A horizontal tube of length \(L\), open at \(A\) and closed at \(B\), is filled with an ideal fluid. The end \(B\) has a small oriffice. The tube is set in rotation in the horizontal plane with angular velocity \(\omega\) about a vertical axis passing through \(A\). Show that the efflux velocity of the fluid is given by $$ v=\omega l\sqrt{\frac{2L}{l}-1}$$ where \(l\) is the length of the fluid.

**Solution:**

Consider a mass element \(dm\) of the fluid at a distance \(x\) from the vertical axis. The centrifugal force on \(dm\) is

$$ Df=dm\omega^2x$$ $$=dm\frac{dv}{dt}$$ $$=dm \frac{dv}{dx}v$$

$$ vdv=\omega^2 xdx$$

$$ \int vdv=\omega^2 \int xdx$$ $$

\frac{v^2}{2}=\frac{\omega^2}{2} \int_{L}^{L-l}

$$

So,

$$ v=\omega l\sqrt{\frac{2L}{l}-1}$$

## No comments, be the first one to comment !