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Homework Help: Work done over two paths

  1. Jun 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle in the [itex]xy[/itex] plane is attracted toward the origin by a force of magnitude [itex]F=k/y[/itex]
    (a) Calculate the work done when the particle moves from point [itex](0,a)[/itex] to [itex](2a,0)[/itex] along a path, which follows two sides of a rectangle consisting of a segment parallel to the [itex]x[/itex] axis from [itex](0,a)[/itex] to [itex](2a,a)[/itex], and a vertical segment from the latter point to the [itex]x[/itex] axis.
    (b) Calculate the work done by the same force when the particle moves along an ellipse of a semi-axes [itex]2a,a[/itex] Hint: Set [itex]x=2a\sin\theta, y=a\cos\theta[/itex]


    2. Relevant equations

    [tex]W=\int\mathbf{F}\cdot d\mathbf{r}=\int F\cos\theta dr[/tex]


    3. The attempt at a solution

    For the part that is parallel to the [itex]x[/itex] axis, I get

    [tex]W_{\alpha}=\int_0^{2a}\frac{k}{y}\cdot\frac{y}{\sqrt{x^2+a^2}}dx=\left.k\sinh^{-1}\left[\frac{x}{a}\right]\right|_{x=0}^{x=2a}=k\sinh^{-1}\left[2\right][/tex]

    Then the second path gives a similar answer:

    [tex]W_{\beta}=\int_0^{2a}\frac{k}{y}\cdot\frac{y}{\sqrt{4a^2+y^2}}dy=\left.k\sinh^{-1}\left[\frac{y}{2a}\right]\right|_{y=a}^{y=0}=-k\sinh^{-1}\left[\frac{1}{2}\right][/tex]

    Then the total work is the sum of the two paths. My problem comes in setting up the next part, I thought to use the hint in this way:

    [tex]W_\gamma=\int\frac{k d\theta}{r}=k\int\frac{d\theta}{\sqrt{4a^2\sin^2\theta+a^2\cos^2\theta}}=\frac{k}{a}\int\frac{d\theta}{\cos\theta\sqrt{1+4\tan^2\theta}}[/tex]

    but this doesn't look correct to me. Any suggestions??
     
  2. jcsd
  3. Jun 1, 2010 #2

    vela

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    You're using the wrong component of the force.
    Probably just an oversight, but the limits on the integral are wrong. Your answer is right though.
    How did you come up with the integrand k/r?
     
  4. Jun 2, 2010 #3
    I was only given the magnitude, so I figured I would have to use [itex]F\cos\theta dr[/itex] instead of [itex]\mathbf{F}\cdot d\mathbf{r}[/itex].

    :biggrin: This is what happens when you copy the previous line & change only the answer.

    I had used k/r in the previous two parts, but I am thinking I need to use

    [tex]F\cos\theta dr=F\cos\theta \frac{dr}{d\theta}d\theta[/tex]

    which would give me

    [tex]\frac{dr}{d\theta}=6a^2\cos\theta\sin\theta[/tex]

    [tex]W=\int_0^{\pi/2}\frac{6ka^2\sin\theta\cos\theta}{\sqrt{4a^2\sin^2\theta+a^2\cos^2\theta}}d\theta=\left.k\ln\left[4a^2\sin^2\theta+a^2\cos^2\theta\right]\right|_{\theta=0}^{\theta=\pi/2}=k\ln\left[4\right][/tex]

    which doesn't look much at all like the previous part (assuming my math is correct).
     
  5. Jun 2, 2010 #4

    vela

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    That's correct. You only want one component of F. My point is your expression is for the wrong one. You're moving along the x-axis, so you want the x-component of F.
    You got k/r because you used the wrong component (in the first integral) and because you were integrating along a horizontal or vertical line. For the elliptical path, it's not going to be the same.
    You need to be careful because the θ in [itex]\textbf{F}\cdot d\textbf{r} = Fr\cos\theta[/itex] isn't the same variable as the θ in [itex]x = 2a\cos\theta[/itex] and [itex]y=a\sin\theta[/itex]. Use a different variable for the angle between F and dr, and try drawing a picture. I also suggest you use [itex]\textbf{F}\cdot d\textbf{r} = F_x dx + F_y dy[/itex] instead.
     
    Last edited: Jun 2, 2010
  6. Jun 2, 2010 #5
    http://en.wikipedia.org/wiki/Work_(physics)#Force_and_displacement" It says here (as well as my text books)

    [tex]W=\mathbf{F}\cdot\mathbf{d}=Fd\cos\theta[/tex]

    I have no [itex]x[/itex] component of [itex]F[/itex]. I have only the magnitude [itex]\left|\mathbf{F}\right|=k/y[/itex]

    I don't think I am using that [itex]\theta[/itex] is the same as the angle in [itex]x = 2a\cos\theta[/itex] and [itex]y=a\sin\theta[/itex]. I am merely using the first [itex]\theta[/itex] (which I can call [itex]\alpha[/itex] for reducing confusion) to get [itex]\cos\alpha=y/\sqrt{x^2+y^2}[/itex].

    And again, I don't have [itex]F_x[/itex] and [itex]F_y[/itex], I only have [itex]\left|\mathbf{F}\right|=k/y[/itex].
     
    Last edited by a moderator: Apr 25, 2017
  7. Jun 2, 2010 #6

    vela

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    The component of F which does work is the one that's parallel to the displacement. The magnitude of that component is F cos θ, where θ is the angle between F and d, so the work done is W = (F cos θ)d. You need to learn how to interpret combinations that appear in equations so you can get a better understanding of what the equations really mean.

    Your mistake is in your expression for cos θ. It's easy to see that it's wrong. For instance, when you're at the point r=(0, a), the force points in the -y direction, and the displacement is in the +x direction. The angle between the two is therefore 90 degrees. Your expression for cos θ, however, gives

    [tex]\left \frac{y}{\sqrt{x^2+a^2}}\right|_{(0, a)} = \frac{a}{\sqrt{0^2+a^2}} = 1 \ne \cos 90^\circ = 0[/tex]

    Also, despite what I said earlier, the work you calculated for the second leg is wrong. It should be positive.
    Well, you certainly used the same variable to represent two different angles when you wrote

    [tex]F\cos\theta dr=F\cos\theta \frac{dr}{d\theta}d\theta[/tex]

    But, yes, I see now how your integral is the result of a different set of errors.
    But you can calculate them in terms of the other quantities in the problem.
     
    Last edited by a moderator: Apr 25, 2017
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