Expert: Daniel Mazur - 2/3/2009

Question
QUESTION: Hi! I have the a question. Its not academic, but not so easy, and not a homework. I'm studying in a secondary school, and I've found this interestin question in one of my books, and i can't solve it.

There is a slope, and on the slope a cylinder, wich has an elliptical cross, and the axes: a=1, b=2. How can I give the biggest x, by that the cylinder stays at rest?
a picture: http://www.upload2world.com/pic110/upload2world_d1d1d.jpg

ANSWER: Hello Alma,

firstly, we need to understand, what determines an angle, at which the ellipse (I will talk about 2D side-view cross section all the time here, because the solution lies in geometry) rolls down the hill. Physically, this is when the center of mass (coincides with ellipse center) encounters no potential minima on its way. In other words, the center can move sideways and down, but not up. An ellipse with large eccentricity needs to move the mass center a long way up from its balanced position on leveled surface in order to pass the maximum and roll over. Tipping the plane of motion lowers the barrier needed to overcome and at some tipping angle phi0 there will be no barrier at all.

Let's start with the equation of ellipse. The points on the circumference have their distance R from center dependent on angle phi this way:

R = a*b/sqrt(a^2-(a^2-b^2)*cos(phi)^2)

You can get this from the standard equation of ellipse in polar coordinates (Wikipedia) and some identities of elliptical geometry. Now see that if you roll this ellipse on a horizontal straight line, the height of its center above the line will be described exactly by this equation for R. If you try to draw the corresponding curve, you will get something like a shortened cycloid (it may really be a shortened cycloid, I didn't like spending the time proving this). Now if you tip the line with the cycloid on it, the cycloid minima will become smaller and smaller until at some angle the minima will disappear altogether. This is the angle phi0 we seek.
I don't know, if you are familiar with the derivative operation and its use in mathematical analysis, but I cannot see, how your problem can be solved without them. So, mathematically, the condition of flattening minima at angle phi0 translates as tipping the cycloid by the angle corresponding to the maximal derivative of the cycloid (of R above). After we calculate the first derivative dR/dphi, we look for its maximum. The location of the maximal dR/dphi is where d^2R/dphi^2 (second derivative) is zero and the third derivative d^3/dphi^3 is negative.
Here is the expression for the first derivative of R(phi):

dr/dphi = -a*b*(a^2-b^2)*sin(2*phi)/((a^2-(a^2-b^2)*cos(phi)^2))^(3/2)

As you can see, it gets complicated for writing here. Then we need to calculate the second derivative and set it equal to zero, d^2R/dphi^2=0:

d^2R/dphi^2 = [3*a*b*(a^2-b^2)^2*sin(2*phi)^2/(4*(a^2-(a^2-b^2)*cos(phi)^2)^(5/2))]-[a*b*(a^2-b^2)*cos(2*phi)/(a^2-(a^2-b^2)*cos(phi)^2)^(3/2)] = 0

You don't be put of by the long equation, AllExperts.com pages are simply not suitable for this sort of thing. The angle(s) then extracted from this equation is(are) then phi0 "candidates". While the derivative can be calculated analytically, it is quite likely that the equation d^2R/dphi^2=0 is implicit in phi and, hence, the phi0 candidates cannot be found analytically, only numerically or (with reduced precision) geometrically.

One way or another you will have the phi0 candidates. The ellipse is simple enough curve so that there should be only 4 candidates per phi interval (0,2*pi) and that they all can be expressed using one principal angle phi0, e.g. {phi0, pi-phi0, pi+phi0,2*pi+phi}. Of these 4 candidates 2 correspond to maxima of the cycloid and 2 are minima. You pick the maxima by considering your choice of coordinates (i.e. if phi is measured from axis a or axis b of the ellipse), draft a sketch, if in doubt. In a pure analytical fashion you ought to calculate the third derivative, substitute the phi0 candidate values into it and see, for which the d^3R/dphi^3(phi0) is less than zero. Either way, this will be your result.

As you can see, the initial consideration in this solution is physics, but everything else is pure geometry.

Cheers,
Daniel

---------- FOLLOW-UP ----------

QUESTION: Hi Daniel! Thanks for the complex answer. I wont have problems with the derivate operation, but at first i couldn't get R = a*b/sqrt(a^2-(a^2-b^2)*cos(phi)^2)  from the equation of ellipse in polar coordinates, what I've found on Wikipedia. Can you show me it in a paar of steps?

thanks a lot

Answer
Certainly, Alma,

the equation in polar coordinates in Wikipedia is given in terms of r (or rather r1, following their drawing) and theta. To get equation R=R(phi) (R being again the distance from the center) one needs to use the cosine rule extensively. Let c be the distance between the focal points. Then for distances of circumference from the focal points 1 and 2 we get, respectively:
r1^2 = R^2+c^2/4-c*R*cos(pi-phi) = R^2+c^2/4+c*R*cos(phi)
r2^2 = R^2+c^2/4-c*R*cos(phi)

Then by using the ellipse identity
r1+r2 = 2*a
we get
2*a^2 - c*R*cos(phi) = r2 = sqrt(R^2+c^2/4-c*R*cos(phi))
and this reduces into
R = a*sqrt((1-(c/(2*a))^2)/(1-(c/(2*a))^2*cos(phi)^2)).

The ellipse identity has a special case:
theta1 = theta2: r1=r2=r=a -> Pythagorean r^2 = b^2+c^2/4 = a^2
from which we get c^2 = 4*(a^2-b^2)
and this we can plug into the R(phi) equation to obtain

R = a*b/sqrt(a^2-(a^2-b^2)*cos(phi)^2)

Take care.
Daniel