Analysis of the Ground Function: f(x) with $$f''(\bar{x})=0$$

[vinicius_linhares]

Homework Statement

(This is my first quest in this forum, trying to understand, thanks for the patience :) )

I saw a car going up a hill with format of a gaussian and I wonder:
What is the energy needed for a car leaves the ground in a certain point going up the gaussian hill.
(To this problem I imagine half of a gaussian curve and the object moving from the flat left region to the middle high and smooth region)

Relevant Equations

$$\frac{d^2\vec{r}}{dt^2}=\vec{F_R}$$

If f(x) is the function of the "ground": My first assumption is that in a certain $$\bar{x}$$, $$f''(\bar{x})=0$$, and from that point I will analyse the situation.

The object has initial energy $$E_0=\frac{mv^2}{2}+mgf(x),$$ then
$$v=\sqrt{\frac{2}{m}}\sqrt{E_0-mgf(x)}.$$
In each point the inclination gives:
$$cos\theta(x)=\frac{1}{\sqrt{1+f'(x)^2}};$$$$ sin\theta(x)=\frac{f'(x)}{\sqrt{1+f'(x)^2}};$$
My way of thinking was: in a certain point x' of the curve, if we cut out the right part of the function in that point, the object will travel as with a projectile motion. So in that point x', in this notion, we have a set of projectile paths Z(x;x',E_0) differing by the initial energy E_0.

**For a certain E_i this Z(x;x',E_i) will be greater than f(x) near the x' point, so the cutting off the right part is no longer needed and it will mean that the object leaves the ground at that point.**

So I started calculating.
Projectile motion from x':

$$x(t)= x'+v_x(x')t=x'+v\cos(\theta) t$$
$$z(t)=f(x')+v_z(x')-\frac{gt^2}{2}$$
$$=f(x')+v\sin(\theta)-\frac{gt^2}{2}$$

Isolating t from the x(t) equation and plugging in the sine and cosine definitions above we get:
$$
Z(x;x',E_0)=f(x')+f'(x')(x-x')-\frac{mg}{4}\bigg(\frac{1+f'(x')^2}{E_0-mgf(x')}\bigg)(x-x')^2
$$
Now for leaving the ground the condition I made and it is the part of the reasoning Im not sure is that:
$$Z(x'+\epsilon;x',E_0)>f(x'+\epsilon)$$
The projectile path in that point together with expanding the function to second order gives
$$-\frac{mg}{2}\bigg(\frac{1+f'(x')^2}{E_0-mgf(x')}\bigg)>f''(x')$$
But it leaves me with
$$E_0<mgf(x')-\frac{mg}{2}\frac{(1+f'(x')^2)}{f''(x')}$$
And at this point I just think I made some mistake somewhere but I can't get were.

If the visualization don't work for you I can't help in the comments to visualize.

 

[TSny]

You have an interesting approach! I get the same result by using the standard method of finding the point where the normal force vanishes; except the nature of the inequality in your last inequality should be >. Think about the sign of $f''(x)$ once you're past the inflection point of $f(x)$. What happens to the nature of the inequality when you divide both sides by $f''(x)$?

 

[vinicius_linhares]

You have an interesting approach! I get the same result by using the standard method of finding the point where the normal force vanishes; except the nature of the inequality in your last inequality should be >. Think about the sign of $f''(x)$ once you're past the inflection point of $f(x)$. What happens to the nature of the inequality when you divide both sides by $f''(x)$?

Wow nice!! When you get bored can you write your approach? I , of course, tryed this vanishing normal approach but I didn't get much.

 

[vinicius_linhares]

You have an interesting approach! I get the same result by using the standard method of finding the point where the normal force vanishes; except the nature of the inequality in your last inequality should be >. Think about the sign of $f''(x)$ once you're past the inflection point of $f(x)$. What happens to the nature of the inequality when you divide both sides by $f''(x)$?

Yes! I really thought that the inequality was switched but didn't found my error. So you say I should think of $$f''(x')$$ as a negative number since it was the assumption and when I multiply both sides the inequality turns to a meaningful way? I will try it.

 

[vinicius_linhares]

You have an interesting approach! I get the same result by using the standard method of finding the point where the normal force vanishes; except the nature of the inequality in your last inequality should be >. Think about the sign of $f''(x)$ once you're past the inflection point of $f(x)$. What happens to the nature of the inequality when you divide both sides by $f''(x)$?

So it worked out. The very strange thing that it leads, assuming only positive velocity, is that the velocity for a object leaves the ground at $x'$ is independent of the mass of the body!!! Do you follow that conclusion.

 

[TSny]

Yes! I really thought that the inequality was switched but didn't found my error. So you say I should think of $$f''(x')$$ as a negative number since it was the assumption and when I multiply both sides the inequality turns to a meaningful way? I will try it.

The car can only leave the hill at points where the hill is concave downward; i.e., $f''(x)< 0$. So, it might be less confusing when you're working with inequalities to write $f''(x) = - |f''(x)]$.

 

[TSny]

So it worked out. The very strange thing that it leads, assuming only positive velocity, is that the velocity por a object leaves the ground at x' is independent of the mass of the body!!! Do you follow that conclusion.

Yes, it's not surprising that the mass cancels out. Here's an outline of the normal force approach.

The radius of curvature at a point $\left(x, f(x)\right)$ of the hill is given by $$R =\frac{ \left[1+f’(x)^2\right]^{3/2}}{|f’’(x)|}.$$ At this point, the car may be considered as moving along a circle with this radius. Then Newton’s second law gives $$mg \cos \theta – N = \frac{mv^2}{R}$$ where $\theta$ is the angle satisfying $\tan \theta = f’(x)$. So, $\cos \theta = \large \frac 1 {\left[1+f’(x)^2\right]^{1/2}}$. Note that $m$ will cancel when $N$ goes to zero.

You have your energy expression for $v$. Substituting these expressions for $R, \cos \theta$ and $v$ into the 2^nd law equation, you can get $N$ in terms of $E_0$, $f(x)$, $f’(x)$, and $f’’(x)$. Setting $N = 0$ gives the relation between $E_0$ and the point $x$ at which the car leaves the surface of the hill.