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I was working on a problem studying for a classical mechanics class and came across an idea I'm not sure about concerning the formalism of Lagrangian mechanics concerning constraint problems.

https://puu.sh/CLVDd/da4a39bad2.png

To be specific, in the problem of a particle constrained to a hemisphere (picture above), I did some research after getting stuck on what the constraint (constant of motion) should be for this particle and found out that instead of being able to write this problem entirely in terms of the angle $\theta$, you need to include the radial direction. (stack exchange post as reference: https://math.stackexchange.com/questions/1347302/particle-on-a-hemisphere-lagrange_)

Why must we include the radial motion of the particle if we know that it is constrained to not move radially? Is there not some other way to enforce the constraint in terms of theta?

My apologies if this is the wrong place to ask such questions as I am new to the site.

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  • $\begingroup$ There is no longer a file at puu.sh/CLVDd/da4a39bad2.png $\endgroup$ Commented Sep 24, 2022 at 20:40
  • $\begingroup$ The other stack exchange post I reference has the same diagram and question - I unfortunately don't have access to the original question anymore. $\endgroup$
    – Purplehats
    Commented Sep 30, 2022 at 16:30

2 Answers 2

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If the particle is constrained to not move radially, this means that its radial component is fixed, i.e. $r = R$ where $r$ is the radial coordinate and $R$ is the hemisphere radius.

If you think about, if a quantity is constrained, it means that there is a condition (equation, inequality, something) that the quantity must obey. So it shouldn't be surprising that the radial coordinate of the particle should feature in some way in the constraint.

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  • $\begingroup$ So for general problems, if you're dealing with a constraint on some aspect of a system (here on r) then you want to make the action stationary with respect to that variable, even if you might already have an intuition that the Lagrange equation for that variable shouldn't vary in time. In other words, in the Lagrange formalism, you need to consider all possible motion of the system and the formalism will bear out the intuition. Thanks! $\endgroup$
    – Purplehats
    Commented Feb 14, 2019 at 4:52
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Is there not some other way to enforce the constraint in terms of theta ?.

yes there is.

The NEWTON equations of motion are:

$$m\,\vec{\ddot{u}}=\vec{f}_a+\vec{f}_z\tag 1$$

where

$\vec{{u}}= \begin{bmatrix} {x(\vec{q})}\\ {y(\vec{q})}\\ \end{bmatrix}\quad$

and

$\vec{q}$ the generalized coordinates

$\vec{f}_a$ applied forces include conservative forces

$\vec{f}_z$ constrained forces

the time derivative of the vector $\vec{u}$ is:

$\vec{\dot{u}}=J\,\vec{\dot{q}}$ and $\quad \vec{\ddot{u}}=J\,\vec{\ddot{q}}+\dot{J} \vec{\dot{q}}$

with:

$J=\frac{\partial \vec{u}}{\partial \vec{q}}\quad$ and $\quad \dot{J}=\frac{\partial \left(J\vec{\dot{q}}\right)}{\partial \vec{q}}$

so equation (1) map to:

$$m\,J\,\vec{\ddot{q}}=\vec{f}_a+\vec{f}_c+\vec{f}_z\tag 2$$

with:

$\vec{f}_c=-m\,\frac{\partial \left(J\vec{\dot{q}}\right)}{\partial \vec{q}}\,\vec{\dot{q}}\quad $ Coriolis and centrifugal forces

To eliminate the constrain forces $\vec{f}_z$ from equation (2), we can write the constrained forces with a distribution matrix $C_z$ and generalized constrained forces $\vec{\lambda}$.

$\vec{f}_z=C_z\,\vec{\lambda}$

$$m\,J\,\vec{\ddot{q}}=\vec{f}_a+\vec{f}_c+C_z\,\vec{\lambda}\tag 3$$

according to d'alembert approach the product $J^T\,C_z$ must be zero

multiply equation (3) with $J^T$ yields to:

$$\boxed{J^T\,m\,J\,\vec{\ddot{q}}=J^T\left(\vec{f}_a+\vec{f}_c\right)}\tag 4$$

multiply equation (3) with $C_z^T$ yields to:

$$\boxed{\mathbb{0}=C_z^T\left(\vec{f}_a+\vec{f}_c\right)+C_z^T\,C_z\,\vec{\lambda}}\tag 5$$

equation (4) are the equations of motion for the generalized coordinates $\vec{q}$, with equation (5) you get the generalized constrained forces $\vec{\lambda}$ .

your example

generalized coordinate is $\theta$ and

$\vec{u}=\begin{bmatrix} a\sin(\theta)\\ a\cos(\theta)\\ \end{bmatrix}$

$\Rightarrow$

$J=\begin{bmatrix} a\cos(\theta)\\ -a\sin(\theta)\\ \end{bmatrix}$

Applied force

$f_a=\begin{bmatrix} 0\\ -m\,g\\ \end{bmatrix}$

with equation (4) we can calculate the equation of motion for the generalized coordinate $\theta$

to calculate the constrained force we have to construct the matrix $C_z$ that fulfill the equation $J^T\,C_z=\mathbb{0}$

$\Rightarrow$

$C_z=\left[ \begin {array}{c} a\sin \left( \theta \right) \\ a\cos \left( \theta \right) \end {array} \right] $

with equation (5) we can calculate the generalized constrained force $\lambda$

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    $\begingroup$ I think this approach is interesting and I appreciate the answer, but my hold-up was using Lagrangian Formalism to represent the constraint on the problem - my apologies if I was unclear. Besides that, may I clarify some of your notation? Does the quantity $J=\frac{\partial \vec{u}}{\partial \vec{q}}\quad$ represent a Jacobian matrix? I haven't seen this notation used before. $\endgroup$
    – Purplehats
    Commented Feb 14, 2019 at 20:02
  • $\begingroup$ Yes this is the Jacobian matrix. I don’t think you can find solution to your problem using lagranian formalism $\endgroup$
    – Eli
    Commented Feb 14, 2019 at 20:36

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