Finding where a ball loses contact with an exponentialy modled hill given there is no friction.

Question

I have an answer to the following question, but I have no easy way to verify it and would like some feedback, I'll go through my whole process.

Imagine a perfectly smooth hill moddled as a $y = -e^x$ curve. We want to study the movement of an object placed at $x = 0$ assuming no air resistance.

Notice that since the curve is exponential, the object will eventually lose contact will the hill and enter parabolic motion, we have to find the x-coordinate of that point.

Firstly I wanted an equation that describes the speed of the object as a function of x. Which is a good time to mention that for all my calculations the value of g will be positive 9.81 and I will add negatives signs when appropriate. $$-mg\Delta h = \frac 12 mv^2$$ $$-2g\Delta h = v^2$$ We know that $\Delta h$ is final y - initial y which will be $$\Delta h = -e^x - (-e^0)$$ $$\Delta h = 1 -e^x$$

So returning to our expression from before $$v^2 = -2g(1 - e^x)$$ $$v^2 = 2g(e^x - 1)$$ or $$v = \sqrt{2g(e^x - 1)}$$

Now comes the hard part of the question. Anything I say from here on might be wrong as it is all my reasoning and I haven't wanted to check it yet (wouldn't be fun).

In order for an object lose contact and go into parabolic motion, the floor under it must 'disappear' by having a decrease in gradient which is greater than that of a parabola. So I conjecture that the ball will lose contact when $\frac {d^2 y}{dx^2}$ of the equation that describes the surface it is rolling on is less than (greater value but negative) that of the parabola $y=ax^2$ that would result from the parabolic motion of the object, where 'a' will be negative and depend on its horizontal velocity and nothing else.

Now we need a way to find 'a' so we can compare it to our equation $y = -e^x$ Through some simple but tedious mechanics I arrived at this expression: $$a = \frac {-g}{2v^2}$$ where v is the horizontal velocity. Notice that since there is only downwards acceleration in parabolic motion, the vertical speed is completely irrelevant as having a different vertical speed will simply describe a different point on the same parabola.

Finally, we have to find the horizontal speed of the object as a function of x. We already have an expression for the magnitude of the velocity $$v = \sqrt{2g(e^x - 1)}$$ so in order to find the horizontal component all we need is some simple trigonometry. The gradient of the velocity vector with respect to the coordinate axis will be the same as that of the curve while it is still in contact, and the gradient of the curve is $\frac {dy}{dx} = -e^x$ as it is exponential. The gradient is the vertical component divided by the horizontal, in other words it is $tan(\alpha)$ where $\alpha$ is the acute angle with the x-axis.

If we let i be the horizontal component of velocity and j be the vertical component, then while the object is still in contact with hill.

$$e^x = \frac {|j|}{i}$$ $$|j| = e^x i$$

By Pythagoras

$$i^2 + j^2 = v^2$$ $$i^2 + e^{2x} i^2 = v^2$$ $$i^2 = \frac {v^2}{e^{2x} + 1}$$

Which we then substitute our original expression for v into to get $$i^2 = \frac {2g(e^x-1)}{e^{2x}+1}$$

Using our equation for a: $$a = \frac {-g}{2i^2}$$ $$a = \frac {-g}{2\frac {2g(e^x-1)}{e^{2x}+1}}$$ $$a=\frac {-(e^{2x}+1)}{4(e^x-1})$$

Finally for the last step we combine all our equations and compare the second derivatives. We want to find the point at which the second derivate of the curve of the hill will be equal to the second derivative of the parabola that would be formed if the object were to enter parabolic motion with its current horizontal speed. $$\frac {d^2}{dx^2}(-e^x) = \frac {d^2}{dx^2}(ax^2)$$ $$\frac {d^2}{dx^2}(-e^x) = \frac {d^2}{dx^2}(\frac {-(e^{2x}+1)}{4(e^x-1})x^2)$$ $$-e^x = \frac {-(e^{2x}+1)}{2(e^x-1)}$$ $$2e^{2x} - 2e^x = e^{2x} +1$$ $$e^{2x} - 2e^x - 1 = 0$$ Now we solve the quadratic and we get $$e^x = 1\pm \sqrt 2$$ And since we are only interested in the positive answer $$x = \ln{(1 + \sqrt2)}$$

This answer looks very nice and seems pretty, but my problem is that I simply have no way to verify it. I don't know how I feel about the fact that g cancels out at the end, which means that the force of gravity was irrelevant from the start (assuming it's non-0 and downwards)

If anyone spots any mistakes in my logic or can provide a simulation to visualise it I would be very grateful.

Answer 1

Your calculations and intuition are correct, given the setup of a constant gravitational field with no friction. I am going to demonstrate that for a general hill of the form $y=f(x)$, the condition for the ball to stay on the surface boils down to the question of whether a specific equation has a solution. Finally, I will show that this equation is identically satisfied for parabolas, which solidifies the status of parabolas as edge cases: anything steeper than a parabola will result in a finite exit time for the particle, while anything flatter should yield an infinite exit time (aka the particle stays on the surface for arbitrarily long times).

For convenience we will have to use the Lagrangian formalism. This problem is a typical Lagrangian mechanics problem with a constraint; a correct Lagrangian density for the description takes the Lagrange multiplier form

$$\mathcal{L}=T-V+\lambda C=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-mgy-\lambda(t)(y-f(x))$$

where $T$ is the kinetic energy, $V$ the potential energy and $C$ is the constraint equation, in this case $C=y-f(x)=0$. The Lagrange multiplier $\lambda$ has a physical meaning as well; it is proportional to the magnitude of the normal force induced on the particle in order to keep it on the hill. Therefore we may note that if the normal force is zero at any point, the constraint is no longer satisfied necessarily. For many curves of the above form, this condition (the first zero of the normal force) signifies that the particle has exited the hill. It may very well be the case that for certain hill shapes the normal force is only momentarily zero and then the particle is allowed to continue its motion along the surface), but we will disregard such an eventuality. The equations of motion for this Lagrangian read

$$m\ddot{x}=\lambda f'(x)\\m\ddot{y}=-mg-\lambda$$

which shows that the Lagrange multiplier is related to the normal force by

$$\lambda(t)=-\frac{N(t)}{\sqrt{1+f'(x)^2}}$$

Hence we can focus on the zeroes of the Lagrange multiplier from now on. To calculate the Lagrange multiplier we need to use a first integral of motion so we turn to the energy, which is conserved. It takes the simple form

$$\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+mgy=E$$

The constant $E$ can be determined from initial conditions (taking the constraint into account) to be

$$E=mgf(0)+\frac{1}{2}mv_0^2(1+f'(0)^2)$$

Also, substituting the constraint into the energy equation allows us to find an expression for the velocity component in the x direction:

$$\dot{x}^2=2\frac{E/m-gf(x)}{1+f'(x)^2}$$

Taking a derivative of this we can find $\ddot{x}$ and hence allows us to solve for the Lagrange multiplier, which reads

$$-\lambda(1+f'(x)^2)^2=mg(1+f'(x)^2)+2f''(x)(E-mgf(x))$$

A zero occurs if and only if the RHS has a zero for some positive $x$. In your case, $E=-mg~,~ f=f'=f''=-e^x$ and therefore the equation reads

$$e^{2x}-2e^x-1=0\Rightarrow x=\ln(1+\sqrt{2})$$

which represents the exit point for the particle. Finding the exit time requires an integration of the velocity , which is not impossible yet tedious, but it is clear that that time is finite (there are no singularities in the integrand). Finally we will ask the question: Are there any hills for a given set of initial conditions that allow the particle to travel freely? We know that a parabolic hill following exactly the unconstrained motion of the particle should be the answer, but we can verify this is the only solution by explicitly solving the equation on the RHS: First slightly rearrange to obtain

$$f'^2-2f''f=-\frac{2E}{mg}f''$$

Dividing by $f'^3$ provides the correct integrating factor for this equation, which reads after integration

$$f=\frac{E}{mg}+A^{-2}f'(x)^2$$

which can be separated and integrated to yield

$$f(x)=\frac{E}{mg}+\frac{A}{2}(x-x_0)^2$$

Here, $A, x_0$ may be obtained by the initial conditions $f(0)=y_0, f'(0)=\frac{v_{0y}}{v_{0x}}$ and hence the resulting curve is unique, and is a parabola. To conclude this analysis, we note that hills that allow for $f'^2(x)-2f''(x)f(x)+\frac{2E}{mg}f''(x)>0$ in the beginning of the trajectory will have nonzero exit time, while those that don't, will force the particle to exit the curve immediately.

Answer 2

Calling $p=\{x(t),y(t)\}$ we have the lagrangian

$$ L(\dot x,\dot y, x, y,\lambda)=\frac 12 m \|\dot p\|^2-m g y+\lambda(y+e^{x}) $$

and the movement equations are obtained from

$$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0 $$

or

$$ \cases{ m \ddot x(t)-\lambda e^{x(t)}= 0\\ g m-\lambda +m \ddot y(t)=0 \\ e^{x(t)} \ddot x(t)+e^{x(t)} \dot x(t)^2+\ddot y(t) = 0} $$

the last equation is obtained from $y + e^{x} - 1=0$ and after solving for $\{\ddot x, \ddot y, \lambda\}$ we have

$$ \cases{ \ddot x(t)= -\frac{e^{x(t)} \left(g+e^{x(t)} \dot x(t)^2\right)}{e^{2 x(t)}+1}\\ \ddot y(t)= \frac{e^{x(t)} \left(g e^{x(t)}-\dot x(t)^2\right)}{e^{2x(t)}+1}\\ \lambda = -\frac{m \left(e^{x(t)} \dot x(t)^2-g\right)}{e^{2 x(t)}+1}\\ \dot x(0) = x(0) = \dot y(0) = 0, y(0)=-1 } $$

this ode system does not have a closed form. Here $\lambda$ is associated to the normal force against the restriction path. Solving for $g = 9.8, m = 1$ we obtain $\lambda(t)$. As $\lambda$ is proportional to the normal force which maintains the ball attached to he path, a signal inversion indicates a detachment at $t_d\approx 0.6713$[s]. Attached a plot showing $x(t)$ in blue, $y(t)$ in red and $\lambda(t)$ in black.

Now to find the coordinate where the material point losses contact we proceed as follows. Considering $f(x) = 1-e^{-x} = y$ and knowing that the curvature radius for $f(x)$ is given by

$$ R(x) = \frac{(1+f'(x)^2)^{\frac 32}}{f''(x)} $$

the material point losses contact when

$$ m\frac{v^2}{R}\ge m g\cos\theta $$

and here

$$ \cases{ v^2 = 2gy\\ \cos\theta = \frac{1}{\sqrt{1+f'(x)^2}} } $$

After substitutions we have

$$ \frac{2 e^x (\sinh (x)-1)}{\left(e^{2 x}+1\right)^{3/2}}\ge 0 $$

or

$$ x \ge \ln\left(1+\sqrt{2}\right)\approx 0.881374 $$

now considering the detachment time $t_d$ we have $x(t_d)\approx 0.881374$