Centrifugal force on water in a glass

AI Thread Summary
The discussion revolves around calculating the height difference of water in a rotating cylindrical glass. The water's surface shape is determined to be parabolic due to centrifugal force, with the pressure at the center being lower than at the rim. Participants engage in deriving the equation for the water surface, using parameters like angular velocity and gravitational acceleration. A free body diagram is suggested to visualize the forces acting on the water. Ultimately, the calculated height difference is approximately 4.5 cm, confirming the expected outcome of the problem.
srecko97
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Homework Statement


There is a cyllindrical glass (r=0.05 m) full of water. It is rotating around its vertical geometric axis with 3 turns per second. How many centimetres higher is water at the rim than in the centre of the glass . Water rotates together with the glass.

Homework Equations


p=ρah
ω=2πγ
a=ω2r

The Attempt at a Solution


All I know is that the shape of water is parabolic. In the centre there is no additional pressure from rotation, the pressure on the glass is p=ρah, a=ω2r
total pressure on any point is px + py (vectorial sumation)
 
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srecko97 said:
All I know is that the shape of water is parabolic.

That's right. If we let the vertex be at (0,0) then you just need to find the Y value when X=0.05m.
Start by drawing a free body diagram for a piece of water on the surface.
 
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probably is y=Cx2 ... but I do not know what is C, maybe a/g ... a=17.8 m/s^2, g=10 m/s^2 i do not know, honestly just guessing, as i do not know how to solve it
 
billy_joule said:
Start by drawing a free body diagram for a piece of water on the surface.

Post your FBD and we'll see where you've gone wrong.
 
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I do not know what free body diagram is. Sorry. Please explain to me and I will try
 
srecko97 said:
I do not know what free body diagram is. Sorry. Please explain to me and I will try
Consider a small parcel of water at the surface, at some radius x. If the surface shape is described by y=f(x), what is the slope there? Draw a diagram of the parcel of water and the forces acting on it. Treat the rest of the water as a frictionless solid.
 
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20170129_121122.jpg
 
srecko97 said:
20170129_121122.jpg
Ok. You left out a normal force, and you need to answer this question
haruspex said:
If the surface shape is described by y=f(x), what is the slope there?
I.e., in terms of x and y.
 
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tanφ =ω2r / g
 
  • #10
and dy/dx = ω2r / g
so: y= ω2r x /g ... for r=x I get y= ω2x2 /2g
 
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  • #11
and I have y if i put r=x!
 
  • #12
srecko97 said:
and I have y if i put r=x!
Right. So what answer do you get?
 
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  • #13
4,4cm
 
  • #14
srecko97 said:
4,4cm
Using g=10m/s2, yes, but 4,5cm would be a bit closer.
Well done.
 
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