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    With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..
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    (Original post by MathsAstronomy12)
    With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..
    Tension.
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    (Original post by Zacken)
    Tension.
    Then how come :

    https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf (question 7b)

    Mark scheme :

    https://8a40d6c38bafca75cc407741c0f3...%20Edexcel.pdf
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    bump
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    I think maybe one is a necessary condition and the other is necessary and sufficient or something like that? I'm not entirely sure myself. Student403 will know.
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    (Original post by MathsAstronomy12)
    With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..
    The speed has to be > 0.
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    (Original post by EricPiphany)
    The speed has to be > 0.
    But then the particle could be moving freely under gravity, not necessarily remaining in circular motion..?
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    (Original post by MathsAstronomy12)
    With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..
    (Original post by Zacken)
    Tension.
    Depends on what the particle is attached to. If it is on the end of a rod or in a tube, it can't fall out of the circular path, so it is the speed that must remain positive (not to mention real). If it is on the end of a string or going round inside a hollow sphere, so that it can fall out of the circular path, it is the tension (or normal reaction) that must remain > = 0 (think the Simpsons movie).
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    (Original post by tiny hobbit)
    Depends on what the particle is attached to. If it is on the end of a rod or in a tube, it can't fall out of the circular path, so it is the speed that must remain positive (not to mention real). If it is on the end of a string or going round inside a hollow sphere, so that it can fall out of the circular path, it is the tension (or normal reaction) that must remain > = 0 (think the Simpsons movie).
    Aaaahh, that makes a ton of sense, thank you!
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    (Original post by MathsAstronomy12)
    But then the particle could be moving freely under gravity, not necessarily remaining in circular motion..?
    Already answered above by tiny hobbit.
    Accept that I would say that the speed > 0 is a condition that needs to be fulfilled either way.
    The speed can never be < 0, if the maths says so, it'll never reach there.
    And tension by definition is greater or equal to zero.
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    (Original post by tiny hobbit)
    Depends on what the particle is attached to. If it is on the end of a rod or in a tube, it can't fall out of the circular path, so it is the speed that must remain positive (not to mention real). If it is on the end of a string or going round inside a hollow sphere, so that it can fall out of the circular path, it is the tension (or normal reaction) that must remain > = 0 (think the Simpsons movie).
    Oh, so if it was at the end of a string, T > 0? That makes sense ish as in the question above, the particle is attached to a rod, the sneaky bugger

    EDIT : having reread your post, you've already answered that ^. Thank you!
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    (Original post by MathsAstronomy12)
    With questions involving a particle moving in a complete vertical circle, is it the tension that has to be > 0 or v^2 that has to be > 0 at the top? I'll post an example if required..
    Yes. tiny hobbit explains it well..


    Here is a summary from the Edexcel textbook. Basically the tension in a rod is irrelevant

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    (Original post by Student403)
    Yes. tiny hobbit explains it well..


    Here is a summary from the Edexcel textbook. Basically the tension in a rod is irrelevant

    Okay, thank you
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    (Original post by Student403)
    Yes. tiny hobbit explains it well..
    I'm glad that all those years of teaching it have proved useful!
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    (Original post by tiny hobbit)
    I'm glad that all those years of teaching it have proved useful!
    I wish I had you to teach me M3 :laugh: Self teaching was a nightmare..
 
 
 
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