Rising of a Liquid During Rotation
A 20-cm-diameter, 60-cm-high vertical cylindrical container, shown in Fig. 3–66, is partially filled with 50-cm-high liquid whose density is 850 kg/m³. Now the cylinder is rotated at a constant speed. Determine the rotational speed at which the liquid will start spilling from the edges of the container.
A vertical cylindrical container partially filled with a liquid is rotated. The angular speed at which the liquid will start spilling is to be determined.
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 The bottom surface of the container remains covered with liquid during rotation (no dry spots).
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
z_{s}=h_{0}-\frac{\omega^{2}}{4 g}\left(R^{2}-2 r^{2}\right)
Then the vertical height of the liquid at the edge of the container where r = R becomes
z_{s}(R)=h_{0}+\frac{\omega^{2} R^{2}}{4 g}
where h_{0}=0.5 m is the original height of the liquid before rotation. Just before the liquid starts spilling, the height of the liquid at the edge of the container equals the height of the container, and thus z_{s}(R)=H=0.6 m. Solving the last equation for ω and substituting, the maximum rotational speed of the container is determined to be
\omega=\sqrt{\frac{4 g\left(H-h_{0}\right)}{R^{2}}}=\sqrt{\frac{4\left(9.81 m / s ^{2}\right)[(0.6-0.5) m ]}{(0.1 m )^{2}}}=19.8 rad / s
Noting that one complete revolution corresponds to 2π rad, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) as
\dot{n}=\frac{\omega}{2 \pi}=\frac{19.8 rad / s }{2 \pi rad / rev }\left(\frac{60 s }{1 min }\right)=189 rpm
Therefore, the rotational speed of this container should be limited to 189 rpm to avoid any spill of liquid as a result of the centrifugal effect.
Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. We should also verify that our assumption of no dry spots is valid. The liquid height at the center is
z_{s}(0)=h_{0}-\frac{\omega^{2} R^{2}}{4 g}=0.4 m
Since z_{s}(0) is positive, our assumption is validated.