# Calculate curl of rotating bucket of water

## Homework Statement

A bucket of water is rotated slowly with angular velocity w about its vertical axis..When a steady state has been reached the water rotates with a velocity field v(r) as if it were a rigid body. Calculate div(v) and interpret the result. Calculate curl (v). Can the flow be represented in terms of a velocity potential such that v = grad phi? If so, what is phi?

## The Attempt at a Solution

Not sure how to do this...
I think somewhere in my notes it says that v here should = (-wy,wx,0)

Is this right?

In which case, div(v) = 0 - what is the interpretation?? No sinks or sources??
curl v = (0,0,2w)...

Can I write it as the gradient of a scalar function? If so, what is the function?

## Answers and Replies

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I just don't see how to write it as the gradient of a scalar function basically..

anyoneee?

I just don't see how to write it as the gradient of a scalar function basically..
The most simple approach here would be to draw a diagram showing the velocity vector, and how it's always perpendicular to the radius vector. (Every droplet goes around in a circle with its energy constant)

This yields:
$$\vec v= -\omega y \hat x +\omega x \hat y$$

Taking the divergence of this function, does indeed get you 0. That means no sources or sinks are present.

Taking the curl, however, shows you that this vector field does have a curl!

$$\nabla \times \vec v= 2\omega \hat z$$

A vector field that can be represented as the gradient of some scalar function must always have 0 curl. (Some googling brought up this: http://en.wikipedia.org/wiki/Conservative_vector_field)

Note that it is a mathematical identity that $$\nabla\times\nabla\phi=0$$ making it impossible for a gradient to have non-zero curl.
If a vector field can be represented as the gradient of a scalar function, then it MUST have 0 curl.
However, if a vector field has 0 curl, the converse is not true in general, you may be able to represent it as the gradient of a scalar function, and you might not be able to do so.

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