convincing 2 cents :) But what is the problem with the energy then?

I think the problem is with the integration W = \int F\,dx. For any particular bit of the chain that is correct, but each bit of the chain goes a different distance so the force acts on each bit for a different distance, thus doing a different amount of work. I haven't yet figured out the correct integral, though. I know I'm not being abundantly clear, but I think the answer is in this direction.

At any rate, if the energy is right, what is wrong with the F = dp/dt solution? Since that is basically the definition of force, it, too, must be right. Picking the energy to be right because dm/dt is "magic" is not really good enough.

 

[ehild]

F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else non-conservative force does also work.

ehild

 

[TSny]

Regarding energy, I suspect that when each link is jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!

 

[ehild]

Regarding energy, I suspect that when each link in jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!

But why?

ehild

 

[tms]

F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else non-conservative force does also work.

I'm confused, too. I still don't see anything wrong with the F = dp/dt solution.

 

[tms]

Regarding energy, I suspect that when each link in jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!

But there is nothing in the problem or solution(s) about the internal working of the "chain". The solution(s) apply just the same to the anchor chain of an aircraft carrier or to a piece of string.

 

[Dick]

It is very convincing 2 cents :) But what is the problem with the energy then? ehild

It's what DH noticed. Suppose the chain has length L. Then at the moment it all lifts off the platform it's center of mass is at L/2 so the potential energy is hmg=(L/2)(Lρ)g. It's kinetic energy is (1/2)mu^2=(1/2)(Lρ)u^2, but if you integrate F you get ρu^2L+(L/2)(Lρ)g. So you put more energy into lifting the chain then there is in the center of mass kinetic energy plus potential energy of the chain. Physically you can wave your hands and say it goes into sound and heat due to damping. If you idealize the chain to be silent and friction free, as DH suggested, and think of an otherwise realistic chain, then the answer would be that the links must be oscillating without any damping. So the extra kinetic energy must be in the wiggling chain. The equation doesn't tell you exactly where the extra energy went, but it does say that it has to happen.

 

[D H]

Regarding energy, I suspect that when each link in jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!

No, it won't. At least not to the extent suggested by the erroneous F=dp/dt solution. To see why this is nonsense, suppose we reverse the process. Instead of raising the chain, we'll lower it. Pile the chain up in one big lump, carry it up en masse to some height, and lower the chain. The F=dp/dt solution yields an over unity energy gain. Free energy! Or maybe it's just a mistake.

 

[Dick]

No, it won't. At least not to the extent suggested by the erroneous F=dp/dt solution. To see why this is nonsense, suppose we reverse the process. Instead of raising the chain, we'll lower it. Pile the chain up in one big lump, carry it up en masse to some height, and lower the chain. The F=dp/dt solution yields an over unity energy gain. Free energy! Or maybe it's just a mistake.

I don't think the dp/dt solution is erroneous. It sounds like a reasonably accurate model of actually lifting a chain. If we unmagic your magical universe picture so we have a machine on the platform attaching links to the chain as it rises then act of attaching each link is basically an inelastic collision between the chain and link. Energy must be lost from the kinetic+potential sum.

 

[TSny]

But why? ehild

Suppose we have just 2 links each of mass m on a horizontal frictionless surface as shown. The right link is already in motion with speed u while the left link is still at rest. Let F Δt be the impulse that the right link gives to the left link to get the left link moving also at speed u. So, F Δt = mu. There will be a reaction impulse of equal magnitude acting to the left on the right link which we arrange to be exactly balanced by an external impulse (from the blue force) so that the right link maintains speed u. Then the work done by the blue force is F Δx = F(uΔt) =(F Δt) u = mu u = mu² = twice the increase in KE of the system. So, more work was done by the blue force than shows up as KE.

 

[D H]

It's what DH noticed. ... The equation doesn't tell you exactly where the extra energy went, but it does say that it has to happen.

You completely missed my point.

I'll be very explicit. Using an F=dp/dt approach and attributing all of that change in momentum to the action by the hoist yields a nonsense answer. This is one of many cases where naively using F=dp/dt is flat out wrong. It's F=ma, not F=dp/dt.

The heating implied by that F=dp/dt approach is ludicrous. Even more ludicrous is the cooling that would result from reversing the process were this approach correct. Even yet more ludicrous is the fact that this erroneous approach leads to an over unity device.

 

[Dick]

Suppose we have just 2 links each of mass m on a horizontal frictionless surface as shown. The right link is already in motion with speed u while the left link is still at rest. Left F Δt be the impulse that the right link gives to the left link to get the left link moving also at speed u. So, F Δt = mu. There will be a reaction impulse of equal magnitude acting to the left on the right link which we arrange to be exactly balanced by an external impulse (from the blue force) so that the right link maintains speed u. Then the work done by the blue force is F Δx = F(uΔt) =(F Δt) u = mu u = mu² = twice the increase in KE of the system. So, more work was done by the blue force than shows up as KE.

Now that I agree with. Picking up the links is an inelastic process.

 

[D H]

Now that I agree with. Picking up the links is an inelastic process.

Of course it is. The correct answer is that F\ge \rho(gx+\frac 1 2 u^2), with equality resulting only if the process is reversible (i.e., picking up the chain is an elastic process).

There's not one thing in the F=dp/dt approach that suggests an irreversible process. In fact, this erroneous approach yields a reversible process. Simply apply this exact same concept but with the chain being lowered rather than raised.

 

[tms]

This is one of many cases where naively using F=dp/dt is flat out wrong. It's F=ma, not F=dp/dt.

What!? F = dp/dt is the definition of force. F = ma only applies when the mass is constant. In this problem, since the speed is constant, there is no acceleration, and, according to F = ma, there is no force.

 

[TSny]

We can replace the linked chain by an “elastic” system. Two masses (each m) are connected by a spring as shown. Initially the spring is unstretched with the mass on left at rest and the mass on right moving to the right with speed u. Also we have arranged a variable force F acting on the right mass that will always have the same magnitude as the spring force. Thus, the right mass always moves with the same speed u. The force F acts between t = 0 and t = t_f where t_f is chosen to be the instant when the velocity of the center of mass (cm) of the two-mass system is u.

The total momentum at t_f will be $(M_{system})V_{cm} = (2m)u = 2mu$. Since we started at time t = 0 with total momentum $mu$, the change in momentum of the system between t = 0 and t = t_f is $mu$. This change in momentum must come from the impulse of the applied force:

$\int{Fdt} = mu$.

The energy of the system at t = 0 is $E_o =mu^2/2$.

The energy of the system at t = t_f is $E_f = KE_{system} + PE_{spring}$.

We can always write the KE of the system as KE due to motion of the cm plus KE relative to cm. The KE due to motion of the cm is

$(1/2)(2m)V_{cm}^2 = mV_{cm}^2 = mu^2$ at t_f.

The KE relative to cm is just the “vibrational” kinetic energy associated with the rate of separation of the two masses. We combine the vibrational KE with the spring PE into a total “internal energy”. So, our energy at time t_f is $E_f = mu^2 +E_{int}$.

Hence, the change in energy from t = 0 to t = t_f is

$\Delta E = mu^2/2 + E_{int}$.

The work done by the applied force betwen t = 0 and t = t_f is

$W = \int{Fdx} = \int{F\frac{dx}{dt}dt} = \int{Fudt} = u\int{Fdt} = u(mu) = mu^2$

Suppose for some reason we didn’t know about the “internal energy”. (Maybe the spring is hard to see and it’s very stiff so that the vibrational amplitudes are two small for us to notice.) Then we would only be aware of the energy due to overall motion of the center of mass. Thus, we would claim that the change in energy of the system between t = 0 and t = t_f would be just the term $mu^2/2$ in the expression for $\Delta E$.

So, it would appear that the work done was twice the increase in energy of the system in apparent violation of the work-energy theorem. But we see that in fact the other half of the work went into “hard-to-see” internal energy (i.e., vibrational KE and PE of the spring).

 

[Dick]

Of course it is. The correct answer is that F\ge \rho(gx+\frac 1 2 u^2), with equality resulting only if the process is reversible (i.e., picking up the chain is an elastic process).

There's not one thing in the F=dp/dt approach that suggests an irreversible process. In fact, this erroneous approach yields a reversible process. Simply apply this exact same concept but with the chain being lowered rather than raised.

When the chain is being lowered the force is not the same as when the chain is being raised. It's not reversible. This isn't even a uniquely confusing problem. Pouring sand onto a conveyor belt gives the same sort of result. Even the standard air table experiment with pucks and velcro shows it. As Tsny points out, you can replace the velcro with locking springs and get the same result. Then you can see where the kinetic energy went. It becomes internal to the two puck system instead of dissapated into friction. This is the same sort of problem.

 

[Saitama]

Thanks everyone for taking your time to explain the things so nicely. I am trying to understand each and every single post in this thread. This is probably going to take some time at my part. This thread is really interesting. Please continue your further discussions. I am following this thread, please don't think that I have left this thread. Its just that I am an average student and it takes me some (or a lot of) time to understand the things.

 

[Dick]

Thanks everyone for taking your time to explain the things so nicely. I am trying to understand each and every single post in this thread. This is probably going to take some time at my part. This thread is really interesting. Please continue your further discussions. I am following this thread, please don't think that I have left this thread. Its just that I am an average student and it takes me some (or a lot of) time to understand the things.

That's good news! If people with a lot of experience with stuff like this can disagree, no reason it shouldn't take you time to digest it. Feel free to chip in.

 

[Saitama]

One more question.

@D H: Then what should be the answer to the second part of the question mentioned in the pdf you posted on the first page. Please don't take me wrong but the pdf you posted is hosted under IIT Kanpur's (a reputed institution here) website and its very unlikely for them to do such mistakes (or maybe they did this time). Were you able to find the solution key of that problem sheet?

 

[haruspex]

Only just found this intriguing thread. Can't resist chipping in.
Yes indeed, TSny, it's the inelasticity. (It could be made even harder to comprehend by dealing with an inelastic string instead of a chain of links, but the same applies.) In time δt, a length uδt, mass ρuδt, accelerates from 0 to u. Note that this is an unbounded acceleration, so we're dealing with impulses; energy cannot be taken to be conserved. Momentum is conserved, so the force is (ignoring gravity - this would also work in free fall) ρu². The KE gained in δt = ρu³δt/2, while the work done, as the force moves distance uδt, is twice that.
With an elastic string, each uδt would take time to reach speed u, and would thereafter oscillate. Now that would make one helluva spring problem!

 

[Dick]

Only just found this intriguing thread. Can't resist chipping in.
Yes indeed, TSny, it's the inelasticity. (It could be made even harder to comprehend by dealing with an inelastic string instead of a chain of links, but the same applies.) In time δt, a length uδt, mass ρuδt, accelerates from 0 to u. Note that this is an unbounded acceleration, so we're dealing with impulses; energy cannot be taken to be conserved. Momentum is conserved, so the force is (ignoring gravity - this would also work in free fall) ρu². The KE gained in δt = ρu³δt/2, while the work done, as the force moves distance uδt, is twice that.
With an elastic string, each uδt would take time to reach speed u, and would thereafter oscillate. Now that would make one helluva spring problem!

Right. But energy is always conserved. What's not conserved is the center of mass kinetic energy of a composite system. It can hit something and absorb it into internal vibration. That was the wiggling chain I was referring to earlier.

 

[ehild]

I got an idea while sleeping . We assumed the situation on the left in the pic, that the chain stays at the same place while lifted, as if it was in a tube. But that involves that the parts on the floor have to move to that tube somehow, which involves some motion.

Imagine that the chain is straight on the floor and the person who lifts the chain moves his hand always above the end of the horizontal part. That means, the vertical part moves not only up but also sideways. The length of the chain is constant, so both components of the velocity are the same u. That means
KE=1/2 m (v²(vertical)+v²(horizontal))=1/2 (ρx)(2u²)
E = PE+KE=1/2 ρx²g+ρxu², dW=F(lift)dx=(ρxg+ρu²[/B])dx. F(lift)=(ρxg+ρu²) from work energy considerations
Anyway, it can not be assumed that the chain does not have any horizontal component of velocity. Either that part which is on the floor, or that part which is in the air or both. The KE has to include both terms. But the time derivative of the vertical component of momentum is equal to the vertical force. ehild

 

[D H]

OPlease don't take me wrong but the pdf you posted is hosted under IIT Kanpur's (a reputed institution here) website and its very unlikely for them to do such mistakes (or maybe they did this time). Were you able to find the solution key of that problem sheet?

No, I wasn't able to find the solution key.

Don't take something written by reputed institutions as sacrosant. MIT makes mistakes. So does Caltech. So do highly reputable publishers. That's why they all write erratum sheets. Those mistakes typically don't propagate in countries with strong copyright laws. There's a problem in India. India either has weak copyright law or has rather weak enforcement of them. Here's a google book search for the exact phrase "A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration" that illustrates what happens in India: http://www.google.com/search?q=A+pa...ius+r+such+that+its+centripetal+acceleration"

Aha! Pranav-Arora, it turns out you were involved in the thread that instigated this search: [thread]616190[/thread].

What!? F = dp/dt is the definition of force. F = ma only applies when the mass is constant.

Where did you get that idea? It's certainly not anywhere in Newton's Principia. (Then again, neither is F=ma. Newton's Principia is almost entirely calculus-free.)

There's a big problem with F defined via F=dp/dt: The \dot m v term makes force a frame dependent quantity in the case of variable mass systems. On the other hand, defining force defined via F=ma means that force is the same quantity in all reference frames since both mass and acceleration are invariant quantities in Newtonian mechanics. This is why aerospace engineers almost inevitably choose F=ma rather than F=dp/dt.

In this problem, since the speed is constant, there is no acceleration, and, according to F = ma, there is no force.

Speed isn't constant. The links at rest on the platform change velocity by u as they are picked up. There is a problem is with F=dp/dt, however. The momentum of the rising part of the chain is identically zero from the perspective of a reference frame moving upwards at a constant speed u with respect to the platform. In this frame, F as calculated with dp/dt is zero if you are looking only at the momentum of the rising part of the chain.

There is a way around this mess, and that's not to use variable mass. Look at what happens to the chain as a whole. Now F=dp/dt and F=ma yield the same answer because \dot m is zero. Unfortunately that will not solve the problem. The problem is that this is a kinematic view of force rather than a dynamic view of force. We need to attribute this kinematical net force to root causes to convert this kinematical POV to a dynamical POV. Naively attributing all of this force to the hoist, or whatever is lifting the chain, is incorrect.

To see that it must be incorrect, simply reverse the process: Lower the chain rather than raising it. To keep the chain moving downward at a constant velocity we'll let connect the upper end of the chain to some physical device and let the lowering chain do work on that device. How much work is performed? If we attribute all of the change in momentum to work done by that the device we get an over unity system. More work will be performed than is allowed by the conservation laws. That doesn't make sense.

One way out of this morass (and the way out of the raising morass as well) is to attribute some of the change in the momentum of the chain to the platform. For example, one could look at the transients as a link is picked up from the platform. Yech. That's getting into finite element analysis, something that is far beyond the scope of an introductory physics course.

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices) if all forces are conservative. All bets are off if the system is not conservative. You aren't going to get a nice simple answer in the case of nonconservative forces. Nonconservative forces almost inevitably yield ugly solutions.

 

[sankalpmittal]

Pranav, from wherever you posted the question, what answer it gives ? I got (D) as my answer and then only can the work energy theorem be justified. Else , it appears magic.

If this is not the correct answer, please look at posts 20 and 21 and tell, where I went wrong.

@DH: As far as the other problem https://www.physicsforums.com/showthread.php?t=616190 is concerned that textbook answer which OP there posted is likely wrong. My textbook too has the question but its answer is way different...

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices) if all forces are conservative. All bets are off if the system is not conservative. You aren't going to get a nice simple answer in the case of nonconservative forces. Nonconservative forces almost inevitably yield ugly solutions.

Well, I always use this equation if non conservative forces are involved.

W_nc=E_f-E_i

W_nc:Work done by non conservative forces.
E_f: Final Mechanical Energy.
E_i: Initial Mechanical energy.

 

[haruspex]

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices)

... and which is wrong. The correct answer is achievable by consideration of conservation of momentum and can be done without involving the dp/dt formulation. Conservation of momentum and conservation of work energy (as opposed to thermal energy) lead to different answers - one must be flawed.

 

[ehild]

Any comment about my post#46?

ehild