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Suppose that we have a bead $p$ of mass $m$ following a fixed wire described by a graph of a smooth function $f:\mathbb{R} \to \mathbb{R}$ without friction under the force of gravity, i.e. $p(t) = (x(t),y(t))$ is constrained to the set $\{(x,f(x)) \mid x \in \mathbb{R}\}$ and the force $G = (0,-gm)$ is acting on it. Using Lagrangian mechanics and the generalized coordinate $q = x$ with the Lagrangian $$L(x,\dot{x}) = \frac{m}{2}(\dot{x}^2+f'(x)^2\dot{x}^2) - gmf(x),$$ it is easy to see from the Euler-Lagrange equation $\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x}$ that the equation of motion for $x$ is given by $$ \ddot{x}(1+f'(x)^2)+\dot{x}^2f'(x)f''(x) = -gf'(x)\,. $$ However, we can take another approach: Let $F(x)$ be the parallel and $N(x)$ be the normal force on the wire in the point $(x,f(x))$ such that $G = F(x) + N(x)$ for all $x \in \mathbb{R}$, i.e. we decompose the force of gravity. Its is easy to see that $$ F(x) = -gm \frac{f'(x)}{1+f'(x)^2} \binom{1}{f(x)} $$ and since the bead is constrained to the wire, only $F(x)$ is acting on it (the normal force $N(x)$ is counteracted by the wire). Hence the equation of motion for the bead is $$ m \frac{d^2}{dt^2} \binom{x(t)}{y(t)} = F(x(t))\,, $$ i.e. in particular $$ \ddot{x} = -g\frac{f'(x)}{1+f'(x)^2}\,, $$ which is clearly a different equation of motion for $x$.

Question: Where does the second approach go wrong (it seems very "intuitive")? What would be the correct way to solve this problem in Newton-mechanics?

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  • $\begingroup$ I don't understand why you say the normal force doesn't act on the bead. $\endgroup$ Mar 28 at 22:40
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since the bead is constrained to the wire, only $F(x)$ is acting on it (the normal force $N(x)$ is counteracted by the wire).

The bead is constrained to move along the wire precisely because the normal force acts on it. The normal force is whatever it needs to be so that the motion follows the wire.

Furthermore, the normal force is not just a function of $x$, but also of the velocity; going faster will cause the constraint force to be larger (think of a bead going around a circle).

Because of the above, unless you are able to reason through what $N$ should be before determining the motion, Newtonian Mechanics won't really help you solve problems like these. Usually to determine $N$ one uses Lagrangian mechanics and then either works backwards to find what $N$ needs to be, or uses techniques such as Lagrange multipliers to determine the constraint force(s).

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  • $\begingroup$ "the normal force is not just a function of $x$, but also of the velocity; going faster will cause the constraint force to be larger" - This is the most convincing and intuitive way to see it, thank you. Still, does the second approach have any actual physical meaning (i.e. if we apply the tangential force $F(x)$ to the (now unconstrained) bead at each coordinate $x$)? $\endgroup$
    – Andrei Kh
    Mar 29 at 6:20
  • $\begingroup$ @AndreiKh If there is no constraint path, what do you mean by "tangential"? $\endgroup$ Mar 29 at 7:21
  • $\begingroup$ Well if we apply the above defined force $F(x)$ to a free particle $p = (x,y)$ and track its motion (i.e. solve $m \ddot{p} = F(x)$), which will, in general, not lie on the wire. Does the resulting motion of the x-coordinate $x$ of $p$ have any physical meaning (some similar setup or something else)? $\endgroup$
    – Andrei Kh
    Mar 29 at 17:25
  • $\begingroup$ @AndreiKh I am still confused by the new system you are asking about. So far you have only defined the force of gravity. If there is no wire, then that is the only force. $\endgroup$ Mar 29 at 17:33
  • $\begingroup$ Yes exactly, I wonder whether this new system with the only force being $F$ (no other forces, including constraint forces) has any meaning/connection related to the original system with the constraints (as solved by the Lagrangian approach). But I guess probably not really $\endgroup$
    – Andrei Kh
    Mar 29 at 21:35

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