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Math Help - A physics proof

  1. #1
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    A physics proof

    I have a differential equations problem that I thought someone here may be able to help with. So here goes:
    In the motion of an object through a certain medium (air at certain pressures is an example), the medium furnishes a resisting force proportional to the square of the velocity of the moving object. Suppose a body falls due to the action of gravity, through the medium. Let t represent time, and v represent velocity, positive downward. Let g be the usual constant acceleration of gravity, and let w be the weight of the body. Use newton's law, for equals mass times acceleration, to conclude that the differential equation of motion is
    (W/g)dv/dt=w-kv^2
    where kv^2 is the magnitude of the resisting force furnished by the medium.
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  2. #2
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    Quote Originally Posted by Sebastian View Post
    I have a differential equations problem that I thought someone here may be able to help with. So here goes:
    In the motion of an object through a certain medium (air at certain pressures is an example), the medium furnishes a resisting force proportional to the square of the velocity of the moving object. Suppose a body falls due to the action of gravity, through the medium. Let t represent time, and v represent velocity, positive downward. Let g be the usual constant acceleration of gravity, and let w be the weight of the body. Use newton's law, for equals mass times acceleration, to conclude that the differential equation of motion is
    (W/g)dv/dt=w-kv^2
    where kv^2 is the magnitude of the resisting force furnished by the medium.
    I would not called physics a "proof" but anyways.

    By the second law: F=ma

    And the equilbrium law: SUM (Forces) = 0 (vec).

    Okay, thus we have that when a object is falling down there are two forces: the force of gravity and the resistance force. By the conditions of the problem the resistance force is proportional to the square of speed, in simple terms, kv^2. But! There is also a downard force, w, the weight. Since the sum of the forces is zero, we have,
    w-kv^2=0 (since it is going in opposite direction, thus we have a negative).
    By Newton's second law the overall force is: a(w-kv^2) where "a" is acceleration. The acceleration in this case is the constant at g. Thus,
    F=g(w-kv^2).
    But there is another way to find the force.
    Again, F=a*m=(dv/dt)*w because m, mass, and weight are presumably the same here. And dv/dt is the acceleration, as you know.
    Thus,
    w(dv/dt)=g(w-kv^2)
    Divide,
    (w/g)(dv/dt)=w-kv^2
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  3. #3
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    I understand the first part, but this second and third parts are getting me:
    b) Solve the differental equation of part A, with the initial conditions that v=v0 when t=0. Introduce the constant a^2=w/k to simpilfy the formulas.

    c) There are mediums that resist motion through them with a force porportional to the first power of the velocity. For such a medium, state and solve problems analogous to parts A) and B), except that for convenience a constant b=w/k may be introduced to replace the a^2. Show that b has the dimensions of a velocity.
    I realize how good this forum is now, and I'll hang around here to learn some math techniques. Thanks for the help if you can provide it.
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  4. #4
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    Quote Originally Posted by Sebastian View Post
    I understand the first part, but this second and third parts are getting me:
    b) Solve the differental equation of part A, with the initial conditions that v=v0 when t=0. Introduce the constant a^2=w/k to simpilfy the formulas.
    The ODE (W/g)dv/dt=w-kv^2 is of variables seperable type so:

    int a^2/(w-kv^2) dv = int dt

    the left hand side may be rewritten:

    k int a^2 / (a^2 - v^2) dv = t + C

    and then partial fractions may be used to integrate the left hand side:

    k (a/2) int [1/(a+v) - 1/(a-v)] dv = t + C

    so:

    k a/2 ln[(v-a)/(v+a)] = t + C

    (v-a)/(v+a) = exp[2 (t+C)/(ka)]

    so:

    v = {exp[2 (t+C)/(ka)] +a}/{1-exp[2 (t+C)/(ka)]}

    The right hand side probably simolifies some more when you put in the initial
    condition etc.

    RonL
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  5. #5
    Forum Admin topsquark's Avatar
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    In case you want it, I'll give you a bit more information about these equations and their solutions. But I'll need to stop home and pick up the appropiate text first. I'll try to post later tonight.

    -Dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    In case you want it, I'll give you a bit more information about these equations and their solutions. But I'll need to stop home and pick up the appropiate text first. I'll try to post later tonight.

    -Dan
    i was able to go to the teacher today and she showed me a simpler way to do it.
    w/g dv/dt=w-kt b=w/k
    w/g/k dv/dt=w-kv/k
    w/gk dv/dt = w/k - v
    bg dv/dt =b-v
    b dv/dt =(b-v)g
    b/(b-v) dv=g dt
    -b/v-bdv =g dt
    intr(-b/(v-b)dv)=intrgdt
    ln|v=b|=-gt/b+c/b
    |v-b|=e^((-gt/b)+c)
    v-b=ce^(-gt/b)
    v=ce^(-gt/b) +b
    v0=ce^0+b
    v0=c+b
    v0-b=c
    v=(v0-b)e^(-gt/b) + b

    Thanks for all your help on this problem guys
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