Step 1: Analyze the Geometry of the Water Surface
The vessel is cylindrical with radius $r$ and height $h$, and it is half-filled with water. When the vessel is rotated and the water is on the verge of spilling, the water surface will tilt. Since the volume of water is constant (half the volume of the cylinder), the critical condition for spilling corresponds to the water surface extending from the top rim of the vessel to the diametrically opposite bottom rim.
Let $\theta$ be the angle the water surface makes with the vertical axis of the cylinder. From the geometry of the cylinder’s cross-section:
$$ \tan\theta = \frac{2r}{h} $$Step 2: Force Analysis
Consider a fluid element of mass $dm$ at the surface. As shown in the diagram, for the water to be on the verge of spilling, the vertical component of the force term $\omega^2 l dm$ must balance the gravitational force $g dm$.
From the force diagram, we have the equilibrium relation along the vertical direction:
$$ (\omega^2 l dm) \cos\theta = g dm $$ $$ \omega^2 l \cos\theta = g $$Step 3: Solve for Minimum Angular Velocity ($\omega$)
Rearranging the force equation for $\omega$:
$$ \omega^2 = \frac{g}{l \cos\theta} = \frac{g}{l} \sec\theta $$We can express $\sec\theta$ in terms of the vessel dimensions $h$ and $r$ using the trigonometric identity $\sec\theta = \sqrt{1 + \tan^2\theta}$. Substituting $\tan\theta = \frac{2r}{h}$ from Step 1:
$$ \sec\theta = \sqrt{1 + \left(\frac{2r}{h}\right)^2} = \sqrt{1 + \frac{4r^2}{h^2}} = \sqrt{\frac{h^2 + 4r^2}{h^2}} = \frac{\sqrt{h^2 + 4r^2}}{h} $$Substituting this back into the expression for $\omega^2$:
$$ \omega^2 = \frac{g}{l} \left( \frac{\sqrt{h^2 + 4r^2}}{h} \right) $$Therefore, the minimum angular velocity required is: