Classical Mechanics (Marion)

[HAMJOOP]

From (Marion 5th ed. Problem 9-15)
A smooth rope is placed above a hole in a table. One end of the rope falls through the hole at t = 0, pulling steadily on the remainder of the rope. Find the velocity and acceleration of the rope as a function of the distance to the end of the rope x.

Ignore all friction. The total length of the rope is L.


I attempt to solve it by lagrangian L = T - U
T = 1/2 (λx) v^2
U = -(λx)g(x/2)

where λ is mass density
But someone said it is not correct. He claimed that if T depends on generalized coordinate explicitly(in this case x), then it violates Newton's Second Law.
So, lagrangian mechanics cannot be used in open system ?



Marion's Solution
F = dp/dt = m(dv/dt) + v(dm/dt) (see Fig)

I learned that we cannot differentiate m (i.e. dm/dt) if the system is open.


So, which one is correct ??
Can anyone explain to me in detail ?

 

[vanhees71]

I don't see what's wrong with the given Lagrangian. The Euler-Lagrange equation give the same equations of motion as given in the book.

 

[Chestermiller]

From (Marion 5th ed. Problem 9-15)
A smooth rope is placed above a hole in a table. One end of the rope falls through the hole at t = 0, pulling steadily on the remainder of the rope. Find the velocity and acceleration of the rope as a function of the distance to the end of the rope x.

Ignore all friction. The total length of the rope is L.


I attempt to solve it by lagrangian L = T - U
T = 1/2 (λx) v^2
U = -(λx)g(x/2)

where λ is mass density
But someone said it is not correct. He claimed that if T depends on generalized coordinate explicitly(in this case x), then it violates Newton's Second Law.
So, lagrangian mechanics cannot be used in open system ?



Marion's Solution
F = dp/dt = m(dv/dt) + v(dm/dt) (see Fig)

I learned that we cannot differentiate m (i.e. dm/dt) if the system is open.


So, which one is correct ??
Can anyone explain to me in detail ?

Bad news. The solution in Merion is incorrect. The entire rope is accelerating, and not just the part that has passed through the hole. This needs to be taken into account. Let F be the tension in the rope at the hole. Then the force balance on the part of the rope below the hole at any time is:
$$σgx-F=\frac{d(σxv)}{dt}$$
The force balance on the part of the rope above the hole is:
$$F=\frac{d(σ(L-x)v)}{dt}$$
If we add these two equations to eliminate F, we get:
$$gx=L\frac{dv}{dt}$$

 

[vanhees71]

Well, then I misunderstood the drawing. I thought the rope is curled up on the table and thus only the part gone through the hole is accelerating. If you have the other case, of course you get another equation of motion.

 

[D H]

I don't see what's wrong with the given Lagrangian. The Euler-Lagrange equation give the same equations of motion as given in the book.

No, it doesn't. You can't use a Lagrangian formulation for this problem because energy is not a conserved quantity in this problem.

Bad news. The solution in Merion is incorrect. The entire rope is accelerating

No, it's not. The rope is coiled just above the hole. The only part of the rope that is accelerating is the part that is falling.

Well, then I misunderstood the drawing. I thought the rope is curled up on the table and thus only the part gone through the hole is accelerating. If you have the other case, of course you get another equation of motion.

Your understanding is correct. The rope is coiled up just above the hole so that the only part that is accelerating is the part of the rope that has already passed through the hole.

 

[vela]

No, it doesn't. You can't use a Lagrangian formulation for this problem because energy is not a conserved quantity in this problem.

Why wouldn't energy be conserved in this problem? The only force doing work is gravity, which is conservative.

 

[vela]

The force balance on the part of the rope above the hole is:
$$F=\frac{d(σ(L-x)v)}{dt}$$

You've omitted the normal force exerted by the table on the coiled part of the rope.

 

[vela]

But someone said it is not correct. He claimed that if T depends on generalized coordinate explicitly(in this case x), then it violates Newton's Second Law.

That's not right. For example, the kinetic energy in polar coordinates, $T = \frac{1}{2}m[\dot{r}^2 + (r\dot{\theta})^2]$, depends explicitly on the coordinate $r$. Changing from Cartesian to polar coordinates certainly doesn't result in a violation of Newton's second law.

 

[Chestermiller]

You've omitted the normal force exerted by the table on the coiled part of the rope.

This was the horizontal force balance on the part of the rope on the table.

 

[Chestermiller]

No, it doesn't. You can't use a Lagrangian formulation for this problem because energy is not a conserved quantity in this problem.


No, it's not. The rope is coiled just above the hole. The only part of the rope that is accelerating is the part that is falling.


Your understanding is correct. The rope is coiled up just above the hole so that the only part that is accelerating is the part of the rope that has already passed through the hole.

Even if it's coiled, part of it is still accelerating, although not nearly as much as if it is straight. But, actually, I didn't notice that it was coiled.

 

[vela]

Ah, okay. I think the intent of the problem is that the rope on the table remains at rest, and only a small piece of length v dt is accelerated from rest to speed v at time t.

 

[D H]

Why wouldn't energy be conserved in this problem? The only force doing work is gravity, which is conservative.

Energy is not conserved because the part of the rope that is falling is constantly bringing new parts of non-falling rope up to speed. One way to look at this is that this acts as a drag on the falling part of the rope. Drag is almost inevitably a non-conservative force.


Hint to HAMJOOP:

Suppose that at some time t the length the rope x(t) that is already falling with velocity v(t) is about to pull a small length of rope Δx up to speed. At some small time t+Δt later, the length of rope x(t)+Δx will be moving at some speed v(t)+Δv. During this short span of time, the only force acting on this segment of rope is gravity. Assume there is no interaction (yet) between the rest of the coiled up rope and that small segment that is about to be brought up to speed. (You might want to justify this assumption.)

I'll leave the rest up to HAMJOOP.

 

[HAMJOOP]

No, it doesn't. You can't use a Lagrangian formulation for this problem because energy is not a conserved quantity in this problem.

So, there is a non conservative force in this problem.

if I am able to formulate the new "potential", is it still possible to use Lagrangian formulation ?

 

[vanhees71]

Indeed, now that I checked the problem more carefully, I also see that this can not be treated via the Lagrange formalism. The reason seems to be that in the motion always new material of the rope is added to the moving part and thus that we have not a closed system.

The correct way is to use Newton's 2nd Law directly:
$$\mathrm{d}_t p=m g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=m g.$$

 

[HAMJOOP]

From wikipedia, differentiate mv with respect to time by product rule gives wrong result.
It claims that F = dp/dt can only be used in closed system.

http://en.wikipedia.org/wiki/Newton_second_law#Variable-mass_systems

 

[vanhees71]

That's true, but in this case the mass elements entering the moving part of the rope have 0 momentum and thus does not contribute an extra momentum. The formula by Marion (quoted in my previous posting) is thus correct. I just checked this in my favorite mechanics book (the good old Sommerfeld Lectures on Theoretical Physics, Vol. 1) :-)).

 

[D H]

The correct way is to use Newton's 2nd Law directly:
$$\mathrm{d}_t p=m g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=m g.$$

The formula by Marion (quoted in my previous posting) is thus correct.

Actually, it isn't correct. Look at the units!

From wikipedia, differentiate mv with respect to time by product rule gives wrong result.

Yes, F_ext=d(mv)/dt can get you in trouble sometimes. But sometimes wikipedia's alternative expression can also be wrong.

This problem can be solved without resorting to variable mass. Look at what is happening to the rope as a whole. Since the rope's mass isn't changing, it is perfectly safe to use F_ext=dP/dt. What is the net external force on the rope? What is the rope's total momentum, and what is it's time derivative?

 

[vanhees71]

But not the whole rope is moving, i.e., not the total mass is to be put into the momentum. I try to get the equation right, again: The moving mass is $m=\lambda x$ and thus
$$\lambda \frac{\mathrm{d}}{\mathrm{d} t} (x \dot{x})=\lambda x g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=g x.$$
This equation is correct, because the material entering the motion is at rest before and thus carries no momentum.

Newton's Law
$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = \vec{F}$$
only holds for the center of mass of a closed system. An example is the rocket equation. Here a gas with relative velocity $u$ (wrt. to the restframe of the rocket) is expelled (mass per unit time $\mu$). The closed system is "rocket + gas", and we can write (for a rocket close to the earth)
$$\frac{\mathrm{d} m v}{\mathrm{d} t}+\mu (v-u)=-m g,$$
where $v$ is the velocity of the rocket, and $m=m_0-\mu t$ is the mass still contained in the rocket.

This leads to the Tsiolkovsky equation (with gravitational force in approximation close to earth),
$$-\mu v+(m-\mu t) \dot{v} + \mu(v-u)=-m g$$
or
$$(m-\mu t) \dot{v}=\mu u - (m-\mu t) g.$$
This leads to
$$\dot{v} = \frac{\mu u}{m-\mu t} -g$$
and can be simply integrated twice to get $x(t)$.

 

[D H]

But not the whole rope is moving, i.e., not the total mass is to be put into the momentum. I try to get the equation right, again: The moving mass is $m=\lambda x$ and thus
$$\lambda \frac{\mathrm{d}}{\mathrm{d} t} (x \dot{x})=\lambda x g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=g x.$$
This equation is correct, because the material entering the motion is at rest before and thus carries no momentum.

I'll address the first sentence later. With regard to the equation of motion, that is the correct equation. The left hand and right hand sides now agree. They didn't agree in what you wrote previously.

Regarding "But not the whole rope is moving, i.e., not the total mass is to be put into the momentum": So? Newtonian mechanics (and it's rewrite as Lagrangian mechanics and Hamiltonian mechanics) can handle a system of particles, each moving with different speeds and each subject to different forces. That's the whole point of the latter half of chapter 2 in Marion, and in just about any other classical mechanics text.

One way to escape the variable mass imbroglio is not to use variable mass. Treating the entire rope as a system, the external force acting on the rope is $F_{\text{ext}} = F_{\text{grav}}+F_{\text{normal}} = \rho L g - \rho(L-x)g = \rho g x$. The momentum is just that of the part that is falling: $p=\rho x v = \rho x \dot x$. Being a constant mass system, there's no problem with using F=dp/dt, yielding $gx = \dot x^2 + x \ddot x$. Same equations as yours, but without the variable mass stuff.

 

[vanhees71]

That's a good point, because it makes the whole business much clearer! Thanks.

 

[HAMJOOP]

But not the whole rope is moving, i.e., not the total mass is to be put into the momentum. I try to get the equation right, again: The moving mass is $m=\lambda x$ and thus
$$\lambda \frac{\mathrm{d}}{\mathrm{d} t} (x \dot{x})=\lambda x g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=g x.$$
This equation is correct, because the material entering the motion is at rest before and thus carries no momentum.

Thanks. Now I understand I can write down dp/dt = m(dv/dt) + v(dm/dt).

But why the net force acting on the rope is only the gravity?
What about the tension due to the rope on the table ?

 

[vanhees71]

But why the net force acting on the rope is only the gravity?
What about the tension due to the rope on the table ?

See D H's posting. That makes it much better understandable!

 

[D H]

What tension?

With the variable mass formulation, you have to assume that there is no interaction between the falling rope and the rope on the table except for some tiny bit of rope that is being uncoiled and is about to start falling. You also need to assume that this tiny bit of rope quickly accelerates and gets up to the speed of the falling rope. With the first assumption, the only force acting on the falling rope is gravity. There is no tension. With the latter assumption you can treat every little bit of the falling rope as having the same velocity.

The latter assumption is still needed with my alternate approach (looking at the rope as a whole), but the first assumption isn't. The tension between the falling portion of the rope and the coiled up portion of the rope, if any, is an internal force. It's only the external forces that come into play in that alternate approach. See chapter 2 of Marion.