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    (Original post by john2054)
    I just don't like it when people come on here and say 'please tell me the answer to this question'. It is true this individual may well be genuine given the insight he has subsequently showed in to the subject matter. But i don't think that it hurts to question their credentials, when they come on here (tsr) doing this. he wouldn't be the first!!
    you must be new on F38
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    I am sure we are all looking forward to seeing a really perceptive mathematical contribution from john 2054 sometime soon.
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    (Original post by TeeEm)
    I tried ....?!

    I made.
    I solved in retrospect that is why I put some numbers after as my solution was all in variables
    Ah, I see. Is r between 1 and 3, by any chance?
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    (Original post by A Slice of Pi)
    Ah, I see. Is r between 1 and 3, by any chance?
    it is...
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    (Original post by TeeEm)
    it is...
    Great. A very nice question btw, I'll post a solution afterwards but I'm guessing you're going to do a much neater one later anyway. I made use of the fact that \frac{d^2r}{dt^2} = \frac{d}{dr}(\frac{1}{2}(\frac{d  r}{dt})^2) which is a neat result that I've only ever seen once before, in a SHM problem.
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    (Original post by A Slice of Pi)
    Great. A very nice question btw, I'll post a solution afterwards but I'm guessing you're going to do a much neater one later anyway. I made use of the fact that \frac{d^2r}{dt^2} = \frac{d}{dr}(\frac{1}{2}(\frac{d  r}{dt})^2) which is a neat result that I've only ever seen once before, in a SHM problem.
    Excellent !!
    I am waiting for Krollo
    (My solution will up around midnight)
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    (Original post by nerak99)
    I am sure we are all looking forward to seeing a really perceptive mathematical contribution from john 2054 sometime soon.
    i know the chaos theory based on random fractals, the basics anyway. failing that and i'm going to have to pass up your offer sorry!
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    (Original post by john2054)
    i know the chaos theory based on random fractals, the basics anyway. failing that and i'm going to have to pass up your offer sorry!
    You crack me up.
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    (Original post by nerak99)
    I am sure we are all looking forward to seeing a really perceptive mathematical contribution from john 2054 sometime soon.
    my major is in sociology and my minor is in theatre studies, how if you have any questions about gramsci's hegemony theory, or south african township theatre (what i did a presentation on today) then please fire away!!!
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    Spoiler:
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    A quick sketch of the situation reveals that the only horizontal force involved in the motion is the tension in the stretched spring. TheBBQ we need not consider the weight as it acts vertically, and the polar coordinates model only takes into account the horizontal forces. By Newton II: \frac{-8(r-1)}{9}=\ddot{r} - r\dot{\theta}^2. Now we should aim to get rid of the angle. To do this, consider the conservation of angular momentum. Initial momentum = momentum at time t so r^2\dot{\theta} = 2. Combining the two results, we get

    \frac{-8(r-1)}{9}=\ddot{r} - 4r^{-3} \implies \frac{-8}{9}r + \frac{8}{9} + 4r^{-3} =\ddot{r}

    Using the marvelous result that \ddot{r} = \frac{d}{dr}(\frac{1}{2}\dot{r}^  2), we see that

    \frac{-8}{9}r + \frac{8}{9} + 4r^{-3} = \frac{d}{dr}(\frac{1}{2}\dot{r}^  2)

    and so

    \int{\frac{-16}{9}r + \frac{16}{9} + 8r^{-3}} dr = \dot{r}^2 + c

    \implies \frac{-8}{9}r^2 + \frac{16}{9}r -4r^{-2} + c = \dot{r}^2
    when r = 1, \dot{r} = 0 \implies c = \frac{28}{9}

    Note that when r is at it's maximum \dot{r}=0, so
    \frac{-8}{9}r^2 + \frac{16}{9}r -4r^{-2} + \frac{28}{9} = 0
    or equivalently
    \frac{-8}{9}r^4 + \frac{16}{9}r^3 + \frac{28}{9}r^2 -4  = 0
    You could use the factor theorem or a sketch to verify that the maximum value of r is indeed 3. Thus 1\leqslant r \leqslant 3
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    (Original post by A Slice of Pi)
    Spoiler:
    Show
    A quick sketch of the situation reveals that the only horizontal force involved in the motion is the tension in the stretched spring. TheBBQ we need not consider the weight as it acts vertically, and the polar coordinates model only takes into account the horizontal forces. By Newton II: \frac{-8(r-1)}{9}=\ddot{r} - r\dot{\theta}^2. Now we should aim to get rid of the angle. To do this, consider the conservation of angular momentum. Initial momentum = momentum at time t so r^2\dot{\theta} = 2. Combining the two results, we get

    \frac{-8(r-1)}{9}=\ddot{r} - 4r^{-3} \implies \frac{-8}{9}r + \frac{8}{9} + 4r^{-3} =\ddot{r}

    Using the marvelous result that \ddot{r} = \frac{d}{dr}(\frac{1}{2}\dot{r}^  2), we see that

    \frac{-8}{9}r + \frac{8}{9} + 4r^{-3} = \frac{d}{dr}(\frac{1}{2}\dot{r}^  2)

    and so

    \int{\frac{-16}{9}r + \frac{16}{9} + 8r^{-3}} dr = \dot{r}^2 + c

    \implies \frac{-8}{9}r^2 + \frac{16}{9}r -4r^{-2} + c = \dot{r}^2
    when r = 1, \dot{r} = 0 \implies c = \frac{28}{9}

    Note that when r is at it's maximum \dot{r}=0, so
    \frac{-8}{9}r^2 + \frac{16}{9}r -4r^{-2} + \frac{28}{9} = 0
    or equivalently
    \frac{-8}{9}r^4 + \frac{16}{9}r^3 + \frac{28}{9}r^2 -4  = 0
    You could use the factor theorem or a sketch to verify that the maximum value of r is indeed 3. Thus 1\leqslant r \leqslant 3
    EXCELLENT!!
    I will post my method now too for completeness
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    (Original post by A Slice of Pi)
    Spoiler:
    Show
    A quick sketch of the situation reveals that the only horizontal force involved in the motion is the tension in the stretched spring. TheBBQ we need not consider the weight as it acts vertically, and the polar coordinates model only takes into account the horizontal forces. By Newton II: \frac{-8(r-1)}{9}=\ddot{r} - r\dot{\theta}^2. Now we should aim to get rid of the angle. To do this, consider the conservation of angular momentum. Initial momentum = momentum at time t so r^2\dot{\theta} = 2. Combining the two results, we get

    \frac{-8(r-1)}{9}=\ddot{r} - 4r^{-3} \implies \frac{-8}{9}r + \frac{8}{9} + 4r^{-3} =\ddot{r}

    Using the marvelous result that \ddot{r} = \frac{d}{dr}(\frac{1}{2}\dot{r}^  2), we see that

    \frac{-8}{9}r + \frac{8}{9} + 4r^{-3} = \frac{d}{dr}(\frac{1}{2}\dot{r}^  2)

    and so

    \int{\frac{-16}{9}r + \frac{16}{9} + 8r^{-3}} dr = \dot{r}^2 + c

    \implies \frac{-8}{9}r^2 + \frac{16}{9}r -4r^{-2} + c = \dot{r}^2
    when r = 1, \dot{r} = 0 \implies c = \frac{28}{9}

    Note that when r is at it's maximum \dot{r}=0, so
    \frac{-8}{9}r^2 + \frac{16}{9}r -4r^{-2} + \frac{28}{9} = 0
    or equivalently
    \frac{-8}{9}r^4 + \frac{16}{9}r^3 + \frac{28}{9}r^2 -4  = 0
    You could use the factor theorem or a sketch to verify that the maximum value of r is indeed 3. Thus 1\leqslant r \leqslant 3
    Maths the beautiful.
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    (Original post by TeeEm)
    EXCELLENT!!
    I will post my method now too for completeness
    Oh, I didn't see that TeeEm. Nice solution.
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    (Original post by Marxist)
    Oh, I didn't see that TeeEm. Nice solution.
    thank you !!!
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    (Original post by john2054)
    or the coursework.

    the fact is that this kid is clearly out of his depth, and essentially with no insight in to the subject matter, is asking us to answer the question for him.

    this is cheating
    "this kid" TeeEm is such a cheat

    :rofl:
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    Yeah, TeeEm is so going to fail his exam as he cannot independently deduce the answer to these simple problems!!!

    No use relying on others TeeEm, they wont be in the exam room with you. I'm sure John will agree with me here.

    Best of luck with your A Levels.
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    If i actually remembered any maths, which i don't, would be a start?!
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    (Original post by john2054)
    If i actually remembered any maths, which i don't, would be a start?!
    Then why bother chirping in?
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    I tell you what, in a weeks time I will comment on gramscis hegemony wotsit and John can give us a short treatise on the conservation of angular momentum balanced with central acceleration as illustrated by Tee Ms original problem.
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    Some of the posts on here are beyond hilarious.


    TeeEm is the ****ing boss.


    End transmission.
 
 
 
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