At 9:46 PM 3/31/5, Robin van Spaandonk wrote:
>In reply to  Horace Heffner's message of Sun, 13 Mar 2005 17:53:41
>-0900:
>Hi Horace,
>
>I'm having some trouble understanding this formula. If it's meant
>to give the relationship between the absolute height of the water
>surface at any radius, then it seems to say that at w=0, h= h0,
>i.e. h0 is the height of stationary water in the tank.
>
>>Correction follows. Sorry!
>>
>>The shape of the final equilibrium surface is:
>>
>>   h = (w^2/2g) x R^2 + h0
>
>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. Therefore, the
>formula either doesn't represent what I thought it was meant to
>represent, or it doesn't describe reality.


I think I misunderstood the problem you note!  The error is not in the
above formula, but in stuff you snipped.  It is merely a sign error.
Following is a repeat derivation of H0 with corrections:

The shape of the final equilibrium surface is:

   h = (w^2/2g) x R^2 + h0

where h is height, w is angular velocity, g = 9.80665 m/s^2, and R is radius.

Using R1 as the radius of the hole, R2 as radius of of tank, we have h = 0
at the radius R1 when equlilbrium is established and no more water can go
down the hole.  So:

   0 = (w^2/2g) x (R1)^2 + h0

   h0 = - 2g/( w^2 x (R1)^2)       <========= Note the sign change!

The final height Hf of the water at the edge of the tank is thus:

   Hf = (w^2/2g) x (R2)^2 - 2g/( w^2 x (R1)^2)   <==== Note subtraction!

The above assumes that the initial angular velocity is small.  If the
angular velocity of the initial condition is high then the initial
condition integration of angular momentum and energy also has to be outside
the boundary established by:

   h = (w^2/2g) x R^2 + h0

It is possible that in the initial condition all the water will be located
outside the radius of the hole, and thus no water can go down the hole at
all.

Regards,

Horace Heffner          


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