Centrifugal force on water in a glass

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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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and dy/dx = ω2r / g
so: y= ω2r x /g ... for r=x I get y= ω2x2 /2g
 
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srecko97 said:
4,4cm
Using g=10m/s2, yes, but 4,5cm would be a bit closer.
Well done.
 
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