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A sealed cylindrical can (height 12 cm, radius 4 cm) contains an incompressible soft drink of specific gravity 0.8 filled to a height of 9 cm; the remaining volume holds air at 10 kPa. The can is taken deep into space (gravity-free) inside a spaceship and spun about its axis at a constant angular velocity of 250/sqrt(3) rad/s. The pressure the liquid exerts on the curved surface of the can is (10000 * N) Pa. Find N.
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Correct answer: 2
Solution
In zero gravity, centrifugal force pushes liquid to the outer wall forming an annular cylinder of inner radius r and outer radius R=0.04 m over full height H=0.12 m. Volume conservation: pi*(0.04²)*(0.09) = pi*(0.04² - r²)*(0.12). Solving: r² = 0.0004, r = 0.02 m. Pressure at outer wall: P(R) = P_air + (1/2)*rho*omega²*(R² - r²) = 10000 + (1/2)(800)(62500/3)(0.0012) = 10000 + 10000 = 20000 Pa = 10000 * 2.
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