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Particle motion in a straight line - find opposing force as a function of velocity

  1. Jul 21, 2011 #1
    Hello everyone,

    This is my first time in Physics forums so forgive me if I stray from the convention for asking questions. Me and a friend are completely stumped on our first assignment for the year and we are not sure where else to turn!

    1. The problem statement, all variables and given/known data

    A particle of mass m can move in a straight line. The only force acting on the particle opposes its motion, and depends only on its speed. The time T taken for the particle to come to rest after it is set into motion with speed u is given by the formula [itex]T=klog(1+au)[/itex] for all u, where k and a are positive constants. Find the force acting on the particle during the interval [itex]0\leq t\leq T[/itex] as a function of its speed v. Find also the distance the particle travels in time T.

    Hint: Recall that the first fundamental theorem of calculus states that [itex]\frac{\mathrm{d} }{\mathrm{d} x} \int_{a}^{x}f(w)dw=f(x)[/itex].

    2. Relevant equations

    [itex]T=klog(1+au)[/itex], [itex]\frac{\mathrm{d} }{\mathrm{d} x}, \int_{a}^{x}f(w)dw=f(x)[/itex]

    [itex]F(v)=m\frac{\mathrm{d}v }{\mathrm{d} t}[/itex]

    3. The attempt at a solution

    As far as I can see the best thing to do is take Newton's second law stated above, and separate the variables to give:

    [itex]m\int \frac{dv}{F(v)}=\int_{0}^{T}dt[/itex]

    Which would give simply T on the RHS, so substituting for T would give:

    [itex]m\int \frac{dv}{F(v)}=klog(1+au)[/itex]

    Now I am not sure how to proceed. I am tempted to try and take F(v) out of the integral to the other side of the equation, but I'm not sure whether this is allowed:

    [itex]m\int \frac{dv}{klog(1+au)}=F(v)[/itex]

    If so, then my answer would simply be [itex] \frac{mv}{klog(1+au)}=F(v)[/itex]. Any ideas as if this is correct?

    For the second part I think I need to modify our starting point with Newton's second law to say:

    [itex]mv\frac{\mathrm{d} v}{\mathrm{d} x}=F(v)[/itex]

    But after that I am a bit lost. If I have the equation for F(v) perhaps I can separate variables again and then express v as dx/dt and integrate again. Unfortunately i cannot see far enough ahead to see how this would work.

    Thanks everyone in advance for taking a look at this!
     
  2. jcsd
  3. Jul 21, 2011 #2
    Re: Particle motion in a straight line - find opposing force as a function of velocit

    You are almost there. You cannot have a definite integral of the RHS, but an indefinite integral on the LHS. So the correct equation is:
    [tex]m\int^{0}_{u} \frac{dv}{F(v)}=\int_{0}^{T}dt = klog(1+au)[/tex]
    Suppose you know [itex]F(v)[/itex], how would you do the integral? You would have to taking anti-derivative of [itex]\frac{m}{F(v)}[/itex]. This should give you a hint of how it is related to the RHS. If still unsure, take derivative with respect to [itex]u[/itex] on both sides.

    As for the second part, since [itex]T(u)[/itex] is the time it takes to stop the particle with velocity u, then at any intermediate time t when the particle is traveling with velocity [itex]v(t)[/itex], it would take [itex]T(v)[/itex] more time to stop it. Therefore, [itex]t + T(v)= T[/itex]. This allow you to invert and find [itex]v(t)[/itex], from which you can integrate to find [itex]x(t)[/itex].
     
    Last edited: Jul 21, 2011
  4. Jul 21, 2011 #3
    Re: Particle motion in a straight line - find opposing force as a function of velocit

    Ah hah!

    So if we were to differentiate both sides with respect to u, the RHS would become ka/(1+au). Then using the theorem given in the hint the LHS would become -1/F(u). So that gives us:

    [itex]F(u)=-\frac{1+au}{ka}[/itex].

    However the question asked for the force in terms of v, so how might I go about that from here?

    Thanks very much for your help so far!
     
  5. Jul 21, 2011 #4
    Re: Particle motion in a straight line - find opposing force as a function of velocit

    Can T=K log(1+au) be rearranged to give us u?

    u=1/a * ((10^(T/K)) -1))

    If that is substituted back into the equation for force...

    F(T)=-(10^T/K) / Ka

    Does this get us anywhere?
     
  6. Jul 22, 2011 #5
    Re: Particle motion in a straight line - find opposing force as a function of velocit

    Let me give you a simpler example here. Suppose [itex]f(x)[/itex] satisfies [itex]\int_0^y f(x) dx = \frac{y^2}{2}[/itex]. I differentiate both side with respect to [itex]y[/itex] and get [itex]f(y) = y[/itex]. What is [itex]f(x)[/itex]?
     
    Last edited: Jul 22, 2011
  7. Jul 23, 2011 #6
    Re: Particle motion in a straight line - find opposing force as a function of velocit

    Hmm, so in your example here, if f(y)=y then it would also be true that f(x)=x. Is then as simple as saying that if:

    [itex]F(u)=-\frac{1+au}{ka}[/itex] then [itex]F(v)=-\frac{1+av}{ka}[/itex]?

    If we substitute this expression for F(v) into the original integral it seems to almost work, apart from the fact that we still have a factor of m:

    [itex]m\int_{u}^{0}\frac{dv}{F(v)}=kmlog(1+au)[/itex]

    Thanks again.
     
  8. Jul 25, 2011 #7
    Re: Particle motion in a straight line - find opposing force as a function of velocit

    If we were to say that:

    [itex]F(v)=\frac{m(1+av)}{ka}[/itex]

    Then the integration seems to work and is equal to [itex]klog(1+au)[/itex]. For now I'll assume this is correct.

    For the second part I have taken the equation t+T(v)=T(u) to give:

    [itex]t+klog(1+av)=klog(1+au)[/itex]

    Rearranging this for v gives:

    [itex]v=\frac{1+au}{a10^{t/k}}-\frac{1}{a}[/itex]

    Integrating this between 0 and T gives:

    [itex]x=-\frac{k(1+au)10^{-t/k}}{a}-\frac{t}{a}[/itex]

    Which is showing a negative value for the position x. I think this is probably incorrect, but I am not sure where I have gone wrong...
     
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