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Liquid pressure and circular motion

  1. Nov 19, 2008 #1
    A tube is sealed at both ends and contains a 0.0212-m-long portion of liquid. The length of the tube is large compared to 0.0212 m. There is no air in the tube, and the vapor in the space above the liquid may be ignored. The tube is whirled around in a horizontal circle at a constant angular speed. The axis of the rotation passes through one end of the tube, and during the motion, the liquid collects at the other end. The pressure experienced by the liquid is the same as it would experience at the bottom of the tube, if the tube were completely filled with liquid and allowed to hang vertically. Find the angular speed (in rad/s) of the tube.

    In this question you have to equate the pressure at the bottom (pressure=(rho)gh) to the centripetal force acting on the string? is that right?
  2. jcsd
  3. Nov 19, 2008 #2


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    No. And what string are you talking about? I think that the pressure at the bottom of the liquid when it's rotating is rho*a*h, where 'a' is the centripetal acceleration and h is the depth of the liquid. Compare that to the pressure at the bottom when the tube is full of liquid. You'll need to use the length of the tube L in the formula. But since they don't give you a value for L, you'd better hope it cancels in the end. Does it?
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