444 APPLIED MECHANICS a. 
vortex is produced. Let A (Fig. 722) represent a fluid particle in a f 
circular vortex whose axis is O. Let AC at. right fo 
angles to OA represent the velocity of A in the free | oan 
circular vortex. Again, let A represent a fluid par- < 
ticle in a current radiating from O. Let AB on OA of Na 
produced represent the velocity of A in the radiating 7 
current. If the two motions be combined, A will Frid. 722 
have, when in the position considered, a velocity a 
represented by AD, a diagonal of the rectangle BC. Since a 
AC-AO=a constant, and AB-AO=aconstant, © 
it follows that AC/AB=a constant=tan 6. Hence the path of the flu 
particle will at every instant make the constant angle @ with the radi 
drawn from the particle to the axis, and this is a property of th 
logarithmic or equiangular spiral. a 
380. Whirling Liquids.—Let a cylinder of radius (Fig. 723), co 
taining a liquid, revolve about its axis YY,, which is vertical, with a 
angular velocity » Let P be a point on i 
the free surface of the liquid, and let w 
be the weight of a very small portion of 
the liquid at P. Consider the forces acting 
on the liquid at P in a vertical plane con- 
taining P and the axis YY,. There is the 
weight w acting vertically downwards, the 
centrifugal force Q acting horizontally, and 
the fluid pressure p, which must be per- 
pendicular to the free surface of the liquid, 
and which must balance the resultant R 
of Q and w. Let the horizontal through P be 
meet the axis at N, and let the line of action of R meet the axis at G. — 
w GN w wg g 
Q= Go°PN, and By = Q~ wo2PN PN’ 7 
Hence GN, the sub-normal of the free surface at P, is a constant, an¢ 
therefore the free surface is a paraboloid, and the section of the free surface 
by a plane containing the axis YY, is a parabola. 
The equation to the parabola, taking the axes as AX and AY, whe 
A is the vertex, is (Art. 11, p. 10) «?=4ay, where a is the focal distane 
of the vertex, PN=2, and AN=y. Now in a parabola the sub-norr 
constant and equal to the semi-latus rectum. But the semi-latus 
is the value of x in the equation a?=4ay when y=a, therefore GN = 2 
Fig. 723. 
wes Eh a 
but GN =5 , therefore 2a= J, , and the equation to the parabola is 
2 2 S 
2g o - 
a= Bh 8 Y= 5 j 
2 
If h is the height of the cup, then haar j 
The volume of a paraboloid is half the volume of the circumse ot vi 
cylinder, hence the volume of liquid in the cylinder above the level AX 
2h : Ja 
“5 If CD (Fig. 724) is the level of the liquid when at rest, th 
a 
= 
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