# A mechanics problem: two masses, one dangling from a table

1. Jan 10, 2010

### quasar_4

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

Consider two point particles, each of mass m, connected by a taut string that passes through a small frictionless hole (idealized as a hole of zero radius) on a frictionless table.

a) for motion in which the top mass remains on the table and the string remains taut, how many degrees of freedom does the system have?
b) Choosing as many coordinates as there are degrees of freedom, write down a Lagrangian for the system.
c) Now restrict the Lagrangian to motion for which the bottom mass moves vertically and is directly below the hole. Write the Euler-Lagrange equations that describe the motion.
d) Which, if any, of the following quantities are conserved?
e) Show the the equations of motion allow the top mass to move in a circular orbit, and find the angular velocity w of the orbit in terms of its radius r0.

There is also a part f and g, but I'd like to make sure I can get to part e first, so I'll leave it at this for now.

2. Relevant equations

Kinetic energy T, potential energy U, Lagrangian is defined as L = T-U
Euler-Lagrange equations

3. The attempt at a solution

First, I'm having trouble interpreting the motion in the system. If the top mass must remain on the table, should I interpret that to mean that it cannot slide at all towards the hole?

It didn't make sense to use this interpretation since part c asks me to restrict the Lagrangian to vertical motion. So, I went ahead and interpreted this as meaning that the top mass can slide towards the hole, but not pass through it.

So... In my analysis it seems that there are only two degrees of freedom. The bottom mass could move vertically, in response to the top mass moving towards the hole. The bottom mass could also swing in one direction (the polar angle, phi), but would not rotate azimuthally in the system without another force to push it in the other direction. (I'm not sure that this is right, either - at first I tried to use all three, and three degrees of freedom, but the Lagrangian got too complicated and so I decided I must be doing it wrong).

So I chose my coordinates to be:

$$x = r cos{\phi}$$
$$y= r sin{\phi}$$
$$\dot{x} = \dot{r} cos{\phi} - r sin{\phi} \dot{\phi}$$
$$\dot{x} = \dot{r} sin{\phi} + r cos{\phi} \dot{\phi}$$

and thus

$$T = \frac{1}{2} m\left(\dot{x}^2+ \dot{y}^2\right) = \frac{1}{2} m \left[\dot{r}^2 + r^2 \dot{\phi}^2 \right]$$

and

$$U = mgh = mgr cos{\phi}$$.

This produces the Lagrangian

$$L = \frac{1}{2} m\left(\dot{x}^2+ \dot{y}^2\right) = \frac{1}{2} m \left[\dot{r}^2 + r^2 \dot{\phi}^2 \right]- mgr cos{\phi}$$.

Is that right?? If so, then for part c, I could set phi and its derivative equal to zero. But this just produces the simple equation of motion [tex] \ddot{r} = -g [\tex]. That doesn't make any sense when looking at part e, which says that the top mass should be able to move in a circular orbit.

Also, I thought that in order for things to be conserved, they have to be cyclic coordinates. But neither is cyclic. Does that mean that nothing is conserved?? I'm so confused!!

So... if you can tell me where I'm going wrong (especially if it's in more than one spot), please help!!

2. Jan 11, 2010

### tiny-tim

Hi quasar_4!

(have a phi: φ )

By setting φ' = 0, you've found the solution for the mass moving straight towards the hole, r'' = -g (but shouldn't it be -g/2?).

Try setting r' = 0.