I have seen two similar problems in Meriam's Dynamics book. I am not asking for solutions here (I have seen the solutions). I am confused as to why we treat energy differently in these two situations.

In both problems, energy loss due to friction is negligible.

In P4/90, energy is conserved for $0<X<L$ and energy is lost at $X=L$ due to whiplash effect. This problem has been discussed in the literature:

  • C. W. Wong, K. Yasui, "Falling chains", American Journal of Physics 74 (2006) 490; and the earlier
  • M. G. Calkin, R. H. March, "The dynamics of a falling chain I", American Journal of Physics 57 (1989) 154.

However, in P4/92 Energy is not conserved for $0<X<L$.

Why? What is the difference between the two?

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  • $\begingroup$ The first one says "also find the total loss Q" so energy is not conserved, you find the loss at the end. The first one says "without frictional resistance or interference" so it almost sounds like there is energy conservation assumed in that problem. Did you say this backwards? $\endgroup$ Commented Feb 27, 2023 at 18:38
  • $\begingroup$ @PoissonAerohead energy is conserved in the first problems (there are several theoritical and experimental articles on this case), energy is lost only at x=L by "whipping" otherwise in 0<x<L energy is conserved. In second problem, energy is not conserved in 0<x<L. $\endgroup$
    – Ebi
    Commented Feb 28, 2023 at 0:22

2 Answers 2


If you have access to the answer and this is incorrect, please let me know. I personally get

$$ v = \sqrt{\frac{2gx}{3}} $$

which is the non temporal solution (v as a function of x) to

$$ g\ddot{x} = \frac{d}{dt}\left( xv \right) $$

which is what I get as the Newton's 2nd Law equation for this system assuming the chain "enters with zero velocity." Please let me know if this is off base.

The resulting $x$, $v$ relation can actually be integrated too, giving the temporal answers:

$$ x = \frac{1}{6}gt^{2} $$


$$ v = \frac{1}{3}gt $$

Clearly the net effect of this looks like the local acceleration $g$ is reduced to a third the true value. Losses can be verified by calculating potential and kinetic energies with distance:

$$ \Delta PE = -gmh_{com} = -g \rho_{L}x * \frac{x}{2} = -\frac{g\rho_{L}x^{2}}{2} $$

$$ KE = \frac{1}{2}mv^{2} = \frac{1}{2}\rho_{L}x * \frac{2gx}{3} = \frac{\rho_{L}gx^{2}}{3} $$

where $\rho_{L}$ is the linear density and $h_{com}$ is the distance to the mass center of the moving portion of the chain.

The losses are then the balance of these changes, since they should sum to zero. For this I get

$$ E_{Loss} = \frac{\rho_{L}gx^{2}}{6} $$

The source of the loss can be conceptually understood by viewing each differential segment of chain at the top entering the falling line of chain as a fully inelastic collision where one body was initially stationary. As you know, when two bodies collide and stick together, there is mechanical (kinetic) energy loss. The most energy loss occurs when a moving tiny body collides and sticks to a massive body stationary, since virtually all the kinetic energy is lost. This is like a pebble landing on the earth. Even in the reverse scenario, there are still losses, although less. Thus, the falling chain is the large body, and it "impacts" the "small body" differential segment that "sticks to it" (since they are now moving with the same velocity).

I will play with it a bit, but I bet that these equations of motion and impact analogy can actually predict the same losses.

Note: regarding the first example, there are no losses in that system because the entire chain is released at once and subject to gravitational force. The entire chain begins accelerating immediately and the force from the constraint at the top is smooth and only acts to redirect the motion. When it reaches the bottom, the "whiplash" is basically a big inelastic collision. So you can think of the second example as having "differential whiplashes" at every step of the way, but the those are in the opposite sense. They speed up stationary chain rather than stop moving chain. Nevertheless, they are inelastic collisions.

Further Note: In Johnston and Beer (that one is my personal favorite) they have a problem like this in the Systems of Particles chapter where they compare your second set up to one where the chain is laid out straight (horizontally) on top and moving as a unit. The assumption is probably supposed to be that the chain redirects with no losses at the hole. In this case, the chain enters the system with the same velocity as the system and so does not need to be sped up. According to my description here, there is no "collision" as it enters. In this case, it probably conserves energy, the book does not have a solution to that one but I bet that is the point to having the two set ups.

  • $\begingroup$ your solution to 4/92 is correct. I also agree with your explanation on the energy loss in that problem. However, why can't we use the same reasoning for P4/90 and conclude energy is not conserve in that case too. Can't we say each segment of the right side falling chain (in P4/90) comes to rest by an inelastic impact with stationary chain in left? $\endgroup$
    – Ebi
    Commented Feb 28, 2023 at 3:16
  • $\begingroup$ Wow! I was expecting to make a fool of myself on Stack Exchange! I updated the answer at the bottom. $\endgroup$ Commented Feb 28, 2023 at 3:26
  • $\begingroup$ If the impact is inelastic as you pointed in your note, then energy should be lost. Isn't it? $\endgroup$
    – Ebi
    Commented Feb 28, 2023 at 3:58
  • $\begingroup$ Yes, it is lost in your "Problem 4/92." This is because when the segment chain at the top goes from zero speed to the current chain speed, that is, in effect, an inelastic collision. In "Problem 4/90", there is no such collision, because the whole chain is released at once and falls smoothly as a unit until the bottom. $\endgroup$ Commented Feb 28, 2023 at 4:19
  • $\begingroup$ Basically, as the first one falls, the curved portion has to unwind. If you want a deep answer, you would have to analyze that. I think you already found that in the literature? But that unwind should be smooth on the way down. $\endgroup$ Commented Feb 28, 2023 at 4:30

The simple answer is that in 4/90 you have a curve at the bottom of the chain with a tension of $\frac{1}{4}\rho v^2$. The force on the falling side of the chain is gravity plus this tension force. And the acceleration is $\frac{force}{mass}$. The force is increasing because $v$ is increasing. And the mass is decreasing because the falling side of the chain is getting shorter. So the acceleration increases at an exponential rate. This means that $v$ increases very rapidly towards the end of the fall of the chain like a whip. Since energy dissipation when accelerating mass from 0 to $v$ is proportional to $v^2$, virtually all the energy dissipation occurs at the end when x~L.

However, it's not correct that energy is conserved for 0<x<L. The energy dissipated is just much, much greater near L because the velocity and thus $v^2$ is far higher at that point.

In contrast, in 4/92 there is no loop of chain at the bottom with increasing tension pulling the chain down, so the rate of acceleration is much lower and the velocity does not increase nearly as fast as in 4/90. This means the energy dissipation is not concentrated at the very end of the fall, but is more spread out over the entire motion. In addition, the chain is not fixed at its end so a lot of the chain's energy will remain kinetic and not be dissipated at all.


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