# Can I apply Bernoulli's equation to rigid body rotation?

1. Oct 4, 2015

### RAP1234

1. The problem statement, all variables and given/know

Say I have a can of water, and I am rotating it about its central axis at a constant angular rate. The water in the tank should make a 3D almost parabolic curve as it touches the the walls of the tank. Can I use Bernoulli's equation along y=0, r =0 ( starting from the minima of my parabola) to the radius (r=R) to solve a problem involving this system? Is P = (1/2) + rho*v^2 = const valid?

2. Relevant equations

P = (1/2) + rho*v^2 = const

3. The attempt at a solution

I would think P = (1/2) + rho*v^2 = const may be valid because

if the fluid is rotating with the tank as a rigid body, assuming we are looking at it after it has been spun up and is rotating with constant angular rate, then it is at steady state?

The density of the fluid I am assuming is constant.

I am assuming viscosity is negligible

Are the streamlines just circles and I'm going normal to them?

2. Oct 4, 2015

### Staff: Mentor

Yes, the streamlines are circles, and you're going normal to them.

Chet

3. Oct 4, 2015

### rcgldr

I had the impression that the inner fluid has a faster rate of rotation than the outer fluid. Also streamlines aren't really streamlines if they are curved. Curvature requires a pressure gradient perpendicular to the direction of flow, and Bernoulli doesn't deal with that. Viscosity is going to have an effect, since different parts of the fluid are moving at different speeds.

4. Oct 4, 2015

### Staff: Mentor

Once the system reaches steady state, the fluid will be rotating like a rigid body, and the angular velocity will be constant throughout the fluid. Of course the circumferential velocity will be equal to the angular velocity times the radial location.
What could possibly have made you think this? It simply is not correct.
Viscosity won't be a factor once the system reaches steady state and the fluid is rotating like a rigid body.

Chet

5. Oct 4, 2015

### rcgldr

I thought this was mentioned in one or more prior threads about wings and lift. Maybe it was in reference to how lift calculations are based on streamline velocities just outside the boundary layer of a wing and the fact that the streamlines had to be broken up into small componennts along the wing chord to deal with the effects of curvature. I think the issue was that the flow past a cross section of a streamline was supposed to be constant, but if the streamline is curved, the pressure is lower and the flow rate faster on the inner part of the streamline.

Back to the spinning bucket with water. Assuming the angular velocity is constant, then the lowest pressure and lowest speed occurs near the center of the bucket, which would violate Bernoulli, so some other method would need to be used to find a mathematical relationship between the pressure gradient and the speed of the water.

Last edited: Oct 4, 2015
6. Oct 5, 2015