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    This problem reminds me of polar coordinates problems. Here the only force is radial (tension in spring), and so the transverse component of the acceleration is zero, and the radial acceleration can be linked to the force by F = ma. Angular momentum conservation may also come in handy?
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    (Original post by A Slice of Pi)
    This problem reminds me of polar coordinates problems. Here the only force is radial (tension in spring), and so the transverse component of the acceleration is zero, and the radial acceleration can be linked to the force by F = ma. Angular momentum conservation may also come in handy?
    You're right. See my half-baked back-of-envelope solution.
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    (Original post by atsruser)
    Set up a rectangular coord system with the spring aligned along the x-axis at t=0. Let the particle have polar coords r, \theta. Then the Lagrangian of the system is:

    L=T-V = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2)-\frac{1}{2}\lamda(r-l)^2

    with associated Euler-Lagrange equations \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}) - \frac{\partial L}{\partial q} = 0 for each q \in \{ r,\theta \} giving:

    \frac{d}{dt}(m \dot{r}) -mr \dot{\theta}^2 + \lambda(r-l)=0 \Rightarrow m\ddot{r}= mr \dot{\theta}^2 +\lambda(l-r)
    \frac{d}{dt}(m r^2 \dot{\theta}) = 0 \Rightarrow m r^2 \dot{\theta} = C

    where the second eqn expresses conservation of AM.

    So we have:

    m\ddot{r}=C+\lambda(l-r) = K -\lambda r with K=C+\lambda l

    Now write \rho =\frac{\lambda r-K}{\lambda} to give m \ddot{\rho} = -\lambda \rho

    which is SHM in \rho. I got bored at this point, so I'll let someone else finish it off/correct my errors.
    I did mentioned earlier that it can be done with Lagrangian/Hamiltonian mechanics but it can be done with A level methods.
    I am waiting for the "comoving_frame" to show his method ... as he made a rapid exit ...
    (by the way I need no help ... I have already written a full solution as this is a question to go into my example books)
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    (Original post by A Slice of Pi)
    This problem reminds me of polar coordinates problems. Here the only force is radial (tension in spring), and so the transverse component of the acceleration is zero, and the radial acceleration can be linked to the force by F = ma. Angular momentum conservation may also come in handy?
    correct route
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    (Original post by TeeEm)
    correct route
    Is this the route you tried?
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    stop trying to cheat. if you can't do the work, you shouldn't be taking the exam!
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    (Original post by john2054)
    stop trying to cheat. if you can't do the work, you shouldn't be taking the exam!
    What exam?
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    (Original post by TeeEm)
    I did mentioned earlier that it can be done with Lagrangian/Hamiltonian mechanics but it can be done with A level methods.
    I am waiting for the "comoving_frame" to show his method ... as he made a rapid exit ...
    (by the way I need no help ... I have already written a full solution as this is a question to go into my example books)
    I decided that it was time to introduce the youngsters to a bit of real mechanics.
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    (Original post by A Slice of Pi)
    Is this the route you tried?
    I tried ....?!

    I made.
    I solved in retrospect that is why I put some numbers after as my solution was all in variables
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    (Original post by Mathemagicien)
    Joined today, knows about TeeEm's booklets, very rude to TeeEm... clearly duplicate account of some troll
    If it's a dupe of a troll then it needs reporting.
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    (Original post by spotify95)
    If it's a dupe of a troll then it needs reporting.
    It somebody with "some mathematical knowledge" which know the maths room well and my personal activity.
    He has done nothing to be reported but I think I know who he is (usernamewise)
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    Is there a way of doing this with polar coordinates? I might try, and inevitably fail, tonight

    Posted from TSR Mobile
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    (Original post by Krollo)
    Is there a way of doing this with polar coordinates? I might try, and inevitably fail, tonight

    Posted from TSR Mobile
    That is the way ...
    I will post full solution before I retire tonight.
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    (Original post by Xenon17)
    What exam?
    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
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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
    you think Teeems cheating?
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    (Original post by A Slice of Pi)
    This problem reminds me of polar coordinates problems. Here the only force is radial (tension in spring), and so the transverse component of the acceleration is zero, and the radial acceleration can be linked to the force by F = ma. Angular momentum conservation may also come in handy?
    Am I missing something on why we can completely omit weight?
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    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!!
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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!!
    shhh
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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
    I am sorry I had to cheat ...
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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!!
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