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## Banks, Coasters and Planets

I continue my own exploration of motion on curves.

## Banks

As Newton’s second law proclaims a body in uniform rectilinear motion will continue so unless acted upon by a force.  Directing frictional force allows us to negotiate curves. Banked curves provides a central force that can allow us to negotiate curves.  The above free body diagram aids in determining the maximum speed that the bank will allow us to negotiate the curve. Let $\theta$ be the angle of the bank, $f$ represent the frictional force ($\mu$ coefficient of friction) ,$N$ the normal force from the surface.

$N \cos\theta= mg + \mu N\sin\theta$

$N=\frac{mg}{\cos\theta-\mu\sin\theta}$

The central force arises from the horizontal component:

$\frac{m v^2}{r}=N\sin\theta + \mu N\cos\theta$

$\frac{m v^2}{r}=\frac{mg (\sin\theta+\mu\cos\theta)}{\cos\theta-\mu\sin\theta}$

Solving for $v$:

$v = \sqrt{\frac{rg (\sin\theta+\mu\cos\theta)}{\cos\theta-\mu\sin\theta}}$

Or equivalently (given $0\leq\theta<\pi/2$),

$v =\sqrt{\frac{rg (\tan\theta +\mu)}{1-\mu \tan\theta}}$

For $\mu =0$ this reduces to:

$v =\sqrt{rg\tan\theta}$

Effect of coefficient of friction for circular arc of radius 50 m and bank angle 10 degrees.

## Roller Coasters

Loops of roller coasters pose challenges for engineers, e.g. forces imposed on those riding on the carriage and speeds required for carriages to remain constrained to the tracks.  Some insights can be obtained by looking at motion around a circular loop. The following free body diagram will assist these considerations. Let $F_c$ represent the centripetal force, $\theta$, the angular position of cart ($\theta =0$ cart at the bottom and $\theta =\pi$ cart at the top) $N$ be the normal or resultant force.

There are three possible outcomes for the car on the circular loop:

• it completes the loop
• it rolls up the loop to a point, stops and then slides down
• it rolls up the loop and loses contact with the track and falls off
The first two scenarios can be explored and between these two bounds the third scenario follows.
Let $v_0$ be the speed the car enters the loop, $v$, the speed at $\theta$ and$r$ be the radius of the circle.

### Completing the loop:

From conservation of energy considerations:
$\frac{mv_0^2}{2} = \frac{mv^2}{2} + mgr(1-\cos\theta)$
For the car to maintain continuity with the track the normal force must be positive:
$F_c=N- mg\cos\theta$
$N=\frac{mv^2}{r}+mg\cos\theta$
$N$ is minimal when $\theta =\pi$, therefore, $N>0$ for the whole loop , yields
$v^2>rg$ at$\theta =\pi$
Therefore, the entry speed must satisfy,
$v_0^2> rg +2rg (1-\cos\pi)=5rg$

### Rolling up, stopping and sliding down

This requires the $v=0$ and $N>0$. This requires: $mg \cos\theta >0$, i.e. $-\pi/2<\theta<\pi/2$. Using the conservation of energey considerations,$v_0^2<2rg$.

### Intermediate Scenario

Therefore, entry speeds between $\sqrt{2rg}$ and $\sqrt{5rg}$ the car will lose contact with the track.

### Clothoid Loop

There are disadvantages to circular loops in terms of speeds needed and accelerations experienced by travellers.
One frequently cited loop is the “clothoid” loop. This is a family of loops where the curvature varies linearly along the loop.
This can  be most readily understood by parametrizing the curve using the arc length parameter. Let $s$ be the arc length parameter. Any curve can then characterized as follows:
$\begin{array}{ccc} \frac{dx}{ds}&=&\cos\theta(s)\\\frac{dy}{ds}&=&\sin\theta(s)\\\frac{d\theta}{ds}&=&\kappa(s) =\text{curvature}\end{array}$
For the clothoid curve,
$\begin{array}{ccc} \frac{d\theta}{ds}&=& bs\\\theta&=& bs^2/2 +\theta_0\\x &=&\int_0^s \cos(bu^2/2 +\theta_0)\,du\\y&=&\int_0^s\sin(bu^2/2+\theta_0)\, du\\\rho&=&\text{radius of curvature} =\frac{1}{bs}\end{array}$
The clothoid loop is obtained by combining a curve from $\theta =0$  to $\theta =\pi$ with  the reflection in the vertical axis.
A clothoid loop for $b=0.002$ with simulated motion and speed and centripetal acceleration is shown in the following animated gif.

Clothoid loop (b=0.002)

## Planets

Finally, using considerations from the post Seeking the centre, the elliptical nature of orbits under the inverse square law can be derived.

$\ddot r-r{\dot\theta}^2=-\frac{GM}{r^2}$ and $r^2\dot\theta =\text{constant}=k$

As $r>0$, the transformation $u =1/r$ provides simplification of the differential equation:

$\begin{array}{ccc} \frac{du}{d\theta}& =& \frac{du}{dt}\frac{dt}{d\theta}=-\frac{1}{r^2\dot\theta}\dot r=-\frac{\dot r}{k}\\\frac{d^2u}{d\theta^2}&=&-\frac{\ddot r}{k}\frac{1}{r^2\dot\theta u^2}=-\frac{\ddot r}{k^2 u^2}\end{array}$

Therefore the differential equation becomes,

$-k^2u^2\frac{d^2u}{d\theta^2} -k^2u^3=-GMu^2$

noting $r{\dot\theta}^2 = (r^2\dot\theta)^2/r^3=k^2u^3$.

$\frac{d^2u}{d\theta^2} +u=\frac{GM}{k^2}$

This is the equation for a harmonic oscillator with constant forcing function. The solution is:

$u = \frac{GM}{k^2}+ A\cos(\theta +\phi)$

Without loss 0f generality, let $\phi =0$ and substituting for $r$ yields:

$r = \frac{1}{\frac{GM}{k^2} + A\cos\theta}$

Rearranging this into a more canonical form yields,

$r =\frac {q}{1+\epsilon\cos\theta}$

where $q =\frac{k^2}{GM}$ and $\epsilon=\frac{k^2A}{GM}$

This is the polar formula of a conic section.

The polar of plot of $r =\frac{1}{1+\epsilon\theta}$ for varying values of $\epsilon$ is show below (note 0= circle->ellipse->1=parabola->hyperbola):

Comment:

The Newtonian gravitational potential energy  can be derived from the force formula:

$U =\int _{r_1}^{r_2 }-GMm/r^2\, dr = GMm (\frac{1}{r_2}-\frac{1}{r_1})$

By convention, U =-GMm/r where the potential energy at “infinity” is regarded as zero and hence all less than this are negative. Changes in potential difference map as above.

For distances small with respect to the radius of the earth, the usual $U = mgh$ is a good approximation.

Note: for $r<:

$U=GMm (\frac{1}{r}-\frac{1}{r+h})=\frac{GM} {r(r+h)}m h\approx m\cdot \frac{GM}{r^2}\cdot h = mgh$

For$r<, $g$ is approximately constant.

$\frac{GM}{r^2}= \frac{6.7 \times 10^{-11} \text{m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2}\times 6.0 \times 10^{24} \text{kg}}{6.4^2 \times 10^{12}\text{m}^2}=9.8 \text{m}\cdot\text{s}^{-2}$