Oy Sentence Examples
Oy. I have the feeling owing you is not a good thing.
The force X1 in KH with X1 in Ox forms a couple about Oy, of moment z1Xi.
Dealing in the same way with the forces Yi, Zi at P1, we find that all three components of the force at P1 can be transferred to 0, provided we introduce three couples L1, Mi, Ni about Ox, Oy, Oz respectively, viz.
Take, for example, the case of a sphere rolling on a plane; and let the axes Ox, Oy be drawn through the centre parallel to the plane, so that the equation of the latter is 1=cf.
In the critical case of 2BT= I it breaks up into two planes through the axis of mean moment (Oy).Advertisement
The moving axes Ox, Oy, 01 form a rigid frame of reference whose motion at time t may be specified by the three component angular velocities p, q, r.
Now consider a system of fixed axes Ox, Oy, Oz chosen so as to coincide at the instant I with the moving system Ox, Oy, Os.
At the instant t+t, Ox, Oy, Os will no longer coincide with Ox, Oy, Os; in particular they will make with Ox angles whose cosines are, to the first order, I, rot, qOt, respectively.
If L, M, N be the moments of the extraneous forces about Ox, Oy, Os this must be equal to Xl--LOt.
If we now apply them to the case of a rigid body moving about a fixed point 0, and make Ox, Oy, Oz coincide with the principal axes of inertia at 0, we have X, u, v=Ap, Bq, Cr, whence A (B C) qr = L,Advertisement
Which is defined oy my goodness itself was principally and that interracial.
Again, could he but do the accent, Brad Pitt might not do too badly... oy!
Any CREST member who has bought inion oy shares recently and holds them through CREST should contact their counterparty to pass on the documentation.
A brief abstract of Smith's methods and results appeared in the Proc. R oy.
Similarly, parallel to Oy, the increase of effective inertia is NT'/th n th(n-a), reducing to M'/th n=M' (a/b), when a= oo, and the liquid extends to infinity.Advertisement
Or oy could use plain unlined paper to make a sketch book or journal.
Uyea, "the isle," from the Old Norse oy (3), to the south of Unst, from which it is divided by the narrow sounds of Uyea and Skuda, yields a beautiful green serpentine.
For a point in the line OY bisecting the magnet perpendicularly, 0 =42 therefore cos 0 =0, and the point D is at an infinite distance.
Thus if d,/ is the increase of 4, due to a displacement from P to P', and k is the component of velocity normal to PP', the flow across PP' is d4 = k.PP'; and taking PP' parallel to Ox, d,, = vdx; and similarly d/ ' = -udy with PP' parallel to Oy; and generally d4,/ds is the velocity across ds, in a direction turned through a right angle forward, against the clock.
Uniplanar Motion of a Liquid due to the Passage of a Cylinder through it.-A stream-function 4, must be determined to satisfy the conditions v24 =o, throughout the liquid; (I) I =constant, over any fixed boundary; (2) d,t/ds = normal velocity reversed over a solid boundary, (3) so that, if the solid is moving with velocity U in the direction Ox, d4y1ds=-Udy/ds, or 0 +Uy =constant over the moving cylinder; and 4,+Uy=41' is the stream function of the relative motion of the liquid past the cylinder, and similarly 4,-Vx for the component velocity V along Oy; and generally 1,1'= +Uy -Vx (4) is the relative stream-function, constant over a solid boundary moving with components U and V of velocity.Advertisement
When the cylinder r =a is moved with velocity U and r =b with velocity U 1 along Ox, = U b e - a,1 r +0 cos 0 - U ib2 - 2 a, (r +Q 2 ') cos 0, = - U be a2 a2 (b 2 - r) sin 0 - Uib2 b1)a, (r - ¢2 sin 0; b and similarly, with velocity components V and V 1 along Oy a 2 b2 ?= Vb,_a,(r+r) sin g -Vi b, b2 a, (r+ 2) sin 0, (17) = V b, a2 a, (b2 r) cos 0+Vi b, b, a, (r- ¢ 2) cos h; (18) and then for the resultant motion z 2zz w= (U 2 + V2)b2a a2U+Vi +b a b a2 U z Vi -(U12+V12) b2 z a2b2 Ui +VIi b 2 - a 2 U1 +Vii b 2 - a 2 z The resultant impulse of the liquid on the cylinder is given by the component, over r=a (§ 36), X =f p4 cos 0.ad0 =7rpa 2 (U b z 2 + a 2 Uib.2bz a2); (20) and over r =b Xi= fp?
Taking two planes x = =b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P' at opposite ends of the diameter PP', is pdy (U - Ua 2 r2 cos 20 +mr i sin 0) (Ua 2 r 2 sin 2 0+mr 1 cos 0) + pdy (- U+Ua 2 r 2 cos 2 0 +mr1 sin 0) (Ua 2 r 2 sin 2 0 -mr 1 cos 0) =2pdymUr '(cos 0 -a 2 r 2 cos 30), (8) and with b tan r =b sec this is 2pmUdo(i -a 2 b2 cos 30 cos 0), (9) and integrating between the limits 0 = 27r, the resultant, as before, is 27rpmU.
An angular velocity R, which gives components - Ry, Ix of velocity to a body, can be resolved into two shearing velocities, -R parallel to Ox, and R parallel to Oy; and then ik is resolved into 4'1+1'2, such that 4/ 1 -R-Rx 2 and 1//2+IRy2 is constant over the boundary.
Motion symmetrical about an Axis.-When the motion of a liquid is the same for any plane passing through Ox, and lies in the plane, a function ' can be found analogous to that employed in plane motion, such that the flux across the surface generated by the revolution of any curve AP from A to P is the same, and represented by 2s-4 -11'o); and, as before, if d is the increase in due to a displacement of P to P', then k the component of velocity normal to the surface swept out by PP' is such that 274=2.7ryk.PP'; and taking PP' parallel to Oy and Ox, u= -d/ydy, v=dl,t'/ydx, (I) and 1P is called after the inventor, " Stokes's stream or current function," as it is constant along a stream line (Trans.
As before in § 31, the rotation may be resolved into a shear-pair, in planes perpendicular to Ox and Oy.Advertisement