Fluids in Rigid Body Motion 1

Fluids in Rigid Body Motion 1

1. Recall, for the case of rigid body motion, the equation of motion for fluid flow (the Navier-Stokes equation) reduces to

2. There are two cases of rigid body motion to be discussed: uniform linear rigid body acceleration, and rigid body rotation.

   • Uniform Linear Rigid Body Acceleration
     • Consider the case where the fluid is accelerated uniformly in some direction. In other words, each fluid particle in the container feels exactly the same acceleration vector, which is constant in time. In such a case, since both the gravity vector and the acceleration vector are constant, the effective gravity vector, G, must also be constant.
       • then, that the equation of motion is identical to the hydrostatics equation, except that gravity g is replaced by effective gravity G
         • his makes problems of this type no more difficult than simple hydrostatics. In fact, for uniform linear rigid body acceleration, the solution is identical to that of hydrostatics, but with g replaced by G, and with z (parallel to g, i.e. down) replaced by s (parallel to G).
           • One consequence of this is that isobars must be perpendicular to G, i.e. perpendicular to direction s. In fact, the pressure increases linearly with distance s, rather than with z.
             • As an example, consider a glass of water in an elevator which is accelerating up:
               • Here, the effective gravity is still downward, but of greater magnitude than g. The isobars are still horizontal surfaces as in hydrostatics. In fact, everything is identical to hydrostatics except for a larger gravity (pretend for example that the glass is sitting on the planet Jupiter). Our simple hydrostatic pressure relationship still applies, but with g replaced by G, and z replaced by s
                 • All else being equal, the pressure at the bottom of the accelerating glass will be greater than that at the bottom of the stationary glass (because G is greater than g).
                   • Now consider a glass of water in an elevator that accelerates uniformly to the right.
                     • Now, since the effective gravity is tilted at some angle to the lower left, the isobars must be perpendicular to this direction. The isobars are thus titled down and to the right as sketched. Note that the surface remains an isobar of constant pressure pa, and is therefore also tilted as sketched. At some point 1 in the fluid, the pressure can be found from the revised hydrostatic pressure relationship as follows:

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