uniform rod revolving around a vertical axis with given angular velocity and given length of rod A uniform rod of given length and given angular velocity is revolving around a vertical axis. 
Clearly it can do so in a horizontal plane with respect to vertical axis. 
At what other angle can it do so? 
That is, what is the angel of inclination with vertical axis? 
Given that inclination angle is constant. 
Answer in the book is :cosine inverse 3g/(2 times square of angular velocity times length of rod).
Where g is acceleration due to gravity. 

The torque due to gravity and centripetal force must be equal. Let's call the coordinate along the rod $x$, varying from 0 to $L$. A small piece, of length $dx$ has a mass of $\frac{m}{L} dx$. The torque due to gravity of this small piece is $$d\tau_g=\frac{m}{L} dx g x \sin \theta$$ The total torque due to gravity is given by $$\tau_g=\int_0^L\frac{m}{L} dx g x \sin \theta=\frac{1}{2}mgL\sin\theta$$
The centripetal force on the same small piece is given by $$dF=\frac{m}{L} dx \omega^2 x\sin\theta$$ $\omega$ is the angular velocity. If the rod is at an angle $\theta$ with respect to the vertical, the rotation radius of that small piece at distance $x$ along the rod is $x\sin\theta$. the torque due to this force is $$d\tau_F=dF x \cos\theta=\frac{m}{L} dx \omega^2 x^2\sin\theta\cos\theta$$ Integrating it we get 
$$\tau_F=\frac{1}{3}m\omega^2L^2\sin\theta\cos\theta$$
Now from $\tau_g=\tau_F$ we get:
$$ \frac{1}{2}mgL\sin\theta=\frac{1}{3}m\omega^2L^2\sin\theta\cos\theta$$
We have two solutions. $\theta$=0 means that the rod is vertical, rotating around the vertical axis. The other solution is the one you were looking for: $$\cos\theta=\frac{3g}{2\omega^2L}$$
Note that the rod cannot be horizontal, since the gravity will pull it downwards.
