Question:
In this exercise one needs to determine the generalised coordinates. We have a pendulum in a magnetic field $B$. The pendulum rotates around its axis with an angular velocity $ω$ on the circle as shown in the picture below. The pendulum is attached in the point $z_0$ and the mass $m$ and charge $e$ attached on the pendulum is located on the tip of the pendulum.
My ansatz:
I only know that the degrees of freedom are given by $f=3N-k$. I thought, that there are $f$ generalised coordinates $𝑞_1,…,𝑞_𝑓$
which implies $f=5$ generalised coordinates and I have only one constraint: $|(√𝑥2+𝑦2)−𝑅𝑐𝑜𝑠(ψ)|=𝑅$
because $k=1$. I think that there should be only $f=2$ generalised coordinates $q_1$=ψ and $q_2$=Φ, but this would imply $N=1$ and therefore $k=0$ and I would loose the constraint.
What to do?
Best Answer
A point mass can move in 3D space towards $x~\,y~\,,z$
because the length of the rod $~L$ is constant , this give you one constraint equation:
$$ \left(\vec{R}-\vec{Z}_0\right)\cdot \left(\vec{R}-\vec{Z}_0\right)-L^2=0={x}^{2}+{y}^{2}+ \left( z-z_{{0}} \right) ^{2}-{L}^{2}\tag 1$$
with eq. (1) you can i.e. obtain $~z=z(x,y)$ thus you have two generalized coordinates $~x\,,y$
but you can choose other generalized coordinate i.e.
$$ \vec{R}\mapsto \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}=\left[ \begin {array}{c} L\sin \left( \alpha \right) \cos \left( \beta \right) \\ L\sin \left( \alpha \right) \sin \left( \beta \right) \\ L \left( 1-\cos \left( \alpha \right) \right) \end {array} \right] \tag 2$$
you can check that with eq. (2) and $z_0=L$ eq. (1) is fulfilled .
your generalized coordinate are now $~\alpha\,,\beta$
The EOM's are:
$$ \begin{bmatrix} \ddot{\alpha} \\ \ddot{\beta} \\ \end{bmatrix}=-\left[ \begin {array}{c} {\frac {g\sin \left( \alpha \right) }{L}}-\frac 12\,\sin \left( 2\,\alpha \right) {\omega}^{2}-\dot{\beta} \,\sin \left( 2\, \alpha \right) \omega-\frac 12\,{\dot{\beta} }^{2}\sin \left( 2\,\alpha \right) \\ -2\,{\frac {\sin \left( 2\,\alpha \right) \dot\alpha \,\omega}{-1+\cos \left( 2\,\alpha \right) }}-2\,{\frac {\sin \left( 2\,\alpha \right)\dot{\alpha} \,\dot{\beta} }{-1+\cos \left( 2\,\alpha \right) }}\end {array} \right]$$