Register to reply 
Variable mass work problem 
Share this thread: 
#1
Apr3005, 07:03 AM

P: 12

The problem is:
we have a rope of total mass M and length L the rope starts from rest on surface and being pulled up by a force F , it's pulled up with constant speed V until the upper end reaches height x . Find the power (P) needed the pull the rope up as a function of X . What I did was : P=dw/dt so if w = Fdx we have P=FV so F is M/L*xg . so P=MgxV/L [j/s] But I feel I need integration somewhere it a variable mass problem I think but where ??? thank you in advance 


#2
Apr3005, 09:36 AM

Sci Advisor
HW Helper
P: 3,031

[tex] W = \int_{??}^{??}Fdx [/tex] How does the force depend on how far you have moved? 


#3
Apr3005, 09:47 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

If I understand this correctly, you have a rope, of length L and mass M, initially lying on a surface. You lift one end of the rope until that end is height X above the table.
Imagine a very small section of the rope, of length dx, at a distance x from the "top" end of the rope (the end being lifted). Since the rope has mass M and length L, its density is M/L and so that little section has mass (M/L)dx (and so weight (M/L)gdx since the velocity is constant, the force F is (M/L)gdx) and is lifted a distance Xx (the first X is the given height, the second x is the distance from the end of the rope). The work done is approximately (M/L)gxdx. ("Approximately" because not all of that section is lifted the same height.) If we add over all such small sections, the total work done is, approximately, [tex]\Sigma (M/L)gxdx[/tex], a Riemann sum. We can make that approximation exact by taking the limit as dx goes to 0 and the sum becomes an integral: [tex]\frac{Mg}{L}\int_0^Zxdx[/tex]. "Z" here is the smaller of X and L if L< X then we lift the entire rope, if X< L then part of the rope remains lying on the surface. 


#4
Apr3005, 10:04 PM

P: 11

Variable mass work problem
Beacause the speed is not changing,so the accelerate is always zero for the in air part.
You ditribute it into two part. The force1 at every specific time is just to counteract the gravitation of the inair part of the rope. Since the mass of this part is m(x)=M*(x/L); So its gravitation is fg(x)=(Mgx)/L; Then f1(x)=fg(x)=(Mgx)/L; The force2 is to accellerate the left part. In a very short time dt,there will be a total mass of dm=(vt/L)M rise from the desk.As fdt=dmv,then f2=v*dm/dt=v^2M/ L; So I think the answer will be f(x)=f1(x)+f2(x)=(Mgx)L+M(v^2)/L; In fact,there is a fumular about this kind of thing,but I can't remember it clearly.May be I can tell you later. If you still have any question,contact me at wangkehandsome@hotmail.com,it will be my pleasure to answer it for you. 


#5
May105, 10:37 AM

Sci Advisor
HW Helper
P: 3,031

[tex] F  mg = \frac{{dp}}{{dt}} = \frac{{dmv}}{{dt}} = v\frac{{dm}}{{dt}} = v\lambda \frac{{ d\ell }}{{dt}} = \lambda v^2 [/tex] [tex] F = mg + \lambda v^2 = \lambda \ell g + \lambda v^2 = \lambda \left( {\ell g + v^2 } \right) [/tex] [tex] F = \frac{M}{L}\left( {\ell g + v^2 } \right) [/tex] [tex] W(t) = \frac{M}{L}\int_0^{x(t)} {\left( {\ell g + v^2 } \right)} d\ell [/tex] Of course this only applies until x = L. It does seem reasonable that a greater force will be required to raise the rope at a faster speed. Without the velocity squared term the force would depend only on how much rope had been raised, independent of how fast. After integrating and finding the rate, the power is [tex] P = \frac{Mv}{L}\left[ {gx(t) + v^2 } \right] [/tex] 


#6
May105, 10:53 AM

HW Helper
PF Gold
P: 1,197

Dan, what about the normal force which is exerted by the ground?



#7
May105, 01:14 PM

Sci Advisor
HW Helper
P: 3,031




#8
May205, 04:00 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

First, if we want to solve this problem properly with respect to a "variablemass" system, then it is totally wrong to do so by assuming that such a system obeys
F=dp/dt. This is readily seen, because with variable mass, this equation is not Galilean invariant, as it must be. If you want to do it this way, you might want to look at my thread: http://www.physicsforums.com/showthr...334#post538334 In this thread, I'll follow another another approach, namely expressing the work done by F to the mechanical energy the WHOLE rope has gained from its position of rest&zero potential energy. The first thing we need to clarify however, is what approximation we need to establish in order to find a "solution". Clearly, the real situation is basically impossible to find a solution to, where the rope curves a bit between the part lying at rest on the table, and the part moving upwards with velocity V due to F. Thus, we simplify the situation by saying that as time goes, a tiny piece of rope which was at rest INSTANTANEOUSLY begins moving upwards with velocity V, i.e, joining the common motion of the upwardsmoving part. We have therefore split the rope into TWO parts: the part at rest, and the part moving upwards with velocity V. Before proceeding with our analysis, we really should determine under which condition this approximation can be regarded as a GOOD approximation: Let us split the rope into 3 parts: 1. The part moving with velocity V, of length x(t) 2. The part of the rope strictly at rest, with length r(t) 3. The JOINING curve C, whose vertical velocity varies from 0 to V, and whose length can be called c. Thus, we have: r(t)=Lxc, where L is the total rope length. Now, if we look at the segment with length r(t)+c (i.e, the part of the rope not moving with uniform vertical velocity V), when is it a good approximation to say that the center of mass of this segment has a NEGLIGIBLE vertical velocity? Evidently, since the vertical velocity at C is less than or equal to V, we have that the center of mass' vertical velocity obeys the inequality: [tex]0<{v}_{C.M}<\frac{\rho{c}V}{\rho{(Lx)}}=\frac{c}{Lx}V[/tex] Requiring [tex]v_{C.M}<<V[/tex] yields: [tex]c<<Lx\to{x}<<Lc\to{x}<<L[/tex] that is, we can only expect us to find a reasonable approximation if the vertically moving part of the rope is a tiny fraction of the whole rope! With this in mind, let us state the mechanical energy equation between the time when F has done no work on the whole rope to the time when the upper end has reached position "x": [tex]W=\oint_{R}\rho\vec{g}\cdot{d}\vec{s}+K[/tex] where W is F's work, the first term on the righthand side is the total rope's gain in gravitional potential energy, and K is the total kinetic energy gained by the rope, conveniently written in parametric form as: [tex]K=\int_{s_{0}}^{s_{1}}\frac{\rho}{2}v^{2}(s)ds, s_{0}\leq{s}\leq{s}_{1}[/tex] Now, only the vertical part of the rope has any potential&kinetic energy, and we'll let the turning point of the table be at s=0, and the upper end at s=x. However, due to the discontinuous velocity jump happening at the "lower" side of 0, we'll need to be careful: At time t, the particle p situated at [tex]s=0^{}[/tex] has [tex]v_{p}=0[/tex], whereas at time t+dt, that particle has moved to [tex]s=0^{+}[/tex], with new velocity [tex]v_{p}=V[/tex] Thus, our workenergy relation reads: [tex]W=\int_{0}^{x}\rho{gs}ds+\frac{\rho}{2}v_{p}^{2}dx+\int_{0^{+}}^{x}\fra c{\rho}{2}V^{2}ds[/tex] Differentiating with respect to time we get, by remembering [tex]\frac{dx}{dt}=V[/tex] everywhere we find it, the power is found by: [tex]P=\frac{dW}{dt}=\rho{g}xV+\frac{\frac{\rho}{2}(V^{2}0^{2})dx}{dt}+\frac{\rho}{2}V^{3}=\rho{V}(gx+V^{2})[/tex] (the force is therefore, of course [tex]F=\rho(gx+V^{2})[/tex]) ([tex]\rho\equiv\frac{M}{L}[/tex] is the density) Now, this happens to coincide with the result you'd get out of the flawed F=dp/dt (flawed, that is, in the variablemass system analysis). The only reason why that method happens to get the correct result, is that we are looking at the problem in the table's rest frame. 


#9
May205, 12:30 PM

Sci Advisor
HW Helper
P: 3,031

One approximation to this problem is that you have a rope of length x above the table moving with constant velocity, and there is an army of nanites frantically adding mass to the bottom of the rope to keep the bottom of the rope in contact with the table. As long as they do their job properly the mass of the rope is changing at a constant rate. The agent pulling the upper end of the rope has no idea what is going on at the other end of the rope, or even if it is happening at the end of the rope instead of somewhere else along its length. All it knows is the momentum of the rope is changing, even though it is moving with constant speed. In the absence of a net force (force in excess of the weight of the rope), the momentum of the rope would be constant and it would slow down as the mass increases. To preserve the velocity, an excess force must be applied to provide the momentum change at the correct rate. It makes no difference where on the rope the mass is being added. It makes a great deal of difference how that mass is being added. If the nanites were doing the work to bring the velocity of the added mass up to v as well as adding that mass to the rope, then the agent at the top would only have to increase the force to compensate for the added weight. However, if the nanites are adding the mass with no upward momentum, then the agent pulling the rope must provide the force needed to give that added mass its upward momentum. An ideal model of this problem would be to assume a massless string extending through a hole in the table. As the string is raised at speed v, mass is painted onto the string at the surface of the table by a ring spraygun painting a uniform layer of mass horizontally onto the string. The mass being attached to the string has no vertical momentum except for that being provided by the force agent at the top. An equivalent problem would be to have the string passing through an extended cylinder spraying mass onto the string uniformly from the walls of the cylinder. In this case the mass is being added all along the length of the string, but it is still being added with only horizontal momentum. The vertical momentum it acquires must come from the force agent pulling the string. There is no problem with Galilean invariance. The problem arises when initial momentum of added mass, or final momentum of discarded mass is not properly taken into account. A horizontal conveyer belt with mass dropping vertically onto it is another example where F = vdm/dt works just fine, because the mass hits the belt with no horizontal momentum. But if you then start thinking about mass dropping off the belt moving at constant speed v and calculate F = v dm/dt you are in trouble because you are neglecting the fact that the dropped mass continues to move horizontally with speed v and retains its horizontal momentum. Thus momentum is conserved and no external force is involved. 


#10
May205, 01:25 PM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Forces act on material particles (in the classical view), and material particles have constant mass (also in the classical view). A system which do not contain the same material particles over time is not a material system, and hence, the rate of change of momentum within that system do not necessarily correspond to the forces acting upon a material system coincident with the system we've arbitrarily chosen to watch. But it is for material systems F=dp/dt works; to gain a proper estimate of F in the opensystem case, we need to include the momentum flux term in addition to the rate of change of momentum within our system. Thus, if you had written [tex]\vec{F}=\frac{d\vec{p}}{dt}\frac{dm}{dt}*\vec{0}[/tex], I wouldn't have had any problems with your analysis. It wouldn't have taken you too much space to say something like "Since the momentum flux is zero, we have..". Since you didn't, it was necessary to point out how flawed your analysis was. 


#11
May205, 04:00 PM

Sci Advisor
HW Helper
P: 3,031

That is what is going on in this problem. Material particles that had no vertical momentum are being given vertical momentum by an applied vertical force. The rate of increase in vertical momentum must be equal to the net force applied in the vertical direction. That is all F = dp/dt means in this problem, and unless I have made two counteracting errors in my analysis, the application of that basic law leads to the correct result. If I have made two errors, show me where they are. I did include the "momentum flux" term in the second discussion. I went to some length to note that the mass being added to the string or rope is being added with no momentum in the vertical direction, and that it makes a big difference when you fail to account for momentum being added or taken away. What have you added to the problem by introducing a null vector? Zero is zero. It contributes nothing to this problem, except to show that your more elegant generalized formalism includes such a term and can be used to correctly include such a flux when it is present. That is certainly worthwhile, and would have been a useful observation to make about the approach I presented instead of dismissing it as totally wrong. It was quite clear in the initial problem that the rope lying on the table had no vertical momentum. Sure I could have said more about it, and I will concede that I should have, but I was not writing a textbook. I was outlining an approach that might be understood by an undergraduate physics student to solving a classical problem where particles that had zero initial vertical momentum were being given vertical momentum by an unbalanced vertical force. 


#12
May305, 04:25 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

All right, OlderDan, I was too touchy on this.
However, this is not the same as to say I agree with you. Forces are always related to the rate of change of the momentum of a MATERIAL system; how that rate of change appears in a particular choice of a geometric system is irrelevant. Whether the rate of change of the material system you've chosen to watch will appear as rate of change of momentum in your geometric system or as momentum flux out of that geometric system depends on the particular choice of geometric system. For example, in our case, fix a geometric control volume with CONSTANT height (the height defined by the coincident top of the rope at the instant we're interested in), and at rest respective to the table, such that there is no net increase of mass within your control volume. Our MOMENTARILY COINCIDENT material system, however, remains the same as in your case: the part of the rope already moving with V, and that set of particles which are going to experience a velocity jump. In this particular case, the rate of change of momentum in the GEOMETRIC system is zero; the term [tex]\rho{V}^{2}[/tex] appears as momentum flux (i.e, from the particles at the top leaving with velocity V). Thus, the forces predicted to act upon the material system momentarily coincident with our geometric control volume are the same as with your choice of control volume**, as it should be. This is an equally valid analysis, but brings to light that: a) Having chosen which material system you want to predict forces for, it is, physically speaking, irrelevant which geometric control volume you choose to work with. It is a matter of convenience, for example to choose that system for which where the maths is simplest. b) But it also means that we shouldn't regard the dp/dtterm for our geometric control volume as anything more fundamental than our momentum flux term. c) Thus, in my view, the crucial thing which a student ought to learn, is that: The rate of change of momentum for a MATERIAL system equals the rate of change of momentum for a GEOMETRIC system PLUS the rate at which the momentum of the material system leaves the geometric system (i.e, the momentum flux) However, precisely because the rate of change of momentum for a MATERIAL system is that which equals the forces upon the system, I think (and, from teaching experience know) )that many students think of the momentum flux term as something weird and unnatural, and tend to forget it (or get it wrong). In my view, therefore, particular care needs to make the student understand the fundamental conceptual difference between an open and a closed system, and that all laws of physics are, in their fundamental form, formulated with respect to the closed system. Having understood that point, a student will more easily accept that opensystem laws are essentially derived by comparing the open system with a momentarily coincident closed system, and that none of the individual terms in the opensystem law are more fundamental than the others. **(I am fully aware of that the geometric system you chose to work with is nonunique: You've basically chosen some geometric control volume which at all times encloses the whole upwardsmoving part of the rope, and whose lower surface is at rest with respect to the table. Which is perfectly fine with me..) EDIT: I am 100% certain that if you had happened to choose the control volume I mentioned above, then you would not have failed to state that the rate of change of momentum within it was 0, and the force given wholly by the momentum flux. So, why the asymmetric treatment? Why is it justified to neglect to specify that the momentum flux is 0 (as in your original case), whereas it is unjustified to neglect to specify that the rate of change of momentum is 0 (in the example I gave)?? 


Register to reply 
Related Discussions  
Work Problem w/variable force Did I do it right?  Introductory Physics Homework  2  
Need Timely Help with Unique Work by a Variable Force Problem  Calculus  1  
Mass problem...please check my work.  Calculus & Beyond Homework  1  
Variable charge & mass?  General Physics  0  
Work with Variable Force  General Math  1 