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**1. The Question**

Two particles of mass m are connected by a light inextensible string of length l. One of the particles moves on a smooth horizontal table in which there is a small hole. The string passes through the hole so that the second particle hangs vertically below the hole. Use the conservation of energy and angular momentum to show that:

[tex]

{\dot{r}}^2 = \frac{gl + {v}^2}{2} - \frac{{l}^2{v}^2}{8r^2} -g r

[/tex]

where r(t) is the distance of the first particle from the hole, A and B are constants and g is the acceleration due to gravity. [Hint: Use polar coordinates in the plane of the table with the origin at the hole].

Given that the particle on the table is a distance l/2 from the hole and is moving with a speed v, directed perpendicular to the string, find the condition that the particle below the table does not pass through the hole. [Ans: [tex] \inline {v}^2 \leq 4gl/3 [/tex]]

What would happen if: [tex] \inline {v}^2 \leq gl/2 [/tex]?

**2. My attempt at a solution**

I managed to get a similar expression. Assuming the mass is travelling at a speed v on the table:

[tex] E_i = \frac {1}{2} m {v}^2 [/tex]

and the angular momentum is:

[tex] L_i = m v \frac {l}{2}[/tex]

I'm not sure if this is valid, but it seemed to work.

Then I got the final energy as:

[tex] E_f = 2(\frac{1}{2}m {\dot{r}}^2) + \frac{{L}^2}{2 m {r}^2} -mg(l-r) [/tex]

Then substituting for the angular momentum and setting the initial and final energies equal I get:

[tex] {\dot{r}}^2 = gl + \frac{{v}^2}{2} - \frac{{(vl)}^2}{8{r}^2} -gr [/tex]

So I'm out by a half for the g*l term. I'm sure this has something to do with the intial energy but I can't explain why.

For the second part I assume the orbit must be bounded, so the Energy must be less than zero, but I cannot seem to get the answer stated.

Thanks for your help.