The fluid flows into the sprayer through point O, so the fluid has no initial angular momentum. Therefore, the sprayer reaches its constant angular speed when the fluid flows out radially. In order to find the condition, it will be easier if first we observe from sprayer's rotating frame.We define a new variable, r, which equals to the distance from point O to end of pipe.

Now observe it from ground frame. Velocity of the fluid at ground frame is a resultant of and Because velocity of fluid must be radial at ground frame, tangential component of v must be equal to ωr,

Substitute v from equation (2) to equation (1), and we get

Substitute r, and finally we get

That's my solution.

If you have any question, correction, or comment, please leave a comment below.

Thank you.

## 5 comments

in the step going to eqtn 1 why isnt their any term for pressure in thr RHS of the equation?

I do the question using angular momentum.. it produce a different answer..

I don't know how to write my answer in latex here.. but, my answer is

w = 6/5 (2P/(rho) a^2)^0.5

I don't know why the answer is different..

by the way, why the fluid velocity must be tangential in ground frame?? why not tangential like usual in simple circular motion?

sorry, correction for above..

by the way, why the fluid velocity must be radial in ground frame?? why not tangential like usual in simple circular motion?

@1st commenter:

The pressure term in RHS of the equation is for atmospheric pressure. And it has been mentioned at the problem's text that P is much larger than atmospheric pressure.

@2nd commenter:

Hmmm... I don't know, too. But could you send your solution to my email, please?

@3rd commenter:

In order to keep angular momentum of the fluid zero when it enters and flows out from the sprayer.

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