FIGURE 6.11. (a) Water placed in a rotating tank and insulated from external forces (both mechanical and thermodynamic) eventually comes into solid body rotation, in which the fluid does not move relative to the tank. In such a state the free surface of the water is not flat but takes on the shape of a parabola given by Eq. 6-33. (b) Parabolic free surface of water in a tank of 1 m square rotating at Q = 20 rpm.

FIGURE 6.11. (a) Water placed in a rotating tank and insulated from external forces (both mechanical and thermodynamic) eventually comes into solid body rotation, in which the fluid does not move relative to the tank. In such a state the free surface of the water is not flat but takes on the shape of a parabola given by Eq. 6-33. (b) Parabolic free surface of water in a tank of 1 m square rotating at Q = 20 rpm.

Fig. 6.11a). Hence the depth of the fluid h is given by:

where r is the distance from the axis or rotation. Thus the free surface takes on a parabolic shape: it tilts so that it is always perpendicular to the vector g* (gravity modified by centrifugal forces) given by g* = -gz -Q x Q x r. If we hung a plumb line in the frame of the rotating table it would point in the direction of g*, which is slightly outward rather than directly down. The surface given by Eq. 6-33 is the reference to which H is compared to define n in Eq. 6-20.

Let us estimate the tilt of the free surface of the fluid by inserting numbers into Eq. 6-33 typical of our tank. If the rotation rate is 10 rpm (so Q ~ 1 s-1), and the radius of the tank is 0.30 m, then with g = 9.81 m s-2, we find Hf - 5 mm, a noticeable effect but a small fraction of the depth to which the tank is typically filled. If one uses a large tank at high rotation, however (see Fig. 6.11b in which a 1 m square tank was rotated at a rate of 20 rpm) the distortion of the free surface can be very marked. In this case

It is very instructive to construct a smooth parabolic surface on which one can roll objects. This can be done by filling a large flat-bottomed pan with resin on a turntable and letting the resin harden while the turntable is left running for several hours (this is how parabolic mirrors are made). The resulting parabolic surface can then be polished to create a low friction surface. The surface defined by Eq. 6-32 is an equipo-tential surface of the rotating frame, and so a body carefully placed on it at rest (in the rotating frame) should remain at rest. Indeed if we place a ball bearing on the rotating parabolic surface and make sure that the table is rotating at the same speed as was used to create the parabola, then we see that it does not fall into the center, but instead finds a state of rest in which the component of gravitational force, gH, resolved along the parabolic surface is exactly balanced by the outward-directed horizontal component of the centrifugal force, (Q2r)H, as sketched in Fig. 6.12 and seen in action in Fig. 6.13.

GFD Lab V: Visualizing the Coriolis force

We can use the parabolic surface discussed in Lab IV in conjunction with a dry ice "puck" to help us visualize the Coriolis force. On the surface of the parabola, $ = constant and so V$ = 0. We can also assume that there are no pressure gradients acting on the puck because the air is so thin.

FIGURE 6.12. If a parabola of the form given by Eq. (6-33) is spun at rate Q, then a ball carefully placed on it at rest does not fall in to the center but remains at rest: gravity resolved parallel to the surface, gH, is exactly balanced by centrifugal accelerations resolved parallel to the surface, (Q2 r)H.

FIGURE 6.12. If a parabola of the form given by Eq. (6-33) is spun at rate Q, then a ball carefully placed on it at rest does not fall in to the center but remains at rest: gravity resolved parallel to the surface, gH, is exactly balanced by centrifugal accelerations resolved parallel to the surface, (Q2 r)H.

Furthermore, the gas sublimating off the bottom of the dry ice almost eliminates frictional coupling between the puck and the surface of the parabolic dish; thus we may also assume F = 0. Hence the balance Eq. 6-31 applies.10

We can play games with the puck and study its trajectory on the parabolic turntable, both in the rotating and laboratory frames. It is useful to view the puck from the rotating frame using an overhead co-rotating camera. Fig. 6.14 plots the trajectory of the puck in the inertial (left) and in the rotating (right) frame. Notice that the puck is ''deflected to the right'' by the Corio-lis force when viewed in the rotating frame, if the table is turning anticlockwise (cycloni-cally). The following are useful reference experiments:

1. We place the puck so that it is motionless in the rotating frame of reference. It follows a circular orbit around the center of the dish in the laboratory frame.

2. We launch the puck on a trajectory that crosses the rotation axis. Viewed from the laboratory the puck moves backward and forward along a straight line (the straight line expands out in to an ellipse if the frictional coupling between the puck and the rotating disc is not negligible; see Fig. 6.14a). When viewed in the rotating frame, however, the particle is continuously deflected to the right and its trajectory appears as a circle as seen in Fig. 6.14b. This is the ''deflecting force'' of Coriolis. These circles are called inertial circles. (We will look at the theory of these circles below).

3. We place the puck on the parabolic surface again so that it appears stationary in the rotating frame, but is then slightly perturbed. In the rotating frame, the puck undergoes inertial oscillations consisting of small circular orbits passing through the initial position of the unperturbed puck.

Inertial circles It is straightforward to analyze the motion of the puck in GFD Lab V. We adopt a Cartesian (x, y) coordinate in the rotating frame of reference whose origin is at the center of the parabolic surface. The velocity of the puck on the surface is urot = (u, v), where urot = dx/dt, vrot = dy/dt, and we have reintroduced the subscript rot to make our frame of reference explicit. Further we assume that z increases upward in the direction of Q.

Rotating frame Let us write out Eq. 6-31 in component form (replacing D/Dt by d/dt, the rate of change of a property of the puck). Noting that (see VI, Appendix A.2.2)

2Q x urot = (0, 0, 2Q) x (urot, Vrot, 0) = (-2QVrot,2Qurot,0), the two horizontal components of Eq. 6-31 are:

Urot dx dt

If we launch the puck from the origin of our coordinate system x(0) = 0; y(0) = 0

10A ball bearing can also readily be used for demonstration purposes, but it is not quite as effective as a dry ice puck.

(chosen to be the center of the rotating dish), with speed urot (0) = 0; vrot (0) = vo, the solution to Eq. 6-34 satisfying these boundary conditions is urot (t) = vo sin 2Qt; vrot (t) = vo cos 2Qt x (t) = — - — cos 2Qt; y (t) = — sin 2Qt. w 2Q 2Q ^v 7 2Q

The puck's trajectory in the rotating frame is a circle (see Fig. 6.15) with a radius of .It moves around the circle in a clockwise direction (anticyclonically) with a period n/Q, known as the ''inertial period.'' Note from Fig. 6.14 that in the rotating frame the puck is observed to complete two oscillation periods in the time it takes to complete just one in the inertial frame.

Inertial frame Now let us consider the same problem but in the nonrotating frame. The acceleration in a frame rotating at angular velocity Q is related to the acceleration in an inertial frame of reference by Eq. 6-27. And so, if the balance of forces is Durot/Dt = -2Q x urot these two terms cancel out in Eq. 6-27, and it reduces to11:

duin

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