 #1
vinicius_linhares
 4
 0
 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
$$v=\sqrt{\frac{2}{m}}\sqrt{E_0mgf(x)}.$$
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:
$$
Z(x;x',E_0)=f(x')+f'(x')(xx')\frac{mg}{4}\bigg(\frac{1+f'(x')^2}{E_0mgf(x')}\bigg)(xx')^2
$$
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
$$\frac{mg}{2}\bigg(\frac{1+f'(x')^2}{E_0mgf(x')}\bigg)>f''(x')$$
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.
The object has initial energy $$E_0=\frac{mv^2}{2}+mgf(x),$$ then
$$v=\sqrt{\frac{2}{m}}\sqrt{E_0mgf(x)}.$$
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:
$$
Z(x;x',E_0)=f(x')+f'(x')(xx')\frac{mg}{4}\bigg(\frac{1+f'(x')^2}{E_0mgf(x')}\bigg)(xx')^2
$$
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
$$\frac{mg}{2}\bigg(\frac{1+f'(x')^2}{E_0mgf(x')}\bigg)>f''(x')$$
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.