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Consider a particle starting from the origin with some initial velocity and sliding along the paraboloid $z=−a(x^2+y^2)$ subject to a uniform gravitational force pointing in the direction of $-\hat{e}_z$.

Using the symmetry of the problem we can consider a two dimensional version. Namely, a particle sliding down a parabola $z=-ax^2$. Using basic equations of motion one can arrive at the conclusion that for any $$u_0>\sqrt{\frac{g}{2a}} \ , \ \ \text{where} \ u_0 \ \text{is the initial velocity}$$ the motion of the particle is free-fall. How could it be shown that for any $u_0$ satisfying the above inequality, the force due to the constraint is zero throughout the motion?

I attempted to derive equations of motion for the two-dimensional case as follows: $$\hat{\mathcal{L}}=\frac{1}{2}m\dot x^2-mgz-\lambda(-z-ax^2)$$ $$\frac{\partial\hat{\mathcal{L}}}{\partial x}-\frac{d}{dt}(\frac{\partial\hat{\mathcal{L}}}{\partial \dot x})=2\lambda ax-\frac{d}{dt}(m\dot x)=0$$ $$\frac{\partial\hat{\mathcal{L}}}{\partial z}-\frac{d}{dt}(\frac{\partial\hat{\mathcal{L}}}{\partial \dot z})= \ \ \ \ \ \ \ \ \ \ -mg+\lambda =0$$ I had hoped that I would be able to show that $u_0=\sqrt{\frac{g}{2a}}$ is the maximum accepted value. However, the solutions to this differential equation are of exponential nature. Where have I gone wrong in my method?

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  • $\begingroup$ Do you mean that for any $u_0$ satisfying that inequality, the particle loses contact with the parabola? $\endgroup$
    – Sal
    Oct 13 at 19:59
  • $\begingroup$ Yes, that means that the force applied by the constraint is zero. I do not know how to show this mathematically though. $\endgroup$
    – benmcgloin
    Oct 13 at 20:02
  • $\begingroup$ Once the particle leaves the parabola, you're dealing with a new system (where there is no constraint force). If you try to find the constraint force for the motion 'off the parabola' with the current equations, you'll get a nonzero answer- the constraint (and Lagrangian) don't know about the particle leaving the parabola at all. You can find a point in the trajectory where the constraint force vanishes- this is where it leaves the parabola $\endgroup$
    – Sal
    Oct 13 at 20:12
  • $\begingroup$ As I understand it, the particle leaves the parabola at the origin provided the $\hat e_x$ component of velocity is large enough. Is this not the point where the constraint vanishes due to the particle not being in contact with the parabola anymore? $\endgroup$
    – benmcgloin
    Oct 13 at 20:15
  • $\begingroup$ Sure, that's the point where it leaves. As soon as it does, the motion is governed by a new Lagrangian $\endgroup$
    – Sal
    Oct 13 at 20:16
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Assuming that the ball starts rolling at the top with some initial velocity $v_0>0$, the correct Lagrangian for the system is given by

$$\mathcal{L}=\frac{1}{2}m(\dot{X}^2+\dot{Z}^2)-mgZ-\lambda(t)(Z+aX^2)$$

The Lagrange multiplier here enforces the constraint and it is equal to the magnitude of the normal force exerted by the parabolic hill, for as long as the ball rolls on it. Writing down the equations of motion and solving for $\lambda$ we obtain the form

$$\lambda(t)=\frac{2ma\dot{X}^2-mg}{1+4ma^2X^2}$$

Suppose there is a time $T$ for which the ball detaches from the hill, then at that point it must be the case that $\lambda(T)=0$ and thus we obtain unsurprisingly that detachment happens when the horizontal velocity exceeds

$$\dot{X}_\max=\sqrt\frac{g}{2a}$$

To compute the ejection point given the initial condition we can use the conservation of energy (energy is of course conserved because the unconstrained Lagrangian does not depend on time!):

$$\frac{1}{2}m(\dot{X}^2+\dot{Z}^2)+mg Z=\frac{1}{2}m v_0^2$$

Because of the constraint it is true that at $\dot{Z}=-2aX\dot{X}$ and substituting everything in we obtain

$$\dot{X}^2=\frac{v_0^2+2gaX^2}{1+4a^2X^2}=\frac{g}{2a}-\frac{g/2a-v_0^2}{1+4a^2X^2}$$

Now if we assume that $v_0^2<\frac{g}{2a}$ we readily see that $\dot{X}^2\to g/2a$ as $X\to\infty$ and because the function on the RHS is monotonically increasing we conclude that the ball never leaves the parabolic slope, and therefore the exit time is $T=\infty$. However, trivially we note that if $v_0^2\geq g/2a$ the exit time is $T=0$ and the ball performs unconstrained motion thereafter. This interesting effect of the ball never leaving the surface for a non-trivial interval of initial velocities owes it's appearance to the fact that we are using free-fall trajectories to constrain a freely falling object. If the ball was rolling on a different curve falling faster than $-ax^2$ like a circle for example, there would be a finite exit time.

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  • $\begingroup$ It is interesting to me that the velocity tends towards a constant as $x \to \infty$. Even with the monotonic nature of the paraboloid, I would think that given the absence of friction, the velocity should tend towards infinity regardless of the trajectory picked? $\endgroup$
    – benmcgloin
    Oct 14 at 7:17
  • $\begingroup$ Also, how can it be asserted that $\lambda$ is the normal force? Is there some way to prove this analytically? $\endgroup$
    – benmcgloin
    Oct 14 at 9:54
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    $\begingroup$ My intuition was the same as yours initially, but think about it this way: the parabolic surface becomes very steep as $X$ grows so it shouldn't be exerting any force on the ball for large $X$. The particle has to perform a free fall asymptotically but also adhere to the surface so the only option is for the velocity to tend to rhe liniting value $\sqrt{g/2a}$. $\endgroup$ Oct 14 at 15:06
  • $\begingroup$ Yes just show that $m\ddot{\mathbf{r}}+m\mathbf{g}=\lambda(t)\mathbf{n}$ where $n$ is a normal vector to the parabola (not necessarily normalized). You can show this from the EL equations for any general form of the constraint $f(x,z)=0$. $\endgroup$ Oct 14 at 15:10

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