# From the notes: Two masses on a string going through a hole

Two masses on a string going through a hole.  Yeah, that's the stuff we learn about in physics class.  Anyway, this is the first in a (possibly) series of posts where I work out some problem from my lecture notes or homework.  The reason for doing this is because 1) there's a (small) chance it will be useful or interesting to someone else, and 2) the drawn out process of typing this up and thinking about how to explain it helps me study.

Figure 1, two masses on a string going through a hole.

Consider two masses connected by a string (figure 1).  The ideal string passes through a hole in a plate which is parallel to the x-y plane.  One mass, $m$ is sitting on the plate and is free to rotate about the hole with no friction.  The second mass, $M$, is hanging below the plate and moves only in the vertical z direction under the influence of gravity.   The total length of the string is $l$ and we define $r$ and$s$ to be the distances between the hole and $m$ and $M$, respectively, so that $l=r+s$.

We can use Lagrangian mechanics to explore the system.  In Cartesian coordinates, the total kinetic energy is the sum of the kinetic energy of each mass:

$T=\frac{1}{2}m(\dot{x_{m}}^2+\dot{y_{m}}^2+\dot{z_{m}}^2)+\frac{1}{2}M(\dot{x_{M}}^2+\dot{y_{M}}^2+\dot{z_{M}}^2)$      (1)

where we've used the dot notation for time derivatives.

Considering that $m$ is only rotating about the $z$ axis and $M$ is moving only vertically, it makes sense to consider switching to a more reasonable coordinate system.  We can transform to cylindrical coordinates with:

$x_{m}=r\cos\theta$

$y_{m}=r\sin\theta$

$z_{m}=0$

$x_{M}=y_{M}=0$

$z_{M}=-s=-(l-r)$

Taking derivatives of these with respect to time and inserting them into (1), we can rewrite the kinetic energy in cylindrical coordinates as

$T=\frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2)+\frac{1}{2}M\dot{r}^2$

where $r$ and $\theta$ are now the generalized coordinates of the system.  The potential energy depends only on the height of $M$ since $m$ is confined to sit on the plate, so

$V=-Mg(l-r)$

With $T$ and $V$ we can write the Lagrangian $L=T-V$:

$L=\frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2)+\frac{1}{2}M\dot{r}^2+Mg(l-r)$

We see that $\theta$ does not appear in the Lagrangian, which means that the generalized momentum corresponding to $\theta$ is a conserved quantity:

$p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}=rmv_m=a$     (2)

$a$ can be identified as the angular momentum of mass $m$.

We can also look at the total energy $E=T+V$:

$E=\frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2)+\frac{1}{2}M\dot{r}^2-Mg(l-r)$     (3)

From (2) we see that $\dot{\theta}=\frac{a}{mr^2}$.  Substituting this in (3) gives

$E=\frac{1}{2}m\dot{r}^2+\frac{a^2}{2mr^2}+\frac{1}{2}M\dot{r}^2-Mg(l-r)$

We can separate the constants on the left hand side and use the remaining terms to define an effective potential $V'(r)$:

$\frac{E+Mgl}{m+M}=\frac{1}{2}\dot{r}^2+\frac{1}{2}\frac{a^2}{m(m+M)r^2}+\frac{Mgr}{m+M}$

$V'(r)=\frac{1}{2}\frac{a^2}{m(m+M)r^2}+\frac{Mgr}{m+M}$

Imagine $m$ is rotating about the hole.  This means there will be some centripetal (or centrifugal, depending on your reference system) acceleration which will pull the hanging mass $M$ up.  We want to find $r$ such that the force is an extremum.  In other words:

$\frac{\partial V'}{\partial r}=0$

Setting the derivative equal to zero gives

$\frac{-a^2}{m(m+M)r^3}+\frac{Mg}{m+M}=0$

And solving for $r$:

$r=\sqrt[3]{\frac{a^2}{mMg}}$

which gives the radius of "orbit" of mass $m$ about the hole for a given angular momentum $a$ such that $M$ remains suspended in the air.