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 $Ο‰$ 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 $π‘ž_1,...,π‘ž_𝑓$ which implies $f=5$ generalised coordinates and I have only one constraint: $|(√π‘₯2+𝑦2)βˆ’π‘…π‘π‘œπ‘ (ψ)|=𝑅$ because $k=1$. I think that there should be only $f=2$ generalised coordinates $q_1$=ψ and $q_2$=Ξ¦, but this would imply $N=1$ and therefore $k=0$ and I would loose the constraint. What to do?



2 Answers 2


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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:

$$ \begin{bmatrix} \ddot{\alpha} \\ \ddot{\beta} \\ \end{bmatrix}=-\left[ \begin {array}{c} {\frac {g\sin \left( \alpha \right) }{L}}-\frac 12\,\sin \left( 2\,\alpha \right) {\omega}^{2}-\dot{\beta} \,\sin \left( 2\, \alpha \right) \omega-\frac 12\,{\dot{\beta} }^{2}\sin \left( 2\,\alpha \right) \\ -2\,{\frac {\sin \left( 2\,\alpha \right) \dot\alpha \,\omega}{-1+\cos \left( 2\,\alpha \right) }}-2\,{\frac {\sin \left( 2\,\alpha \right)\dot{\alpha} \,\dot{\beta} }{-1+\cos \left( 2\,\alpha \right) }}\end {array} \right]$$

  1. We need as many generalized coordinates as degrees of freedom (DOF). The number of DOF is in fact ${f=3N-k}$, with $N$ the number of particles and $k$ the number of constraints.

    You have already noticed that we only need two angles to describe the configuration of the system, and that we have one constraint. To solve the apparent riddle, notice that there is only 1 particle. The attachment point of the rod is not a particle, because the rod is exerting some force on it, yet it does not accelerate, so essentially it has infinite mass.

  2. The number of constraints $k$ is actually independent of the number of particles. We cannot say that having only $N=1$ particle necessarily implies $k=0$ constraints and therefore $f=3$ DOF.

    For example, a free particle (not constrained) in 3-dimensional space has ${f=3\cdot1-0=3}$ DOF. We can choose as generalized coordinates, say, its cartesian coordinates ${x,y,z}$, or its spherical coordinates ${r,\theta,\varphi}$.

    But we could choose to constrain the particle to move in a plane, say $z=0$; that's $1$ constraint and therefore ${f=3\cdot1-1=2}$ DOF. Some generalized coordinates for this case could be $x$ and $y$, or polar coordinates $\rho$ and $\varphi$.

    We could go further and constrain the particle to move in a line, say the line \begin{cases}z=0\\y=0.\end{cases} Those two equations defining the line are $k=2$ constraints. Therefore that would give us $f=3\cdot1-2=1$ DOF. The natural generalized coordinate for this case is the coordinate $x$.

    In your example, the rod constrains the particle to move on the surface of an imaginary sphere of radius $R$ centered in the attachment point. That constraint can be expressed mathematically using $1$ equation, hence ${k=1}$ and ${f=3\cdot1-1=2}$.

  • $\begingroup$ Thank you very much for your answer! I am just very confused because in class the professor provided a table to provide the number of degrees of freedom. It looked like that; N=1=>k=0 and f=3; N=2=>k=1 and f=5; N=3=>k=3 and f=6 and so forth with f as the number of degrees of freedom, k constraints and N particles. Now I only have one particle, but this would imply k=0 and f=3 and I would loose the constraint so I kinda need to use the option of N=2 to include the constraint because N=2 implies k=1 and f=5 with ||x1-x2||=R but now I would have too many (f=5) generalised coordinates $\endgroup$
    – user277332
    Nov 30, 2020 at 9:11
  • 1
    $\begingroup$ Hi @NirvanicUniverse, I've edited the answer. $\endgroup$
    – Urb
    Nov 30, 2020 at 15:42
  • $\begingroup$ Thank you very very much @urb now I understand it! $\endgroup$
    – user277332
    Dec 1, 2020 at 17:34

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