2
$\begingroup$

I'm currently studying Goldstein's Classical Mechanics book and I can't get my head around his reasoning in section 2.4. (Extending Hamilton's principle to systems with constraints). I'd like to understand the example he gives. Here it comes:


Consider a smooth solid hemisphere of radius $a$ placed with its flat side down and fastened to the Earth whose gravitational acceleration is $g$. Place a small mass $M$ at the top of the hemisphere with an infinitesimal displacement off center so the mass slides down without friction. Choose coordinate $x, y, z$ centered on the base of the hemisphere with $z$ vertical and the $x$-$z$-plane containing the initial motion of the mass.

Let $\theta$ be the angle from the top of the sphere to the mass. The Lagrangian is $L = \frac{1}{2}\cdot M \cdot (\dot x^2 + \dot y^2 + \dot z^2) - m\cdot g\cdot z$. The initial conditions allow us to ignore the $y$ coordinate, so the constraint equation is $a - \sqrt{x^2 + y^2} = 0$. Expressing the problem in terms of $r^2 = x^2+z^2$ and $x/z = \cos(\theta)$, Lagrange's equations are $$ Ma\dot\theta^2 - M g \cos(\theta) + \lambda = 0$$ and $$ Ma^2\ddot \theta + M g\, a \sin (\theta) = 0$$

Solve the second equation and then the first to obtain $$ \dot\theta^2 = -\frac{2g}{a}\cos(\theta) + \frac{2g}{a}$$ and $$ \lambda = M g (3\cos(\theta)-2)$$ So $\lambda$ is the magnitude of the force keeping the particle on the sphere and since $\lambda = 0$ when $\theta = \cos^{-1}(\tfrac{2}{3})$, the mass leaves the sphere at that angle.


I have the following questions:

  1. Shouldn't it be $x/z = \tan \theta$?

  2. Could it be that he's mixing up $r$ and $a$? My guess is that from "Lagrange's equations are" it should say $r$ instead of $a$. I get confused whether $a$ is a system parameter or a Lagrangian multiplier.

  3. Could you give me a) an explanation or b) a good read on why setting $L' = L + \lambda\cdot f$ gives us an analogue of Hamilton's principle on constraint systems? I don't understand Goldstein's derivation. ($L$ is the original Lagrangian, $f$ is the constraint and $\lambda$ is the Lagrangian multiplier.)

  4. Why can $\lambda$ be thought of as the constraint force?

When I understand 3., I understand the example -- I reverse engineered the supposedly Lagrangian equations to see that $L'$ needs to be of form $$\frac{1}{2}M r^2 \dot\theta^2 - Mrg\cos(\theta) + \lambda \cdot f$$ with generalized coordinates $\theta$ and $r$. Then everything works out just fine.

$\endgroup$
3
$\begingroup$
  1. Yes, it should be $x/z = tan \theta$, this is probably a typo.
  2. The constraint should be $a - \sqrt{x^2 + z^2}=0$ for the argument to make sense. $r$ is a coordinate which is variable but due to the constraint it will always be equal to $a$, so we can use $a$ in the equations instead. ($\dot{a}=0$).
  3. You know that a gradient of $f$ is always perpendicular to the surface of constant $f$, so you can understand the extra term coming from $\partial L'/\partial x_i $ as a force acting perpendicular to the surface of $f=0$ holding the particle on it. However, $\lambda$ has to be solved so that the motion of the system is only along the constant surface. But imagine now a force equal to the solved $\lambda \partial f/\partial x_i$ - it would have the same effect as the constraint, so this is the force by which the constraint actually has to be acting to hold the particle/system. (This also answers 4.)
$\endgroup$
  • $\begingroup$ OK, thanks a lot for the clarifying answer, Void! I just realized that there is one thing I don't quite understand yet: In "regular" Lagrangian equations, using new generalized coordinates incorporates all constraints. Why isn't that enough in the Hamilton approach? (i.e. why do we need Lagrangian $\lambda$s where a coordinate transformation had sufficed before?) $\endgroup$ – Mercury Bench Sep 10 '14 at 16:27
  • $\begingroup$ You could then just use only one of the equations - the one gained by variation of $\theta$ - the derivation of this fact can be done in various ways, one of them is via the d'Alembert principle. $\endgroup$ – Void Sep 10 '14 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.