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I am trying to reproduce a calculation by Carre et al. (1995) in which they calculate the shape of a droplet on an incline.

My issue is in the derivation of the potential energy (essentially the center of mass) of the droplet. The droplet is defined in the coordinate system as shown below, with the droplet base being circular. The function for $z$ is known, but not important for now.

coordinate system for droplet on an incline

The potential energy is defined with respect to the origin of the coordinate system.

The function given in the paper (in Eq. 5) is $$E_p=\int_{-\pi}^{\pi}\int_0^{r_0} \rho g z r \left(\frac{z}{2}\cos\alpha-r\sin\alpha\cos\phi\right) dr\,d\phi$$

I cannot figure out how they get to this. I understand that there should be a $\cos\alpha$ in there due to the inclination. Moreover, I would say the potential energy should be $$E_p=\int_{-\pi}^{\pi}\int_0^{r_0}\int \rho g z dz\,rdr\,d\phi $$ and since I am going to apply the equation in an euler-lagrange like scheme I only need to integrate the $z$ component. However, I am having some doubts here, because the height of the center of mass above the origin would be $h=z \cos\alpha$.

Apart from those 'vague' statements above I can't get to the answer. Could someone please explain (i) whether my equation for the potential energy is correct (or whether I need something in terms of $h$?) and (ii) how I could go about doing the integration.

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1 Answer 1

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The potential energy is calculated from $$E_p=\int_{-\pi}^{\pi}\int_0^{r_0}\int_0^{z_{max}} \rho g \,z_0(z,r,\phi) \;\; dz\,rdr\,d\phi $$ where $z_0$ is the vertical position along the direction of gravity.

We can calculate $z_0$ by doing a change of reference frame. We can use a laboratory frame $(x_0,y_0,z_0)$ where $z_0$ is along gravity direction. Then, the frame of the incline $(x_1,y_1,z_1)$ relates the rotation $$\left\{ \begin{array}{} x_0=x_1\\ y_0=y_1\cos\alpha+z_1\sin\alpha\\ z_0=-y_1\sin\alpha+z_1\cos\alpha \end{array} \right.$$ The cylindrical coordinates in the incline frame are then given by $y_1=r\cos\phi$ and $z_1=z$. You put these together and find $z_0=z\cos\alpha-r\sin\alpha\cos\phi$.

Now, you can integrate $z_0$ from $0$ to $z_{max}$, inject it into the potential energy equation, and you easily get the same answer as the article.

For your second question, you need the value of $z_{max}$ as a function of $r$ and $\phi$, which may be in the article. Then you can proceed trying to calculate the integral.

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Awesome, thanks for this clear and concise explanation! –  Michiel May 25 '13 at 6:49

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