Describe the movement of a string-pendulum with initial speed $v_0=2\sqrt{gL}$

Question

I posted this question here (physics.stackexchange) a few days ago, but our fellow physicians did not seem to like the question. I am giving it another try among the community of mathematicians!

The question was asked in a written test years ago, in a post graduate course. Therefore I assume it is possible to answer without the help of a computer. Nevertheless, all approaches are welcome.

The question is quite straightforward, it's all in the title. Suppose you have a classical pendulum of mass $m$, linked to the frictionless pivot by a weightless string (not the classical rod) of length $L$. The pendulum is initially in vertical position, in its equilibrium position. If you put it in motion with a horizontal speed $v_0=2 \sqrt{gL}$, what happens ? How can the movement be described?

I know for a fact (from a simulation) that I need to find the equations of the trajectory in the picture below. That is, the trajectory is circular (in red) up to a certain height, after which it becomes parabolic (in blue).

Approach #1:

I know the equation describing the movement of a pendulum is $$ \frac{d^2 \theta}{d t^2} + \frac{g}{L}\sin \theta = 0, $$ where $\theta$ is the amplitude of the pendulum. So I suppose one could solve the differential equation numerically with the initial conditions $$\theta(0)=0,\quad \frac{d \theta}{d t}(0) = \frac{2 \sqrt{gL}}{L}=2 \sqrt{\frac{g}{L}}$$ and find an approximate solution. However, there is no closed form possible, and this was probably not the expected approach in the written exam. This being said, I am interested in knowing if the approach is correct from a theoretical point of view.

Approach #2:

It is not very difficult to show that the tension of the string is given by $$ T(\theta) = mg \cos \theta + m L \left(\frac{d \theta}{dt}\right)^2 $$ where again, $\theta$ is the amplitude from the vertical position. Or in terms of $h$, the height of the pendulum (assuming $h=0$ in the initial position): $$ T(h) = mg \frac{h-L}{L}+ 4mg(1-\frac{h}{2L}) $$

The string becomes slack when $T(h)=0$, i.e., when $ h = \frac{5L}{3}$. At this point, the pendulum is only submitted to its weight, and the trajectory becomes parabolic. Would this be correct? How can we find the height at which the pendulum's trajectory becomes circular again (the bottom left part of the red trajectory in the picture).

Any input is welcome. Thanks!

Answer 1

Approach number 1 describes what I would call a classical pendulum, in which a weightless rod holds a point mass $m$ at a fixed distance $L$ from a frictionless pivot.

Approach number 2 is a better description of the case when the mass is suspended from a string. The first equation is written incorrectly; it should be $$ T(\theta) = mg \cos \theta + m L \left(\frac{d \theta}{dt}\right)^2. $$

I suspect this is just an error in the transcription from the formulas you wrote when you were working out this problem, since the next equation you derived appears to be correct.

You now know the exact point in space at which the weight's trajectory becomes parabolic and the exact velocity (speed and direction) of the weight at that instant. From this you can write an equation of the parabolic path in Cartesian $x$ and $y$ coordinates (or $x$ and $h,$ if you like). You can also write the equation of the circle in Cartesian coordinates. Solve the equations simultaneously, and you have the two points of intersection, one where the parabolic segment starts and the other where it ends.

What happens when the string becomes taut again is another question entirely. The idea that the weight would instantaneously resume its circular path with no instantaneous change in speed seems incorrect to me; the force exerted by the string is not perpendicular to the path of the weight at that point, and it seems to me the string must decelerate the weight. The question then is, how elastic is the string? Does it convert the kinetic energy of the weight to heat during the deceleration, or does it store that energy as a spring would and return it to the weight after the weight has stopped moving away from the pivot? Or is there a combination of the two effects? You might see the weight settle quickly into an oscillation along the circular path (but staying near the bottom of the circle), or you might see it "jump" back inside the circle in another parabolic arc. One thing I would definitely not believe is that the weight travels around the circular path to the same height as before and follows the same parabolic segment again.

Answer 2

The dynamic under constraint is \begin{align} m\ddot x &= -\lambda x\\ m\ddot y &= -mg-\lambda y\\ x^2+y^2&=L^2\\ x\dot x+y\dot y&=0\\ -λL^2-mgy+m(\dot x^2+\dot y^2)&=0 \end{align} where the last 2 equations are consequences of the first 3. This is valid as long as $λ>0$. The phase change point happens when $\dot x^2+\dot y^2=gy$. As the conserved energy of the system is $$ E=\frac{m}2(\dot x^2+\dot y^2)+mgy=\frac{m}2(4gL+0)+mg(-L)=mgL $$ this happens when $\frac32y_1=L$ or $y_1=\frac23L$, $x_1=\frac{\sqrt5}{3}L$ with velocities satisfying \begin{align} \sqrt5\dot x_1+2\dot y_1&=0\\ \dot x_1^2+\dot y_1^2&=\frac23 gL=\frac95\dot y_1^2\\ \implies \dot y_1&=\sqrt{\frac{10}{27}gL}\\ \dot x_1&=-\sqrt{\frac{8}{27}gL} \end{align} and the parabolic arc is, setting time to zero there, \begin{alignat}3 x&=x_1+\dot x_1t&&=\frac{\sqrt5}{3}L-\sqrt{\frac{8}{27}gL}\cdot t\\ y&=y_1+\dot y_1t-\frac g2 ⋅t^2&&=\frac23L+\sqrt{\frac{10}{27}gL}\cdot t-\frac g2 ⋅t^2 \end{alignat} For the return point to the circle one has to solve \begin{align} \newcommand{\red}[1]{\color{red}{#1}} \newcommand{\blu}[1]{\color{blue}{#1}} \newcommand{\grn}[1]{\color{green}{#1}} L^2=x_2^2+y_2^2&=\red{x_1^2}+\grn{2x_1\dot x_1t}+\blu{\dot x_1^2t^2}\;+\; \red{y_1^2}+\grn{2y_1\dot y_1t}+\blu{\dot y_1^2t^2} -\blu{gy_1t^2}-g\dot y_1t^3+\frac{g^2}4t^4\\ &=\red{L^2}-g\dot y_1t^3+\frac{g^2}4t^4\\ \implies t&=\frac{4\dot y_1}{g}=4\sqrt{\frac{10L}{27g}} \\ x_2&=\frac{\sqrt5}{3}L-\frac{16\sqrt5}{27}L=-\frac{7\sqrt5}{27}L\\ y_2&=\frac23L+\frac{40}{27}L-\frac{80}{27}L=-\frac{22}{27}L \end{align}

Numerical integration, the red point uses the computed coordinates $(x_2,y_2)$ and is indeed the intersection point