This is essentially a better formatted version of this question which was put on hold (also by myself), mainly for being a HW&E question without any research effort by the OP and poor formatting.

I tried solving the problem, but don't seem to be able to and want to pose it as a new question.

Ball, spring and wall.

A mob of mass $m$ is connected to a massless Hookean spring of length $L_0$ and constant $k$, itself connected by a frictionless hinge at $O$ to a vertical wall. At $t=0$ the bob is released with the spring horizontal and unstressed.

What's the trajectory of the bob and its speed when it impacts the wall?

I tried several approaches without avail:

  1. Cartesian free body diagram and force balances to set up equations of motion in $x$ and $y$-directions. That became very complicated due to the expression of extension of the spring.

  2. Cartesian Hamiltonian: same problem.

  3. Polar coordinates Hamiltonian: that got very complicated after time derivation.

The most promising seems setting up equations of motion in polar coordinates.

$O$ is the origin, the horizontal through it the polar axis and $\theta$ the angle. $R$ is the distance of the bob to $O$.

For the equation of rotation I get:



For translation along the line of the spring, I get:




$\text{Eq.1}$ and $\text{Eq.2}$, if correct, would be a system to two non-linear DEs.

But is it and if so, how to proceed from there?


closed as off-topic by ACuriousMind, user36790, John Rennie, Kyle Kanos, DilithiumMatrix Dec 10 '15 at 16:20

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  • $\begingroup$ You've shown work, but where's the conceptual question? $\endgroup$ – ACuriousMind Dec 9 '15 at 3:32
  • $\begingroup$ @ACuriousMind: perhaps there isn't one. Perhaps it's a case where rotation and translation seem to create a difficult combination. I'm asking to leave this question to stand unflagged for a bit and see if some one can at least throw some light on it. I maybe wrong but easy this is certainly not. One could argue that 'why doesn't Hamiltonian with time derivation not work here?' is conceptual. $\endgroup$ – Gert Dec 9 '15 at 3:38
  • 2
    $\begingroup$ A rigid pendulum with large deflection already has an intractable equation of motion. I can't see it getting better by adding the spring... $\endgroup$ – Floris Dec 9 '15 at 3:46
  • $\begingroup$ @Floris: I have to say that that thought also occurred to me: no small angle approximation here. Does the rigid pendulum with large deflection has numerical solutions that we know of? $\endgroup$ – Gert Dec 9 '15 at 3:48
  • 1
    $\begingroup$ A pretty complete analysis is given at depts.washington.edu/amath/wordpress/wp-content/uploads/2014/01/… (50 pages). I don't vouch for its accuracy - but take a look. $\endgroup$ – Floris Dec 9 '15 at 3:48

The form you have for the equation of rotation holds if $R=$const., otherwise it is necessary to account for $\dot R \neq 0$.

In polar coordinates, as you have, the tangential velocity is $v_\theta = R\dot\theta$, so for tangential motion we have instead $$ mg\cos\theta = m\dot v_\theta = m R\ddot\theta + m\dot R\dot\theta $$

This given, let's check that energy is an integral of motion. Multiply the above by $R\dot\theta$, $$ mgR\dot\theta\cos\theta = mR^2\dot\theta\ddot\theta + m R\dot R \dot\theta^2 $$ and rearrange as $$ mgR\frac{d}{dt}\sin\theta = \frac{1}{2}mR^2\frac{d}{dt}\dot\theta^2 + \frac{1}{2}m\frac{dR^2}{dt}\dot\theta^2\\ \frac{d}{dt}\left(mgR\sin\theta\right) - mg\dot R\sin\theta = \frac{d}{dt}\left( \frac{1}{2}m R^2\dot\theta^2\right)\\ \frac{d}{dt}\left( \frac{1}{2}m R^2\dot\theta^2 - mgR\sin\theta\right) = - mg\dot R\sin\theta $$ Get rid of the rhs by using the radial equation $$ m\ddot R = mg\sin\theta + k(L_0 - R) $$ multiplied by $\dot R$. This gives $$ \frac{d}{dt}\left( \frac{1}{2}m R^2\dot\theta^2 - mgR\sin\theta\right) = - m\dot R\ddot R + k(L_0 - R)\dot R\\ \frac{d}{dt}\left( \frac{1}{2}m R^2\dot\theta^2 - mgR\sin\theta\right) = - \frac{d}{dt}\left( \frac{1}{2}m \dot R^2 + \frac{1}{2}k(L_0 - R)^2 \right) $$ Rearranging again leaves us with $$ \frac{d}{dt}\left[ \frac{1}{2}m \left(R^2\dot\theta^2 + \dot R^2\right) - mgR\sin\theta + \frac{1}{2}k(L_0 - R)^2 \right] = 0 $$ Recognize that $R^2\dot\theta^2 + \dot R^2 = \dot{\bf r}^2 \equiv v^2$ is the instantaneous velocity squared and the above reduces to $$ \frac{d}{dt}\left[ \frac{1}{2}mv^2 - mgR\sin\theta + \frac{1}{2}k(L_0 - R)^2\right] = 0 $$ as expected. Moreover, given that the initial energy is null, we actually have $$ \frac{1}{2}m \left(R^2\dot\theta^2 + \dot R^2\right) - mgR\sin\theta + \frac{1}{2}k(L_0 - R)^2 = 0 $$

At this point I don't think this is an integrable system. I only know that eventually one could obtain an EOM in terms of $R$ only from the energy conservation eq. above and the EOMs. Not sure this would help in regards to something tractable though, but didn't look much into it either.

  • $\begingroup$ Referring to the this video: Swinging spring it appears that after a steady state is reached $R$ is maximum at the equilibrium and minimum at the ends, keeping this in mind and referring to the diagram above, can we make a approximate a relation such as $R=k\theta$, for some constant $k$? $\endgroup$ – Oswald Dec 9 '15 at 10:39
  • $\begingroup$ @udv: thank you for pointing out the error of my ways. Sapere Aude! :-) $\endgroup$ – Gert Dec 9 '15 at 14:40
  • $\begingroup$ @Vishwaas: thanks for the video. What would be worth looking at is the system at low $\theta$. $\endgroup$ – Gert Dec 9 '15 at 14:45
  • $\begingroup$ @Gert Welcome . $\endgroup$ – udrv Dec 9 '15 at 23:46
  • $\begingroup$ @Vishwaas: I can't see anything against $R\sim\theta$, it could work. The case of small $\theta$ mentioned by Gert is worth a look too, but I would actually go for both small $\theta$ and small deviations of $R$ from $L_0$ (as independent variable). Note that even in this case the EOMs remain nonlinear and nontrivial due to the $\dot R \dot\theta$ coupling, so the dynamics could be pretty interesting. $\endgroup$ – udrv Dec 9 '15 at 23:47

Interesting problem. I set up the problem in polar coordinates with $\theta$=0 defined as pointing downward and wrote down the Euler-Lagrange equations. Since udrv already cranked through much of the math, I'll just state the results below for the two equations:

$g m r(t) \sin (\theta (t))+2 m r(t) r'(t) \theta '(t)+m r(t)^2 \theta ''(t)=0$ $-g m \cos (\theta (t))+k (r(t)-\text{$r_0$})+m r''(t)-m r(t) \theta '(t)^2=0$

Mathematica couldn't find an analytical solution for this pair of coupled differential equations, but it was able to numerically solve the equations using the initial conditions r(0)=$r_0$, r'(0)=0, $\theta$(0)=$\pi$/2, and $\theta$'(0)=0 when given particular numerical values for $r_o$, m, g, and k.

  • $\begingroup$ Not sure what you mean by $\theta=0$? Does this mean $\theta=\pi /2$ with the 'old' $\theta$? Also, what was the outcome? Remember, the question really concerns itself with only part of the trajectory: until the mass first crosses $y$, that is 'old' $\theta > \pi /2$. For +10 creds. :-) $\endgroup$ – Gert Dec 10 '15 at 2:51
  • $\begingroup$ Yes, $\theta$=$\pi$/2 corresponds to the 'old' $\theta$=0. I redefined $\theta$ so that it agrees with the definition of $\theta$ commonly used for standard rigid pendulum problems. If I had been able to find an analytic solution, I would have posted that but offhand I don't see any such solution or any way to simplify the problem further. I just tossed the Euler-Lagrange equations up here in case anyone wants to investigate the problem further. $\endgroup$ – Samuel Weir Dec 10 '15 at 5:07

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