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Motion on 2D surface - particle hanging from string

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data
    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.

    Derive the expression

    [itex]\dot{r}^2 = A - \frac{B}{r^2} - gr[/itex]

    where r is the distance of the first particle from the hole, and A and B are constants. Use the equations of motion for each particle in polar coordinates.

    2. Relevant equations
    General acceleration in polar coordinates:

    [itex]\mathbf{\ddot{r}}=(\ddot{r}-r\dot{\theta}^2)\mathbf{\hat{r}}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \mathbf{ \hat{\theta} }[/itex]


    3. The attempt at a solution
    I'm using a plane-polar coordinate system in the plane of the table with the origin at the hole. The forces acting on the first particle (on the table) (in the plane) are just the tension, T. So:

    [itex]-T\mathbf{\hat{r}} = (\ddot{r}-r\dot{\theta}^2)\mathbf{\hat{r}}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \mathbf{ \hat{\theta} }[/itex]

    And for the hanging particle, the forces are the tension and weight, so (aligning the z axis perpendicular to the table):

    [itex](T-mg)\mathbf{\hat{z}} = -m\ddot{r}\mathbf{\hat{z}}[/itex]

    From these equations, I get the following system:

    [itex]\ddot{r} - r\dot{\theta}^2 = -\frac{T}{m}[/itex]

    [itex]r\ddot{\theta} + 2\dot{r}\dot{\theta} = 0[/itex]

    [itex]\ddot{r} = g - \frac{T}{m}[/itex]

    I've tried manipulating these equations for a while now but don't get anything close to the required form. Are these equations incorrect? I would be very grateful for any hints! :)
     
  2. jcsd
  3. Feb 8, 2014 #2
    I believe there is a sign error here. Let ##h## be the distance from the hole to the hanging particle, pointing downward. Clearly, ##h = l - r##. Then ##\ddot h = mg - T ##. ##mg## is positive and ##T## negative because ##h## is chosen to point downward. So ## - \ddot r = mg - T ## or ## \ddot r = T - mg ##.

    The second equation can be integrated. That can also be obtained by recalling conservation laws in a central force field.
     
  4. Feb 8, 2014 #3
    There was actually a previous part of the question asking for the derivation of the same equation using conservation laws, so I can't use that here. I don't understand how that equation can be integrated.
     
  5. Feb 8, 2014 #4
    Multiplied by ##r##, it becomes...?
     
  6. Feb 8, 2014 #5
    ...exact. So this gives:

    [itex]r^2\dot{\theta} = k[/itex]

    where k is a constant.

    Rearranging for [itex]\dot{\theta}[/itex] and substituting into the first equation:

    [itex]\ddot{r} - r\frac{k^2}{r^4} = \ddot{r} - g[/itex]

    where I have also used the other eqn. to eliminate T. EDIT: just realised this part is wrong, now correcting... This gives

    [itex]\frac{k^2}{r^3} = g[/itex]

    [itex]\frac{k^2}{r^2} = gr[/itex]

    Which is close, but no constant term and no mention of [itex]\dot{r}^2[/itex] :frown:
     
    Last edited: Feb 8, 2014
  7. Feb 8, 2014 #6
    So I actually get:

    [itex]\ddot{r} - r\dot{\theta}^2 = -\ddot{r} - g[/itex]

    [itex]2\ddot{r} - r\dot{\theta}^2 + g = 0[/itex]

    Now substituting [itex]\dot{\theta}[/itex], from above post:

    [itex]2\ddot{r} - \frac{k^2}{r^3} + g = 0[/itex]

    [itex]2r\ddot{r} - \frac{k^2}{r^2} + gr = 0[/itex]

    So still not correct.
     
  8. Feb 8, 2014 #7
    That is correct, but not finished. Use the same trick again - but I will let you find the integrating factor this time. The equation you are supposed to derive in the end is a good hint.
     
  9. Feb 8, 2014 #8
    Oh, okay, I've got it now (multiply by [itex]\frac{\dot{r}}{r}[/itex]).

    Thank you so much for your help, voko.
     
  10. Feb 8, 2014 #9
    Have you figured out the part with conservation laws? What law does the integral you found in #5 correspond to?
     
  11. Feb 8, 2014 #10
    Yes, that was conservation of angular momentum, so that [itex]k=\frac{L}{m}[/itex]
     
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