Work done by gravity - what is wrong?

[Zoli]

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:
W_{AB} = \int_{\mathbf{r}_A}^{\mathbf{r}_B} \mathbf{G}\cdot \mathrm{d}\,\mathbf{r}.
In the x-y coordinate system we can write
\mathbf{G} = G(0,-1), \quad \mathrm{d}\,\mathbf{r} = R(\sin\varphi,-\cos\varphi)\mathrm{d}\,\varphi,

therefore the value of the integral becomes W_{AB} = GR(\sin\varphi_B - \sin\varphi_A), 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 W_{AB} = \mathbf{G} and \mathrm{d}\,\mathbf{r} is an acute angle, so the work should be positive.
What have I done wrong?

Thank you, Zoli

 

[PhanthomJay]

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.

 

[athosanian]

I see it. It should be noticed that dr=-R(sinφ,−cosφ)dφ,becasue from A to B dφ is negative.

 

[Zoli]

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.

Yes, I surmised that I changed the path of integration, since \mathrm{d}\,\varphi is negative. I forgot the idea of the potential function. Thank you.