At 4:55 PM 4/1/5, Robin van Spaandonk wrote:
In reply to Horace Heffner's message of Thu, 31 Mar 2005 13:43:54
However when the water rotates, a dip forms at the middle, which
can drop right down to the floor of the tank at sufficiently high
w. However, according to the formula, for any w > 0, h > h0 for
all R, since the first term is always positive.
The h0 above is negative.
If h = (w^2/2g) x R^2 + h0 and h0 is negative, then for w=0, h=h0
and is thus also negative. How does one end up with a negative
height?
As I stated in the last post, the above surface is only meaningful when h >
0. There is no water in the tank above [at] radii where h <= 0.
If w=0 then h<=0 everywhere because no water will stay in the tank. No
angular momentum is involved. Any water in the tank is not in equilibrium
as assumed - it will all run out. Please note again that the coreolis
force is ignored throughout.
Or should the original formula perhaps be:
h = h0 - (w^2/2g) x R^2 ?
(Since the second term in this version is positive, the height
becomes less for higher w and also for smaller R, both of which
make sense).
In short, is h the distance up from the bottom of the tank, or the
distance down from the surface?
The variable h is the distance up from the bottom of the tank. When h=0 or
h<=0 then there is no water above the radius at which h is computed. The
variable h at final equilibrium is a function of R, R1, and w, where R1 is
the radius of the drain hole.
The initial or final surface, assumed to be in equilibrium with w constant
at every radius, is concave upwards. The coefficient of R^2 is thus
positive.
In the initial condition, h0 can be anything depending on how much water is
in the rotating tank. This h0 does not affect the *curvature* of the
surface, however, which is only a function of w, g, and R, assuming the
drain hole is plugged, and w is constant over all radii. The variable h0
changes as the water drains from the tank. The equation describing the
water surface changes as well, but the final surface should return to the
form h = (w^2/2g) x R^2 + h0 for a rotating tank, assuming that viscosity
forces w to be uniform across all radii, where in the *final* equilibrium:
h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2)
and w is the final angular velocity of the water and tank, R1 is the drain
radius, and R is a given radius.
Regards,
Horace Heffner
