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Central Forces Question

  • Thread starter iSY
  • Start date
  • #1
iSY
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Hi, I've spent quite a while trying to figure this out but can't quite seem to get there... any ideas?

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.
 

Answers and Replies

  • #2
454
0
Your expression for the initial energy doesn't include a term for gravitational energy. mg(l-r) is not 0 when r = l/2.
 
  • #3
iSY
2
0
I thought that there would be no gravitational energy initially if both particles are at the plane of the table. Then when one fall through the halls its change in height is -(l-r). Could this be where my assumption is wrong?
 
  • #4
454
0
well it says "Given that the particle on the table is a distance l/2 from the hole". The other particle must be l/2 below the hole. Having the string started in a non-stretched position and the particle falling down until the string is stretched is problematic because the collision after the particle falls is probably unelastic.
 

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