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Incorrect solution to impulsive dynamics problem?

  1. Oct 15, 2011 #1
    Hey, all!

    Browsing around the library for some mechanics books, I happened to come across Manuel Prieto Alberca's Curso de Mecánica Racional. Dinámica. In this book, I managed to find an interesting problem that is also solved, but having talked to other people, I have started to suspect that the reference solution is wrong.

    I will now copy the problem statement:

    Consider a bar [itex]AB[/itex] of length [itex]\ell[/itex] and mass [itex]m[/itex] that falls down due to gravity, starting from rest at [itex]Ox[/itex]. The point [itex]A[/itex] of the bar is attached to [itex]O[/itex] with a massless string of length [itex]h[/itex]. Determine the angular velocity and velocity of the center of mass right after the string has completely stretched out, that is, reached a length of [itex]h[/itex].


    The reference solution goes as follows:

    Let [itex](\xi, \eta)[/itex] be the coordinates of the center of mass of the bar and [itex]\theta[/itex] the angle it forms with the horizontal in an arbitrary position.

    We have

    [tex]T = \frac{1}{2}m(\dot{\xi}^2 + \dot{\eta}^2)+ \frac{1}{2}\cdot\frac{1}{12}m\ell^2 \dot{\theta}^2[/tex]
    Right before the percussion, we have the following values

    [tex]\dot{\xi_0} = 0, ~ \dot{\eta_0} = \sqrt{2gh}, ~ \dot{\theta_0} = 0[/tex]​

    We wish to determine these values right after the percussion.

    We have the constraint

    [tex]\left(\xi - \frac{\ell}{2} \cos \theta\right)^2 + \left(\eta - \frac{\ell}{2} \sin \theta\right)^2 = h^2[/tex]​

    Applying Lagrange's equations we have

    [tex]m \dot{\xi}_1 - 2\lambda \left(\xi - \frac{\ell}{2}\cos \theta \right) = 0[/tex]

    [tex]m \dot{\eta}_1 - m \sqrt{2 g h} - 2\lambda \left(\eta - \frac{\ell}{2}\sin \theta \right) = 0[/tex]

    [tex]\frac{m\ell^2}{12} \dot{\theta}_1 - \lambda\left[ \left(\xi - \frac{\ell}{2}\cos \theta \right) \ell \sin \theta - \left(\eta - \frac{\ell}{2}\sin \theta \right) \ell \cos \theta \right] = 0[/tex]​

    And since the constraint is persistent, we have

    [tex]\left(\xi - \frac{\ell}{2}\cos \theta \right) \left(\dot{\xi}_1 + \frac{\ell}{2} \dot{\theta}_1 \sin \theta \right) + \left(\eta - \frac{\ell}{2}\sin \theta \right) \left(\dot{\eta}_1 - \frac{\ell}{2} \dot{\theta}_1 \cos \theta \right) = 0[/tex]​

    Since at the moment of the percussion we have [itex]\theta = 0[/itex], [itex]\xi = \ell/2[/itex], [itex]\eta = h[/itex], we are left with

    [tex]m \dot{\xi}_1 = 0[/tex]
    [tex]m \dot{\eta}_1 - m\sqrt{2gh} - 2 \lambda h = 0[/tex]
    [tex]\frac{m\ell^2}{12} \dot{\theta}_1 + \lambda \ell h = 0[/tex]
    [tex]\dot{\eta}_1 - \frac{\ell}{2}\dot{\theta}_1 = 0[/tex]​

    System that solved yields

    [tex]\dot{\xi}_1 = 0 ~ ~ ~ \dot{\eta}_1 = \frac{3}{4}\sqrt{2gh} ~ ~ ~ \dot{\theta}_1 = \frac{3 \sqrt{2gh}}{2\ell}[/tex]​

    So, having shown the reference solution, the issue raised by people I have to talked to is the fact that the constraint applied is for a suddenly appearing hard bar.

    The string isn't stiff, even though it doesn't change length during the moment of impact; it allows a velocity component at [itex]A[/itex] that changes the length of the string.

    Thus, an alternative approach suggested by the same person was to include the ceiling the string is attached to at the top (point [itex]O[/itex]), use conservation of energy and momentum and take the limit for infinite mass for the ceiling.

    So, what do you guys think? Is the reference solution wrong? Would the alternative approach seem better?

    I'm interested in knowing what you think. :approve:


    For moderators: I'm not 100% sure if this goes here or in the HW/Coursework section. It is a problem from a textbook, but I was looking for a section where instead of plain academic assistance, I was expecting an engaging conversation with knowledgeable people on why this method doesn't work, why the other may, etc... If it's in the incorrect section, my apologies to the moderation staff and administrators. :)
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 15, 2011 #2


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    The string must be inextensible, otherwise there will be no impulse applied by the string at the instant it becomes stretched, so there will be no instantaneous change in velocity or angular velocity of the bar.

    If the string was extensible, of course its stiffness would affect the motion after it was stretched to length h, but not the motion at the instant when it reaches length h.

    Without using Lagrangian mechanics, it should be clear the string applies a vertical impulse to the rod, and after the impulse the rod is rotating around its end. So the problem can be solved by conservation of angular momentum about the end of the rod.

    Velocity before impact [itex]u = \sqrt{2gh}[/itex]
    Angular momentum before impulse = [itex]mul/2[/itex]
    Angular momentum after impulse = [itex]mvl/2 + ml^2\omega/12[/itex]
    where [itex]\omega = 2v/l[/itex]
    Which gives [itex]v = 3u/4[/itex] as in your solution.
  4. Oct 15, 2011 #3
    Thank you for taking the time to read my message, AlephZero. :smile:

    Another thing that's interesting about the book's solution, or equivalently, the one you presented with conservation of angular momentum, is that there is a difference in initial mechanical energy and final mechanical energy, which hints at a non-conservative force having done work. Is it the impulse force having done work on the rod to change it's kinematic situation?
  5. Oct 15, 2011 #4


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    Yes. The KE changes in any inelastic collision problem. Even in an elastic collision where the total KE of the system is unchanged, the impulsive forces transfer KE between different parts of the system
  6. Oct 15, 2011 #5

    D H

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    The string is infinitely stiff -- but in one direction only. The vertical velocity at the point on the bar where the bar is attached to the string instantaneously goes from [itex]\sqrt{2gh}[/itex] downward to zero at the instant the string becomes taut.

    You have to be careful when applying the conservation principles: What makes you think they do apply? For example, conservation of (mechanical) energy does not apply in the case of an inelastic collision. In fact, this is an inelastic collision. There is no difference between the string suddenly becoming taut a distance h below the release point and the very end of a bar hitting a solid, non-moving object a distance h below the release point.

    While conservation of angular momentum does apply to this problem, neither mechanical energy nor linear momentum are conserved. So why is angular momentum conserved? The answer lies in your Lagrangian. The Lagrangian that you formulated is what allows you to use conservation of angular momentum in the first place.
  7. Oct 15, 2011 #6
    I see, makes sense. :smile:

    Hm, very good points. :biggrin: I shall take them into account and see what the other parties think.
  8. Oct 15, 2011 #7


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    I'm not arguing against the general point you are making, but for this specific problem (and speaking as an engineer!) I would justify conservation of angular momentum of the bar about its end point (and also non-conservation of linear momentum) simply by drawing a free body diagram for the bar at the moment of impact.
  9. Oct 15, 2011 #8

    D H

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    Sure. From a freshman physics POV, conservation of angular momentum is a consequence of the force being parallel to the string, making r×F identically zero.

    What justifies the use of "force"? It certainly is not a fundamental concept. The Lagrangian / Hamiltonian formulation that the OP is now learning is a much deeper concept. It is the conservation laws that are fundamental in this formulation; forces and torques are derived quantities.

    So that begs the question: Why aren't energy and linear momentum conserved here? Explaining why energy does not appear to be conserved is easy. By looking only at mechanical energy we are only looking at part of the total energy picture. In an inelastic collision, some of the kinetic energy is converted to sound energy, temperature, used to bend or break the components (which may change potential energy), and so on. The underlying law is conservation of energy, not conservation of mechanical energy.

    Situations such as this where conservation of linear momentum appears not to hold are a bit more interesting. Such situations either (a) posit an infinite mass or (b) analyze things from the perspective of a non-inertial frame. The apparent failure of conservation of linear momentum is easily explainable in case (b) by the fact that working in non-inertial frames causes all kinds of havoc. Momentum is conserved in case (a); it's just not obvious that it is conserved.

    Suppose a blob of sticky putty of mass m is moving in free space at a velocity v relative to some target object of mass M and collides inelastically with that target object. In a frame initially at rest with the target object, the combined putty+object system has a post-collision velocity of [itex]\frac{mv}{M+m}[/itex]. The collision transfers [itex]\frac{M}{M+m}mv[/itex] momentum from the putty to the target object. In the limit M→∞, the post-collision velocity is zero. It appears that linear momentum has been lost. It hasn't. All of the momentum has been transferred to the target object but thanks to the use of an infinite mass, there is no outward sign that this transfer took place.
  10. Oct 15, 2011 #9


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    I think the key word there is "learning". This type of problem is certainly a good way to learn about Lagrangian mechanics, and that was evidently the OP's motivation for starting this thread. On the other hand, in the envronment that I work in (where we spend quite a lot of or time considering the real-world consequences of "rotating rods" impacting on other things) anybody who used Lagrangian mechanics to solve a problem as straightforward as this would certainly get some strange looks, and/or have it quietly pointed out that there is a quicker way to get the answer by trading off some of the math against "engineering common sense".

    But of course you need to practice in straightforward situations, so you can be confident in using Lagrangan mechanics where an average amount of "common sense" isn't enough to guarantee a correct solution every time.

    One small comment the solution in the OP: since (as DH already said) the string is only rigid in one direction, shouldn't the constraint equation be an inequality not an equality? Not that it will make any difference to the answer to this particular problem...
  11. Oct 17, 2011 #10
    AlephZero, D H, thanks a lot for the interesting discussion and interesting points you have raised.

    However, one thing still bothers me. AlephZero, you have said that the kinetic energy changes in this problem because it is an inelastic collision. Usually, in an inelastic collision, changes in kinetic energy are due to transfer of kinetic energy to different parts of the system that cause non-elastic deformations, heat, etc... In the instant the string delivers an impulse to the rod, the string doesn't deform, nor does the rigid rod, so it can't be due to non-elastic deformations.

    Heat, maybe? Perhaps where the string is linked to the rod has some heat building up?

    I also thought of some sort of wave motion being established in the string, but if it doesn't have mass, I don't really see how that could be possible.
  12. Oct 17, 2011 #11

    D H

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    One answer is that you essentially are asking here "What do the laws of physics say will happen when I violate the laws of physics?"

    So where are you violating the laws of physics in this setup? I count at least seven: You have (1) a massless string that (2) becomes infinitely rigid as soon as it becomes taut and that is connected with (3) infinite rigidity to some (4) infinitely rigid structure that is connected with (5) infinite rigidity to an (6) infinitely rigid and (7) infinitely massive Earth.

    One way to find where the energy is going is to relax these idealized conditions. A real string is not massless and does not become infinitely rigid when it becomes taut. Some of the energy will be transferred to the string in the form of vibrations and eventually transferred to the environment or to heat. The connections between string and structure and between structure and earth are not perfectly rigid, either. Some energy will be lost there. A real structure will be compressible, and the compression will convert some of that energy to heat. Finally, the earth is neither rigid nor infinitely massive. Some of the energy will be transferred to the Earth itself, making it ring by a tiny bit (and later) make it warm up by a tiny bit.
  13. Oct 17, 2011 #12
    I am well aware that there is no such thing as a body with infinite rigidity. However, within the framework of Classical Mechanics, where perfectly rigid bodies are allowed, this loss of energy is what surprised me. :biggrin:
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