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Cambridge Physics Problems 11

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    This is a part of the "Cambridge Physics Problems" thread, an exercise which I have taken up myself. Please read here for a short introduction on the types of questions featured in threads of the same name. Please search for "Cambridge Physics Problems" for threads featuring past questions.
    I appreciate all of your help on this. If possible, please show me the mathematical workings of the question (since this is what the question emphasises).
    Thank you!

    Question:

    A cable of mass m and area of cross-section A has length L when it is laid out on the ground and measured. It is then suspended vertically on one end.

    By considering a small length \delta x of the suspended cable, show that the total energy stored as a result of the extension of the cable under its own weight is \frac{Lm^2g^2}{6AE}, where E is the Young modulus of the material of the cable. Assume that Hooke's law applies.

    Attempt:

    I am assuming that at different distances from the point of hanging, the wire carries different values of stress (and hence different values of strain). So in this case, the "discs" at different sections of the wire will produce different values of extension, and I right?

    Stress in a section of the cable at a distance x below the point of suspension is expressed as \dfrac{(L-x)mg}{LA}.
    Thus strain in a section of the cable at a distance x below the point of suspension is expressed as \dfrac{(L-x)mg}{LAE}.
    Strain energy stored = 1/2 \dfrac{(L-x)mg}{LA} \dfrac{(L-x)mg}{LAE}

    How do we integrate and find the total energy of the wire as a result of the extension of the cable under its own weight? Different parts will have different extensions, and hence different amount of energy, so am I actually dealing with 2 variables over here?

    These are the information that I manage to get on my own. I am uncertain of my own reasoning at attempting this question. Could anyone please guide me through this? What other questions do you think I should ask myself when I attempt the question?

    Your help is greatly appreciated. Thank you!
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    (Original post by johnconnor92)
    This is a part of the "Cambridge Physics Problems" thread, an exercise which I have taken up myself. Please read here for a short introduction on the types of questions featured in threads of the same name. Please search for "Cambridge Physics Problems" for threads featuring past questions.
    I appreciate all of your help on this. If possible, please show me the mathematical workings of the question (since this is what the question emphasises).
    Thank you!

    Question:

    A cable of mass m and area of cross-section A has length L when it is laid out on the ground and measured. It is then suspended vertically on one end.

    By considering a small length \delta x of the suspended cable, show that the total energy stored as a result of the extension of the cable under its own weight is \frac{Lm^2g^2}{6AE}, where E is the Young modulus of the material of the cable. Assume that Hooke's law applies.

    Attempt:

    I am assuming that at different distances from the point of hanging, the wire carries different values of stress (and hence different values of strain). So in this case, the "discs" at different sections of the wire will produce different values of extension, and I right?

    Stress in a section of the cable at a distance x below the point of suspension is expressed as \dfrac{(L-x)mg}{LA}.
    Thus strain in a section of the cable at a distance x below the point of suspension is expressed as \dfrac{(L-x)mg}{LAE}.
    It's ok up to this point.
    As strain is extension per unit length
    The extension δe of this small section of cable is strain times δx where δx is the length of the small section.

    So you now have
    \delta e = \frac{(L-x)mg}{LAE} \delta x --- (1)

    Next you need the strain energy δV of this small section, this is ½k(δe)² --- (2)
    Finally, k can be found from E
    If you do this you will get k = EA/δx --- (3)

    Sub k from (3) and δe from (1) into (2)

    You will get a formula for δV in terms of x (and δx)
    Integrate this wrt x from L=0 to L=L to find the total strain energy in the wire.

    This will give you the formula required.
    I have done this and it does work.
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    (Original post by johnconnor92)
    Strain energy (PER UNIT VOLUME) stored = 1/2 \dfrac{(L-x)mg}{LA} \dfrac{(L-x)mg}{LAE}
    Is strain energy (PER UNIT VOLUME) useful in this context? How should I go about solving the question with the formula for strain energy (per unit volume)? Will it be more difficult to solve the question?

    How do we integrate and find the total energy of the wire as a result of the extension of the cable under its own weight? Different parts will have different extensions, and hence different amount of energy, so am I actually dealing with 2 variables over here?
    You mentioned that my reasoning is okay up to the point before the quote above. Is my quote here invalid or flawed? How would you advise me to improve my reasoning?



    What further questions would you advise me to ask myself when I'm attempting the question?
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    There's nothing wrong with strain energy per unit volume, but in this question it isn't asked for. If it had been you would have obtained a slightly different formula at the end. You were asked for the total strain energy in the wire. The energy per unit volume is just this divided by LA.
    But that step in your method was not the correct one.

    You ask about reasoning.

    These things come with experience. After you have solved a few similar (integration) problems you start to see the pattern and the logic.
    The steps in the reasoning (for me at least) were
    I need strain energy in the whole wire.
    Hooke's Law obeyed therefore ½ke² gives strain energy. e = extension.
    The wire is stretched under its own weight so it's not linear and needs integration.
    Take a small cylinder section of wire length δx a distance x from one end, and find the extension δe.
    Use δe to find the strain energy in this small cylinder ½kδe²
    I need to integrate wrt x along the whole length of wire so I need the strain energy in terms of x and δx
    δx then becomes the dx in the integral.
    I need k in terms of δx.
    The key to doing these integrals is to get the formula to be integrated in terms of the correct variable.
    Here you have the element of length δx and have the distance x along the wire.
    You know you will need to integrate wrt x along the wire, so you must get the formula for the strain energy in the element in terms of x. The other terms must be constant.
    I'm sure there may be other ways of approaching this, but this is my recommendation.
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    Is k = EA/δx derived from such? E = </span></font>\frac{k \delta e/A}{\delta e / \delta x}<font color="#444444"><span style="font-family: verdana">

    I had trouble determining which symbol to assign the small extension. Please ignore the formatting codes. I tried to eliminate them but I failed. And there's nothing wrong with my LateX codes.
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    Why are we assuming that A remains the same after extension? By right it SHOULD decrease, yes? How do we justify not taking into account the small decrease in cross-sectional area?
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    (Original post by johnconnor92)
    Why are we assuming that A remains the same after extension? By right it SHOULD decrease, yes? How do we justify not taking into account the small decrease in cross-sectional area?
    Ask the exam board.
    The answer they expect is arrived at by not taking into account any change in the cross section area.
    They no doubt consider it a reasonable assumption to make, that gives the correct answer to a good approximation.
    You can find out for yourself if it is a reasonable assumption by checking out the Poisson's Ratio for the material used.
    http://en.wikipedia.org/wiki/Poisson's_ratio

    Young's Modulus = (stress/strain) = (F/A)/(e/L)
    Which in the case of the small section length δx becomes

    E=(F/A)/(δe/δx) --- (1)

    The Hooke's Law constant is by definition k = F/δe (Force per unit extension)

    Sub for F/δe in this equation from equ (1)
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    (Original post by Stonebridge)
    Ask the exam board.
    The answer they expect is arrived at by not taking into account any change in the cross section area.
    They no doubt consider it a reasonable assumption to make, that gives the correct answer to a good approximation.
    You can find out for yourself if it is a reasonable assumption by checking out the Poisson's Ratio for the material used.
    http://en.wikipedia.org/wiki/Poisson's_ratio
    Ah thank you very much. As a matter of fact, the book introduces a question on Poisson's ratio shortly after this one. I shall post that up soon.

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