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    Far easier than the first ....

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    (Original post by TeeEm)
    Far easier than the first ....

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    Are you sure that this is right? I get an extra term of \frac{J^2}{m^2r^2} where J is the conserved AM.
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    (Original post by atsruser)
    Are you sure that this is right? I get an extra term of \frac{J^2}{m^2r^2} where J is the conserved AM.
    I hope so but these days I can never be sure

    this is

    conservation of AM
    shape constraint
    energy = constant
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    (Original post by TeeEm)
    I hope so but these days I can never be sure

    this is

    conservation of AM
    shape constraint
    energy = constant
    This is a rush job, so may be errors.

    The initial AM J=maU.

    Use cylindrical polars. We have z=f(r). Taking z and V(z)=mgz increasing upwards, and V(0)=0, we have

    L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2)-mgz

    and E-L equations for (r,\theta,z):

    (1) \frac{d}{dt}(m\dot{r}) - mr\dot{\theta}^2=0 \Rightarrow \frac{1}{2}\frac{d}{dr}(\dot{r}^  2)- r\dot{\theta}^2=0

    (2) \frac{d}{dt}(m r^2 \dot{\theta}) =0 \Rightarrow m r^2 \dot{\theta} =J \Rightarrow \dot{\theta} = \frac{J}{mr^2}

    (3) \frac{d}{dt}(m\dot{z}) + mg = 0 \Rightarrow \frac{m}{2}\frac{d}{dz}(\dot{z}^  2) + g = 0 \Rightarrow \dot{z}^2 + 2gz = C_1

    Now \dot{z} = \frac{d}{dt} f(r) = f'(r) \frac{dr}{dt} = \dot{r} f'(r) so:

    \dot{r}^2 (f'(r))^2 + 2gf(r) = C_1 (4)

    Substituting (2) into (1) gives:

    \frac{d}{dr}(\dot{r}^2)- \frac{2J^2}{m^2r^3}=0 \Rightarrow \dot{r}^2 + \frac{J^2}{m^2 r^2} =C_2 (5)

    and adding (4) and (5) gives:

    \dot{r}^2(1+[f'(r)]^2) + 2gf(r) +\frac{J^2}{m^2 r^2}  = K
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    (Original post by atsruser)
    This is a rush job, so may be errors.

    The initial AM J=maU.

    Use cylindrical polars. We have z=f(r). Taking z and V(z)=mgz increasing upwards, and V(0)=0, we have

    L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2)-mgz

    and E-L equations for (r,\theta,z):

    (1) \frac{d}{dt}(m\dot{r}) - mr\dot{\theta}^2=0 \Rightarrow \frac{1}{2}\frac{d}{dr}(\dot{r}^  2)- r\dot{\theta}^2=0

    (2) \frac{d}{dt}(m r^2 \dot{\theta}) =0 \Rightarrow m r^2 \dot{\theta} =J \Rightarrow \dot{\theta} = \frac{J}{mr^2}

    (3) \frac{d}{dt}(m\dot{z}) + mg = 0 \Rightarrow \frac{m}{2}\frac{d}{dz}(\dot{z}^  2) + g = 0 \Rightarrow \dot{z}^2 + 2gz = C_1

    Now \dot{z} = \frac{d}{dt} f(r) = f'(r) \frac{dr}{dt} = \dot{r} f'(r) so:

    \dot{r}^2 (f'(r))^2 + 2gf(r) = C_1 (4)

    Substituting (2) into (1) gives:

    \frac{d}{dr}(\dot{r}^2)- \frac{2J^2}{m^2r^3}=0 \Rightarrow \dot{r}^2 + \frac{J^2}{m^2 r^2} =C_2 (5)

    and adding (4) and (5) gives:

    \dot{r}^2(1+[f'(r)]^2) + 2gf(r) +\frac{J^2}{m^2 r^2}  = K
    shouldn't the r2 in the extra term be a2
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    (Original post by TeeEm)
    shouldn't the r2 in the extra term be a2
    Not as far as I can see. a,U have already been absorbed into J, so they won't appear again.
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    (Original post by atsruser)
    Not as far as I can see. a,U have already been absorbed into J, so they won't appear again.
    I cannot see anything wrong with you ... maybe I got it wrong
    will check my solution before I post
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    (Original post by atsruser)
    Not as far as I can see. a,U have already been absorbed into J, so they won't appear again.
    I found what is happening we are both correct ...
    Stupidity on my part ...
    the extra bit is a function or r, which I absorbed into the potential, rellabelling as f(r), forgetting I had already used it to describe the equation of the curve
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    my solution
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