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GCSE physics problem

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  1. Feb 5, 2007 #1
    I am currently studying for my GCSE's but have a keen intrest in physics and maths. Today at school i noticed a question;

    A chain of length a is placed on a frictional-less table with length b of the chain dangling off the edge of the table.

    Using [tex]\\F = ma[/tex]

    derive the formula:

    [tex]\\T= \sqrt{\frac{a}{g}} \\In \frac{\sqrt{ a^2- b^2 }}{b}[/tex]
     
    Last edited: Feb 5, 2007
  2. jcsd
  3. Feb 5, 2007 #2
    What exactly is T? I have some guesses, but better to state it explicitly.
     
  4. Feb 5, 2007 #3
    sorry that should be
    [tex]\\T= \sqrt{\frac{a}{g}} \\In \frac{\sqrt{ a^2-b^2 }}{b}[/tex]
     
  5. Feb 5, 2007 #4
    T is the time it takes for the entire chain to fall off the table

    im sorry i shoudl have made myself clear, but using LaTex for the first time is canny confusing
     
  6. Feb 5, 2007 #5
    I am currently studying for my GCSE's but have a keen intrest in physics and maths. Today at school i noticed a question;

    A chain of length a is placed on a frictional-less table with length b of the chain dangling off the edge of the table.

    Using [tex]\\F = ma[/tex]

    derive the formula:

    [tex]\\T= \sqrt{\frac{a}{g}} \\In \frac{\sqrt{ a^2-b^2 }}{b}[/tex]

    where T= time taken for the chain to slip off the table
    and the acceleration due to gravity is normal

    any solutions would be greatly appreciated as i have gave the question a lot of thought and have some idea of how it should be solved. however i doubt my knowledge of physics/maths will allow me to complete this question.

    cheers in advance
     
  7. Feb 5, 2007 #6
    The forum rules ask what efforts you have made, so maybe here at least a description of your thoughts regarding the acceleration of the chain--is it constant, variable, and why?
     
  8. Feb 5, 2007 #7
    The force exerted by the dangling chain is enough to create a resultant force allowing the rest of the chain to be accelerated, and hence be pulled off the table. The mass of the dangling chain increases(as more of the chain is pulled of the table), and the force exerted by the dangling chain increases, which means that a bigger force is pulling the chain on the table towards the ground. As F increase and mass stays constant doesn't this mean that the rate of aceleration must increase as F increases.

    Im really quite unsure as to how to reach the final equation from F=ma.

    Thanks in advance devendoc.
     
  9. Feb 5, 2007 #8
    Thats how I see it. As the dangling chain gets longer, the purely inertial component on the table decreases. So far starters I would try to set up both as a function of time, maybe let x(t) be the change in length of the supported section, so that A(t) length falling vertical = A+x(t) while the other segment B-x(t). Assume that mass is proportional to length by inserting some constant, then you should have masses of both as a function of time, and from that acceleration as a function of time. Then if my crystal ball is working the math will get a little messy. See what you can come up with.
     
  10. Feb 5, 2007 #9

    Dick

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    Yes, it is kind of messy - involves setting up and solving an ODE with boundary conditions. The answer as a function of time comes out like b(t)=b*cosh(k*t) where k=sqrt(a/g). Inverting and solving for t involves solving a quadratic. And then finally, you notice that the above answer is wrong. If you take b very close to a, this formula would starting giving you negative values of T. Add a/b to the argument of the ln to correct it. Is this really all worth it?
     
    Last edited: Feb 5, 2007
  11. Feb 5, 2007 #10
    Not to me it wasn't. Just understanding the concepts of the problem I thought was, which is not so hard to get to the ODE part, assuming the OP knows a(t)=d'' x(t)/dt^2. From there I planned to go running to my copy of Taylor. ;-D
     
  12. Feb 27, 2007 #11
    Im sorry the formula should be [tex]\\T= \sqrt{\frac{a}{g}} \\ln \frac{\a + sqrt{ a^2-b^2 }}{b}[/tex]

    Having worked out acceleration and intergrating to find velocity then rearranging the formula, i am left with;

    [tex] \\int frac{dx}{\\sqrt{x^2 - b^2}} = \\int \\sqrt{frac {g}{a}} dt [/tex]

    how do i intergrate the left hand side????
    im soo frustrated this question has been annoying me for ages, i would love to complete it!! thank you for any help in advance

    peace
     
  13. Feb 27, 2007 #12
    Having worked out acceleration and intergrating to find velocity then rearranging the formula, i am left with;

    [tex] \\int \\frac{dx}{\\sqrt{x^2 - b^2}} = \\int \\sqrt{\\frac {g}{a}} dt [/tex]

    how do i intergrate the left hand side????
    im soo frustrated this question has been annoying me for ages, i would love to complete it!! thank you for any help in advance

    peace
     
  14. Feb 27, 2007 #13
    ARGHHHH I GIVE UP WITH LATEX!!! sorry to spam this topic

    peace
     
  15. Feb 27, 2007 #14
    how do i intergrate dx/sqrt((x^2)-(b^2))
     
  16. Feb 27, 2007 #15

    Dick

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    Try the substitution x=b*cosh(u). Are you ok with hyperbolic functions?
     
  17. Feb 27, 2007 #16
    noo, im trying to research them but any hints would be appreciated. thanks dick
     
  18. Feb 27, 2007 #17

    Dick

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    They are a lot like trig functions, but even easier. Here's what you need to know.

    cosh(x)=(e^x+e^(-x))/2
    sinh(x)=(e^x-e^(-x))/2

    Now as an exercise, you prove (which you'll need to do the problem):

    d(cosh(x))=sinh(x)*dx
    cosh(x)^2-1=sinh(x)^2

    That makes the LHS integral easy.
     
  19. Feb 27, 2007 #18

    cristo

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    As an aside; here's your LaTex corrected. You just put too many \'s in it!
     
  20. Feb 27, 2007 #19
    haha sorry that doesnt help. i dont have a clue about trig functions either. im only 16 and my knowledge of maths is no where near that need to complete this question. although i have read up outside the syllabus and would consider myself to be very good at math, it would be good to have a bit more focused step by step advice. ive got this far through the question and it would be a shame to have to wait a few years before i had the knowledge to finish it.
    thanks a lot Dick for the help and any future help.
    peace
     
  21. Feb 27, 2007 #20
    thanks a lot cristo, i did go a bit crazy with the \\
     
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