IDENTIFY: Consider the fluid in the horizontal part of the tube. This fluid, with mass ρ Al, is subject to a net force due to the pressure difference between the ends of the tube.
SET UP: The difference between the gauge pressures at the bottoms of the ends of the tubes is
ρg(yL − yR).
EXECUTE: Thenetforceonthehorizontalpartofthefluidis ρg(yL −yR)A=ρAla, or, (yL − yR)= al.
g
(b) Again consider the fluid in the horizontal part of the tube. As in part (a), the fluid is accelerating; the center of mass has a radial acceleration of magnitude arad =ω2l/2, and so the difference in heights