Torque on a point on a sphere in a fluid/finding pressure?

AI Thread Summary
In the discussion about torque on a rotating sphere in a fluid, it is clarified that the torque varies across the sphere's surface depending on latitude. The relationship between the axis of rotation and the direction of travel is confirmed to be the same, which influences the drag experienced by different points on the sphere. The drag equation provided includes fluid density, velocity, drag coefficient, and cross-sectional area. The conversation hints at the Magnus effect, suggesting that differences in shear forces create pressure differences around the sphere. Overall, understanding the torque distribution is essential for calculating the pressure exerted by opposing sides of the sphere on the fluid.
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If there is a rotating sphere (falling through a fluid) a) is the torque the same at every point on the sphere's surface, and b) how would I use said torque to work out the pressure exerted by opposite 'sides' of the sphere on the fluid?



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What equations do you have for drag?
 
drag = 1/2 ρf v^2 Cd A

where pf is the fluid density, v is the velocity of the sphere, Cd is the drag coefficient for the sphere (0.1 for smooth, 0.6 for rough) and A is the reference (cross-sectional) area
 
I forgot to ask what the relationship is between the axis of rotation and the vertical (the direction of travel). Are they the same or orthogonal? If orthogonal, think about the v term of the drag for different points on the surface.
I assume what this is leading to is an explanation of the Magnus effect. Intuitively, I think I see how the difference in shear forces leads to the required pressure difference, but I'm not an expert in this area.
 
They're the same.
 
6283186 said:
They're the same.
In that case, the torque at a point on the surface will clearly depend on its latitude, and you could write the equation down fairly easily. But I've no idea how this connects with pressure.
 
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