# Mass hanging under a table: a problem from Goldstein

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1. Mar 27, 2016

### avorobey

1. The problem statement, all variables and given/known data

This is Exercise 1.19 in Goldstein's Classical Mechanics 2nd edition. Self-study, not for a class.

Two mass points of mass $m_1$ and $m_2$ are connected by a string passing through a hole in a smooth table so that $m_1$ rests on the table and $m_2$ hangs suspended. Assuming $m_2$ moves only in a vertical line, what are the generalized coordinates for the system? Write down the Lagrange equations for the system and, if possible, discuss the physical signiﬁcance any of them might have. Reduce the problem to a single second—order differential equation and obtain a ﬁrst integral of the equation. What is its physical signiﬁcance? (Consider the motion only so long as neither $m_1$ nor $m_2$ passes through the hole).

2. Relevant equations

3. The attempt at a solution

I'm trying to find the Lagrangian as rigorously as I can. It seems that the intent of the problem is to establish the constant length of the string as a constraint and the tension forces on both points as forces of constraint that can be disregarded when drawing up the Lagrangian. But I'm unable to prove that I can disregard them.

Let the origin be at the hole, and $\mathbf{r}_1$,$\mathbf{r}_2$ be the position vectors of the two points. I assume as known that the tension $T$ is equal at both ends of the rope. Then if the polar angle of the first point is $\phi$, the force of tension on the first point is $(F_{1x},F_{1y}) = (-T\cos\phi,-T\sin\phi)$ and on the second point $F_{2z} = -T$, listing only nontrivial coordinates. The holonomic equation of constraint is $|\mathbf{r}|_1+|\mathbf{r}_2| = R$ where $R$ is the constant length of the string.

Now clearly the tension forces are not like the normal force of constraint from the table on the first point, that just straight out vanishes because it's orthogonal to velocity. The tension forces do nonzero virtual work on each particle, but it seems that I should be able to prove that just as with a rigid body, due to Newton's 3rd law they vanish when summed up over the particles. In other words, I must prove that the d'Alambert principle holds in the system: the net virtual work of the forces of constraint is zero.

Let $\delta\mathbf{r_1} = (\delta r_1,\psi)$ (in polar coordinates) and $\delta\mathbf{r_2} = \delta r_2$ be virtual displacements of the two points consistent with the constraints. The total virtual work of the two forces of constraint is then $$-T\cos\phi(\delta r_1\cos\psi)-T\sin\phi(\delta r_1\sin\psi)-T\delta r_2 = T\delta r_1\cos(\phi-\psi) + T\delta r_2 = 0$$
and I must show that this holds if $(\delta\mathbf{r_1},\delta\mathbf{r_2})$ satisfy the equation of constraint, meaning the differences in length between the position vectors must match: $$|\mathbf{r_1}+\delta r_1|-|\mathbf{r_1}| = -(|\mathbf{r_2}+\delta r_2|-|\mathbf{r_2}|)$$
The right side is just the scalar $-\delta r_2$, but the left side doesn't simplify that easily. Using the polar coordinates $(r,\phi)$ it seems to come down to
$$\sqrt{(r\cos\phi+\delta r_1\cos\psi)^2+(r\sin\phi+\delta r_1\sin\psi)^2}-r = \sqrt{r^2+\delta r_1^2 + 2r\delta r_1\cos(\phi-\psi)}-r$$
and now I'm stuck. It seems that I need to prove $-\delta r_2 = \delta r_1\cos(\phi-\psi)$, for the forces to do zero virtual work together. In the special case $\psi=\phi$, when the first point moves in a straight line towards the origin, the square root above simplifies, $r$ vanishes, and I get the desired $\delta r_1 = -\delta r_2$. But I don't see how I can assume that, and in fact it seems physically incorrect if for example the first point has initial velocity in the $y$ direction.

What am I doing/calculating/assuming wrong?

2. Mar 27, 2016

### Orodruin

Staff Emeritus
You are overthinking the problem. The setup is described by two generalised coordinates. Write down the kinetic and potential energies in these coordinates and you will have the lagrangian. The constraint is holonomic and can be implemented directly in the lagrangian. You do not need to worry about the constraining forces.

3. Mar 27, 2016

### avorobey

Thanks you! - but I yearn to understand why I don't need to worry about the constraining forces. If I solve the problem the way you're saying, then I'm ignoring these forces when I'm writing down potential energies; what is my justification for doing so? After all, they're just forces acting on the particles; what prevents me from declaring that some applied force is "actually" a constraint force I can just ignore and not include in my V, and when I do so, why won't I get the correct equations of motion?

My understanding (very possibly faulty) was that the reason I can ignore constraining forces is that their combined virtual work is zero. This condition is used explicitly in Goldstein's Chapter 1 when deriving the Lagrange equations, and I think it's contained implicitly when deriving them variationally, from Hamilton's principle, allowing us to only include applied forces' potentials in L. But if I can't show, in this problem, that the combined virtual work is zero, I don't understand why I can ignore the tension forces, and why I would get the correct equations of motion if I do.