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Work done by gravity - what is wrong?

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  1. Sep 27, 2014 #1
    I would like the determine the work done by gravity on a mass attached to a rod (see the attached image). The rod is assumed to be weightless and rigid.
    I start from the definition of work:
    [tex] W_{AB} = \int_{\mathbf{r}_A}^{\mathbf{r}_B} \mathbf{G}\cdot \mathrm{d}\,\mathbf{r}.[/tex]
    In the x-y coordinate system we can write
    [tex] \mathbf{G} = G(0,-1), \quad \mathrm{d}\,\mathbf{r} = R(\sin\varphi,-\cos\varphi)\mathrm{d}\,\varphi, [/tex] k4xmpv.png
    therefore the value of the integral becomes [itex] W_{AB} = GR(\sin\varphi_B - \sin\varphi_A), [/itex] which is negative since we are in the first quadrant. However, if we look at the attached draft, we may notice that the angle between [itex] W_{AB} = \mathbf{G} [/itex] and [itex] \mathrm{d}\,\mathbf{r} [/itex] is an acute angle, so the work should be positive.
    What have I done wrong?

    Thank you, Zoli
     
  2. jcsd
  3. Sep 27, 2014 #2

    PhanthomJay

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    You tried it the hard way and in so doing, looks like you integrated backwards. The easier way is to note that the work done by gravity is independent of the path taken, and equal to the negative of the potential energy change. You can calculate the potential energy from basic trig to determine the vertical height change. Since the gravity force and it's vertical displacement are in same direction, work must be positive.
     
  4. Sep 27, 2014 #3
    I see it. It should be noticed that dr=-R(sinφ,−cosφ)dφ,becasue from A to B dφ is negative.
     
  5. Sep 28, 2014 #4
    Yes, I surmised that I changed the path of integration, since [itex] \mathrm{d}\,\varphi [/itex] is negative. I forgot the idea of the potential function. Thank you.
     
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