Calculating Added Mass: What is the Correct Approach?

[member 428835]

Hi PF!
I'm calculating the added mass of a sphere accelerating in a fluid, which I found here: http://web.mit.edu/2.016/www/handouts/Added_Mass_Derivation_050916.pdf
My thought process was slightly different from theirs, but I am not getting the same answer. My thoughts are to take the surface normal element of a sphere, in this case ##\vec{dS} = R^2 \sin \theta \, d\theta \, d\phi \hat{r}## and dot this with the direction of the sphere, say in the direction of the zenith angle, ##\hat{y}##. We know ##\hat{r} = \sin \theta \sin \phi \hat{x} + \sin \theta \sin \phi \hat{y} + r \cos \theta \hat{z}##, which implies ##\hat{r} \cdot \hat{y} = \sin \phi \sin \theta##, which means my surface element in the direction of motion would be ##R^2 \sin^2 \theta \sin \phi d\theta d\phi##. now if we integrate ##\phi## from ##[0,2\pi]## the ##\sin \phi## term takes this to zero. Even if it did give me ##2 \pi## like in the link, I still have an extra sine and lack a cosine.

Any ideas on how to amend my approach, and also why it's not working?

Thanks so much!

 

[haruspex]

I don't have clear in my head what your diagram looks like relating the direction of motion to the coordinates. An obvious switch is to take it to be in the z direction instead. Is suspect you have effectively only considered part of the contributions so the element in direction of motion (e.g., shoild be an x term too).
Wrt integration range, you only want the leading hemisphere, so the positive range of the trig functions.

 

[member 428835]

An obvious switch is to take it to be in the z direction instead. Is suspect you have effectively only considered part of the contributions so the element in direction of motion (e.g., shoild be an x term too).

Oops, right, so what I should have had was ##\hat{r} \cdot \hat{z} = \cos \theta##. This implies the surface element toward the direction of motion is ##R^2 \sin\theta \cos \theta \, d\theta \,d \phi##.

Wrt integration range, you only want the leading hemisphere, so the positive range of the trig functions.

So the force from added mass is then $$\int_0^{2 \pi} \int_0^{\pi/2} p R^2 \sin\theta \cos \theta \, d\theta \,d \phi$$ But integrate the leading hemisphere implies integrating ##\phi \in [0,\pi/2]## yet they integrate from ##[0,\pi]##. Can you explain the difference here? It also looks like they integrate the cylinder from ##[0,2\pi]## rather than ##[-\pi/2,\pi/2]##.

Thanks a ton for your reply!