A particle of mass m is on top of a sphere of radius $R$. A small displacement makes the particle slide frictionlessly down the sphere.
a)Set up the lagrange equations of the first kind for cartesian coordinates; determine the absolute value of the constraint force $F_z$ and compute the point at the sphere at which the particle detaches from the sphere and its velocity at that point.
b)Set up the lagrange equations of the first kind for plane polar coordinates; determine the constraint force $F_z$ and compute the angle and angle velocity at which the particle detaches.
We dealt with lagrange equations of the first kind and to be honest the concept didn't stick to me. I mean, I know the formulas but if I were to solve a problem such as this I don't know how to set up the equations and solve them.
Anyway, here were my ideas on how to approach this:
The constraint is $x^2+y^2+z^2-R^2=0=f$, right? But I don't know how to set up the lagrange equation with that. And a stupid question in-between: I'm not yet familiar with the terminology concerning the lagrange equations; does $L$ stand for the total energy of the system?
Anyway, I was thinking do I even need the lagrange equations to solve the second part of the questions, meaning the detachment point/velocity/angle/angle velocity?
Here were my ideas for those, and tell me if I'm wrong:
a) First I was thinking of the path the particle takes down the "road". I was trying at the movement from the side, meaning I was looking at it as if it were a 2D-problem where the particle moves down a circular segment.
Something like this:
My approach to the problem was setting up the radial force components, meaning:
$F_r=F_z+F_{g,r}$, where $F_{g,r}$ is the radial component of the gravitational force.
Since I want the detachment point I was setting $F_z=0$, meaning:
$0=F_z+F_{g,r}$. I won't write down all the transformations I did because they are rather longish.
Anyway, I came to a relation of the angle with the detachment point, or rather the radius and the height $h$ as seen in the figure, giving me an angle of $\alpha\approx 41.81°$, a height $h=\frac{1}{3}R$, meaning the detachment is $(cos(\alpha)\cdot R;\frac{2}{3}R)$ for the sphere centered at the origin.
Now, for the velocity I was thinking: $mgz=\frac{1}{2}mv^2\leftrightarrow v^2=2mgz$. Can I assume that this is the equation for the detachment velocity when $z=\frac{2}{3}R$?
Next, about the angle velocity: I know that $v=\omega r$. So can I just plug in the velocity I got before and solve for $\omega$?
Anyway, at this point I'm not even sure if I came to the right solution since the solution with the lagrange equations could differ.
This is explained quite well in David Morin's book on Intro to Classical Mechanics; you can find this very problem done both ways on the web at
http://www.people.fas.harvard.edu/~djmorin/chap6.pdf
(Your reduction to a 2-D problem is valid.)
On second thought, it is explained but not quite well enough to find the point of departure. The result of his exercise 6.37 finds the constraint force in terms of the position and velocity as $$ F= m \left( \frac{zg}{R} - \frac{(x\dot{z}-z\dot{x})^2}{R^3} \right) $$ but that does not immediately say where $F=0$.
The point (of both Morin and your professor) is to show how much easier this problem is in spherical (or cylindrical) coordinates. If you work it through in Cartesian coordinates to get to the force shown, you will surely appreciate that.
At any rate, your answer that the departure is at $z = \frac{2R}{3}$ is correct.