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    STEP I, 2001 Q11

    A smooth cylinder with circular cross-section of radius a is held with its axis horizontal. A light elastic band of unstretched length 2\pi a and modulus of elasticity ? is wrapped round the circumference of the cylinder, so that it forms a circle in a plane perpendicular to the axis of the cylinder. A particle of mass m is then attached to the rubber band at its lowest point and released from rest.

    (i) Given that the particle falls to a distance 2a below the below the axis of the cylinder, but no further, show that:
    \lambda =\frac{9\pi mg}{(3\sqrt{3}-\pi)^2}

    I've let the tension in the string=T and I have found T=\frac{mg}{\sqrt3}

    I have also found the extension of the band to be 2a(\sqrt{3}-\frac{\pi}{3})

    This question has had me stumped for a while. I keep on getting similar answers in terms of a.

    Any pointers?
    Is it correct to say T=\lambda 2a(\sqrt3 -\frac{\pi}{\sqrt3})? (tried it, couldn't get it to work)
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    If you imagine this scenario in real life, what would happen *after* the particle reaches the lowest point?
    Does this tally with your analysis of the tension?
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    When you say my "analysis of the tension", which part do you mean?
    T=\lambda 2a(\sqrt3 -\frac{\pi}{\sqrt3})?
    I guess what you are getting at is that the mass will "bounce" back up again?
    If this is to do with harmonic oscillators or something like that then I don't know as I haven't really studied those.
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    (Original post by ben-smith)
    When you say my "analysis of the tension", which part do you mean?
    T=\lambda 2a(\sqrt3 -\frac{\pi}{\sqrt3})?
    I guess what you are getting at is that the mass will "bounce" back up again?
    If this is to do with harmonic oscillators or something like that then I don't know as I haven't really studied those.
    Yeah, the particle would bounce up again, so it would have acceleration at the lowest point, so you can't assume that the tension force equals gravity there.

    Instead, use an energy argument.
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    (Original post by DFranklin)
    Yeah, the particle would bounce up again, so it would have acceleration at the lowest point, so you can't assume that the tension force equals gravity there.

    Instead, use an energy argument.
    Do you mean calculating the potential energy gain?
    PE=\frac{1}{2}\lambda x^2where x is the extension.
    PE=\frac{1}{2}\lambda (2a(\sqrt{3}-\frac{\pi}{3}))^2
    I'm still a bit lost
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    Yes, the gain in elastic PE (i.e. the energy in the string) has to equal the loss in gravitational PE (because the particle has moved down).
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    (Original post by DFranklin)
    Yes, the gain in elastic PE (i.e. the energy in the string) has to equal the loss in gravitational PE (because the particle has moved down).
    thanks But I still can't cancel the as.
    \frac{1}{2}\lambda (2a(\sqrt{3}-\frac{\pi}{3}))^2=mga, you get an a^2 on the LHS and an a on the RHS leaving one solitary a.
    Hmmmmm... it appears I have some more work to do.
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    OK, I'm afraid I've had to give in and look at the answer. I've been trying to do this question for literally days and haven't been able to do it. And to top it all off I don't understand the answer given by BrianEverrit, he seems to use some form of hooke's law that I've never seen and I cannot find on the internet.
    He says:
    tension=\frac{2\lambda a(\sqrt3- \pi/3)}{2\pi a}
    Am I being a moron? where did the 2\pi a come from?
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    Sorry, should have noticed this earlier. Tension = extension * elasticity / original_length. I think you have been using just extension * elasticity.
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    (Original post by DFranklin)
    Sorry, should have noticed this earlier. Tension = extension * elasticity / original_length. I think you have been using just extension * elasticity.
    I am not familiar with that formula, is that a special case of hooke's law?
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    It's the definition of modulus of elasticity. (Note: it is entirely possible I'm getting this wrong, as it's about 4 years since I did such a question, and that was just the odd STEP Q - it's 25 years since I learnt this. On the other hand, I did check on Wikipedia).
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    (Original post by DFranklin)
    It's the definition of modulus of elasticity. (Note: it is entirely possible I'm getting this wrong, as it's about 4 years since I did such a question, and that was just the odd STEP Q - it's 25 years since I learnt this. On the other hand, I did check on Wikipedia).
    I looked at the same place and have been studying a similar thing in physics, but all those formulae required you to know the cross sectional area so I thought I couldn't go down that route.
    So I guess you're saying I can bypass this?
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    Actually now that I have done some more reading, in wikipedia it says
    "stress is the restoring force caused due to the deformation divided by the area to which the force is applied".
    Since \lambda=\frac{stress}{strain}, Would it be true to say:
    \lambda=\frac{Tension}{strain}? because I think that gives me what brianeverrit got.
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    The key line in the wiki page is: "the elastic modulus becomes the stress needed to cause a sample of the material to double in length." (I don't recall the definitions of stress and strain as I never used those outside of my Physics A-level many many years ago).
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    In Physics, you typically learn that F=kx, with some Young's Modulus thrown in from time to time. In the Maths mechanics, they use the modulus of elasticity: k=\frac{\lambda}{\ell}, where lambda is your modulus of elasticity and l your initial length. So E=\frac{1}{2}kx^2 becomes E=\frac{\lambda x^2}{2 \ell}. It's worth checking how Maths mechanics compares to Physics mechanics - in general they are the same but in a few situations, like this one, they differ slightly.
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    (Original post by TheDavibob)
    In Physics, you typically learn that F=kx, with some Young's Modulus thrown in from time to time. In the Maths mechanics, they use the modulus of elasticity: k=\frac{\lambda}{\ell}, where lambda is your modulus of elasticity and l your initial length. So E=\frac{1}{2}kx^2 becomes E=\frac{\lambda x^2}{2 \ell}. It's worth checking how Maths mechanics compares to Physics mechanics - in general they are the same but in a few situations, like this one, they differ slightly.
    Thanks - I was hoping someone who had a bit of knowledge about the A-level syllabus would weight in.
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    (Original post by TheDavibob)
    In Physics, you typically learn that F=kx, with some Young's Modulus thrown in from time to time. In the Maths mechanics, they use the modulus of elasticity: k=\frac{\lambda}{\ell}, where lambda is your modulus of elasticity and l your initial length. So E=\frac{1}{2}kx^2 becomes E=\frac{\lambda x^2}{2 \ell}. It's worth checking how Maths mechanics compares to Physics mechanics - in general they are the same but in a few situations, like this one, they differ slightly.
    This is what I was going to say. The difference causes some confusion when you teach elasticity in Maths if the Physics teachers have got there first.

    Note that this question would no longer come up on STEP I. The specification changed in 2002 to fit in better with the Curriculum 2000 A level specifications. Elasticity would now only appear on STEP III.
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    Thanks guys, I've got it now. Two days well spent ehhh?
 
 
 
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