Pressure on a cylindric wall

  • Thread starter nazarian
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  • #1
nazarian
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The problem is the following: we have 2 half's of a cylinder and they are put together using 6 screws, forming a full cylinder. Inside there is water and we are asked to calculate the force on the screws when the cylinder rotates with constant angular velocity Ω. Using the equations of hydrostatics I calculated the pressure as a function of z and r, but here comes the part I cannot yet solve; If I want to calculate te resultant force over some half do I have to integrate [itex]\oint_S p [/itex] or [itex]\oint_S \hat{n} \dot \hat{i}[/itex]? (being S the surface of the semi cylinder). I know the pressure is a scalar but i still have doubts about it. Do I project the force over [itex]\hat{i}[/itex] ? (taking x as the symmetric axis)

Thanks, excuse me if my english is not correct.
 

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  • #2
haruspex
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I'm going to assume that the screw axes are perpendicular to the plane of join of the two half cylinders. Further, that each half cylinder is sufficiently rigid that you do not have to worry about forces that tend to distort (flatten) it. So all you should care about are the forces in the direction of the screw axes.
Does that help you?
 
  • #3
nazarian
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Correct! but do I have to decompose the pressure which I believe is orthogonal to the surface, i.e. is in the [itex]\hat{\rho}[/itex] direction, to the cartesian directions and then integrate? I mean do I have to think the pressure as a vector field (I was told it was a scalar) ?

Let me use a simpler question;
Asume the pressure is constant, H is the height, R is the radius, [itex]\hat{i}[/itex] is the direction perpendicular to de plane of the screws. The resultant force on one half is then [itex]\vec{F}=\pi PHR \hat{i}[/itex] or [itex]2PHR \hat{i}[/itex]?. The 2 comes from integrating [itex]sin\theta[/itex] between 0 and [itex]\pi[/itex], while the first expression comes from integrating without [itex]sin\theta[/itex]
 
  • #4
haruspex
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The pressure is scalar, but the force at each element of area is a vector. Just take the component of that force in the screw axis direction. Components in other directions will cancel by symmetry. So in the simplified question, it's 2, not pi.
Btw, are you supposed to worry about gravity too?
 

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