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Please help. Urgent. A spring with mass

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data

    We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass M, equilibrium length L0, and spring constant k. The work done to stretch or compress the spring by a distance L is 0.5kx^2, where x = L – L0.
    (a) Consider a spring, as described above that has one end fixed and the other end moving with speed v. Assume that the speed of points along the length of the spring varies linearly with distance l from the fixed end. Assume also that the mass M of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of M and v.
    (Hint: Divide the spring into pieces of length dl; find the speed of each piece in terms of l, v, and L; find the mass of each piece in terms of dl, M, and L; and integrate from 0 to L. The result is not 0.5Mv^2, since not all of the spring moves with the same speed.)


    The attempt at a solution

    v = (qL^2)/2 where q is the constant proportionality of v and l
    m = (M^2)/(2 landa) where landa is the linear mass density

    I'm not sure if my current workings are correct. And how to get rid of these constants?

    Pls help. Thanks!!
     
  2. jcsd
  3. Feb 20, 2009 #2

    LowlyPion

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    I'd start by identifying the mass element

    Let λ = M/L, so that means that a mass element at any point can given by λ*dl.

    As noted then the Velocity can be given as V*l/L

    So all they want you to do is construct an expression for the kinetic energy of the mass element and then integrate over the length of l from 0 to L.
     
  4. Feb 20, 2009 #3
    hmm. how come Velocity can be given as V*l/L?
     
  5. Feb 20, 2009 #4

    Delphi51

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    Seems to me you have to work with a bit of length dl on the spring.
    It's speed will depend on its distance l from the end of the spring, say v = kl but the speed at the far end is V = kL, so k = V/L and v = Vl/L, which makes sense at l=0 and l=L so it should in between, too.
    It's mass will be a fraction of the whole: dm = dl/L*m.
    I went on to find dE, the kinetic energy for dm moving at v, then integrated over the length of the spring to get E. It turned out to be a nice fraction of 1/2mV^2.
     
  6. Feb 20, 2009 #5
    means if i work out, velocity = 0.5Lv^2 and mass is (M^2)/(2λ)
     
  7. Feb 20, 2009 #6
    hmm. but i tot the qns sae it is not 1/2mv^2 ??

    anyway i tried to find dE which is 1/2(l*dm/dl)(Vl/L)^2 ?
     
  8. Feb 20, 2009 #7

    Delphi51

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    Are you using dm = dl/L*m and v = Vl/L?
    If so, your expression for 1/2*dm*v^2 will not be dE = 1/2(l*dm/dl)(Vl/L)^2 .
    As a first check, you have one differential (dE) on the left, so you should have one on the right - but you have none (dm/dl is a derivative, not a differential).
    Anyway, just substitute the expressions on the first line into dE = 1/2*dm*v^2.
     
  9. Feb 20, 2009 #8
    so now i have integrate dE wrt dl = integrate 1/2(m/l)(dl)(vl/L)^2 right?
    so when i integrate it, the RHS i no need to write dl again right?
    can i just take the dl out from the eqn?

    erm u know what i mean? :)
     
  10. Feb 20, 2009 #9

    Delphi51

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    Just use the dl in the equation for dE - don't add another one.
    The idea is to write the constants (which don't vary with l) before the integral sign, and the variables after it, with the dl last of all.
     
  11. Feb 20, 2009 #10
    okay. Thanks a lot!! :))

    are u free now? cos i have another qns which i'm stucked at.. =x
     
  12. Feb 20, 2009 #11
    its the spring gun qns i posted here too..
     
  13. Feb 22, 2009 #12
    thanks for the help. i finally got it. its 1/6 mv^2
     
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