## Why doesn't Bernoulli's Equation apply in this problem?

1. The problem statement, all variables and given/known data

A cylindrical bucket of liquid (density ρ) is rotated about its symmetry axis, which is vertical. If the angular velocity is ω, show that the pressure at a distance r from the rotation axis is

$$P = P_0 + \frac{1}{2} \rho \omega^2 r^2$$

where P0 is the pressure at r = 0.

2. Relevant equations

P = F/A

3. The attempt at a solution

I was able to get the correct answer by considering the net force on a mass element dm since it is undergoing centripetal acceleration.

$$P = P_0 - \frac{1}{2} \rho \omega^2 r^2$$
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 Recognitions: Homework Help Bernoulli's equation works along a streamline, which means the path of fluid flow. In this case, you took the radial pressure difference but used the tangential velocity v instead of the radial velocity of 0. If you use 0 and put back the gravitational potential term, because centripetal acceleration is equivalent to gravity, you'll get the right answer.
 I'm sorry. I've been thinking about what you said but I do not understand what to do. I tried to rewrite ρgy as ρ(v2/r) but this isn't working either. Also, to make sure I understand what streamline means, would this be a proper application of Bernoulli's equation? If you consider a small portion of the water dm traveling along its path of flow (in a circle), then v1 = ωr = v2 and y1 = y2. Thus P1 = P2, so it's pressure remains constant.

Recognitions:
Homework Help

## Why doesn't Bernoulli's Equation apply in this problem?

 Quote by azure kitsune I'm sorry. I've been thinking about what you said but I do not understand what to do. I tried to rewrite ρgy as ρ(v2/r) but this isn't working either.
ρgy is meant to represent the difference in gravitational potential. In the rotating reference frame of the liquid, the equivalent "g" is w^2r, and you'll have to find the potential difference between the center of the liquid and the sides of the cylinder using integration. If you haven't learnt integration yet, the potential difference you'd get is (1/2)ρw^2*r^2.

 Also, to make sure I understand what streamline means, would this be a proper application of Bernoulli's equation? If you consider a small portion of the water dm traveling along its path of flow (in a circle), then v1 = ωr = v2 and y1 = y2. Thus P1 = P2, so it's pressure remains constant.
Yup, that would be a proper application of the equation. It doesn't have any practical use, but you've gotten the concept of the streamline.