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Math Help - Newtonian Mechanics - the movement of a particles

  1. #1
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    Newtonian Mechanics - the movement of a particles

    A particle of mass m which moves along a horizontal straight line with a velocity of v encounters a resistance of av + b(v^3), where a and b are constants. If there is no other force beside the resistance acting on the particle and its initial velocity is U , show that the particle will stop after it has moved a distance of m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2). Also show that the velocity is 1/2U after time (m/2a)ln(4a+b(U^2))/(a+b(U^2)))

    to show the distance when the particle stops :

    i use :

    F = ma

    => m (dv/dt) = - (av+b(v^3))

    i let v = y and t = x to make it less confusion ....(1)

    thus m(dy/dx) = -(ay+b(y^3))

    dy/dx = -(ay/m)-(b(y^3)/m)

    dy/dx + (a/m)y = -(b/m)(y^2) ... (2)

    (y^-3)(dy/dx) + (a/m)(y^-2) = (-b/m) ===> this is Bernoulli equation

    let v = (y^-2)

    thus, dv/dx = -2(y^-3) (dy/dx)

    => -1/2(dv/dx) + (a/m)v= -b/m

    searching for integrating factor, miu(x) = exp(integrating(a/m)dx)

    thus, miu (x) = exp(ax/m)

    miu(x) x (2) : d/dx (exp(ax/m).v) = exp(ax/m).(-b/m)

    exp(ax/m).v = (-b/m) integrating (exp(ax/m) dx)

    v = (-b/a) + c(exp-(ax/m))

    but v = (y^-2) = (v^-2)

    thus,

    v^-2 = (-b/a) + c(exp-(ax/m))

    v^2 = (1/c(exp-(ax/m))) - (a/b)

    v = sqrt ((1/c(exp-(ax/m))) - (a/b))

    applying initial condition where v_0 = u, x=0, t=0

    c = 1/u^2 + b/a

    thus v = dx/dt = sqrt ((1/(1/u^2 + b/a)) - a/b)

    when particle stop dx/dt = 0

    then

    0 = sqrt ((1/(1/u^2 + b/a)) - a/b)
    .
    .
    .
    x = m/a ln |(1/ab)+1|

    why i can't get the answer given which is m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2)???????

    where i'm going wrong??????

    please help me!!!!!!!!!!!!!!!!!

    p/s : i don't know to use SYNTAX. sorry =p
    Last edited by bobey; March 3rd 2010 at 09:23 PM.
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  2. #2
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    Quote Originally Posted by bobey View Post
    A particle of mass m which moves along a horizontal straight line with a velocity of v encounters a resistance of av + b(v^3), where a and b are constants. If there is no other force beside the resistance acting on the particle and its initial velocity is U , show that the particle will stop after it has moved a distance of m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2). Also show that the velocity is 1/2U after time (m/2a)ln(4a+b(U^2))/(a+b(U^2)))
    Hi Bobey,

    This problem is not so difficult as it seems. It takes some care in the calculation, so let's go over it with some large steps. I assume you will do the intermediate calculations yourself. The DE you obtained is correct, it is:

    m\frac{dv}{dt}=-av-bv^3

    It can be solved using the method of Bernoulli but it is easier to use the method of separation of variables. This gives:

    \frac{dv}{v\left(a+bv^2\right)}=-\frac{dt}{m}

    Partial fractions is the way to solve the first integral and this gives after the integration:

    \frac{1}{a}ln(v)-\frac{1}{2a}ln\left(a+bv^2\right)=-\frac{t}{m}+C_1

    With C1 the integration constant which can be determined from the initial condition at t=0 is v=U, plugging in gives now:

    \frac{\frac{v}{U}}{\sqrt{\frac{a+bv^2}{a+bU^2}}}=e  ^{-\frac{a}{m}t}

    From this we can obtain the answer from the second question. Indeed, setting v=U/2 and derive t from this gives:

    t=\frac{m}{2a}ln\left(\frac{4a+bU^2}{a+bU^2}\right  )

    To answer the first question, we need to rewrite the solution of the DE as follows:

    v=\frac{dx}{dt}=\frac{\sqrt{a}}{\sqrt{\left(\frac{  a+bU^2}{U^2}\right)e^{\frac{2a}{m}t}-b}}

    This can be integrated and the result is:

    x+C_2=\frac{m}{\sqrt{ab}}atan\left(\sqrt{\frac{a+b  U^2}{bU^2}e^{\frac{2a}{m}t}-1}\right)

    Using the boundary condition at t=0 is x=0 and plugging in we get:

    x=\frac{m}{\sqrt{ab}}\left[atan\left(\sqrt{\frac{a+bU^2}{bU^2}e^{\frac{2a}{m}  t}-1}\right)- atan\left(\frac{\sqrt{a}}{\sqrt{b}U}\right)\right]

    The particle will come to rest for t going to infinity as can be seen from the resulting equation from the DE. Plugging this in gives:

    x_{\infty}=\frac{m}{\sqrt{ab}}\left[\frac{\pi}{2}- atan\left(\frac{\sqrt{a}}{\sqrt{b}U}\right)\right]

    which can be rewritten using pi/2=atan(T)+atan(1/T) as:

    x_{\infty}=\frac{m}{\sqrt{ab}}atan\left(\frac{\sqr  t{b}U}{\sqrt{a}}\right)

    The solution you were looking for. It is an exercise requiring careful calculations. Take your time for this and make sure you go over every step before going to the next one. As a final remark the DE can indeed be solved by the method of Bernoulli, it gives exactly the result as with the proposed method. I checked it :-)

    Hope this helps and do come back if you need more info.

    Coomast
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