Question

QUESTION 4. (4a) Show that mi DF - 2mw xi – mü x (w x M) where is the total force acting on an object of mass m measured by an inertial observer and is the acceleration of the object measured in a frame rotating with angular velocity (4b) Determine the horizontal and vertical components of the Coriolis acceleration expe- rienced by an airplane flying due East with a velocity of 750 km/hr at latitude 50° N (4c) Find the angle through which it must tilt its wings to compensate for the horizontal component of the Coriolis force, assuming that it experiences a lift force perpendicular to its wings (40) Estimate the ratio of the centrifugal force to the coriolis force in this problem (take the radius of the Earth to be 6370 km).

Answer 1

YA 0 • Р F S Х O S Z Z Carro .. r fixed fixed since the directions X', Y', Z go on changing continuously with rotation and so do the 4.2 Rotating frame of reference We shall now derive expressions for the velocity, acceleration and force on a particle in a uniformly rotating frame in terms of the corresponding quantities in an inertial frame relative to which the rotating system rotates. Let a fixed inertial frame S be represented by the coordinate axes X, Y, Z and a non-inertial rotating frame S' by X', Y', Z'. S rotates relative to S with an angular velocity ū about an axis passing through the common origin O of the frames (Fig. 4.1). Physically, a vector A is the same in both the frames, but the components will differ as the unit vectors in the two frames point Fig. 4.1 differently. Let ņ be the position vector of a point P. But, Exit yi t zk (4.2.1) where (x, y, z) are the coordinates of P and i, j,  the unit vectors along X, Y, Z- directions in S-frame. The position vector r in S'-frame is represented as = x² + y + z' (4.2.2) where (2', y', z") are the coordinates of P and î', Î', â' the unit vectors along X', Y', Z'- directions respectively in S'-frame. The time derivative of rfixed and Frot must also be equal as they are one and the same physical vector. d d (Tfixed) dt dr dxi + dy dz From equation (4.2.1) j + -k, dt dt dt since the unit vectors i, į and  are constants. d (Trot) (x'î' + y'ſ' +z'Â'), ising (4.2.2) dt dt dz' + +y dt dt dt dt dt r Trot at Trot) dt / fixed dt ) fixed dylit. I da de dil = C dil + + 2 dt unit vectors i', Î' and Ấ'.

dr dr ,dî + x dt + y +z S dt S dt The first three terms in r.his is denoted by (df / dt)rot for this is precisely the time derivative of 7 as measured by an observer in the rotating frame S'. dhi dh' (4.2.3) dt dt But (dpdt)s = Ū, the velocity of the particle P relative to the inertial S-frame and (df/dt)s = ✓", the velocity of P relative to the rotating S'-frame. Since the unit vectors i', Î' and Â' rotate with an angular velocity w, di Eū xi'; dî di w x '; = w x k dt dt So, the equation (4.2.3) can be written as under. dr dr dt +ū * (z'î' + y'î' + z'Ã') = tw x 5 (4.2.4) dt dt (a) s = dr S' S or, ū= ' + ū x = v' + Vo (4.2.5) where ūo is the linear velocity of the particle due to the rotation. Thus, the velocity of the particle in the inertial frame is the vector sum of its velocity in the rotating frame ū' and the linear velocity due to rotation vo, the value of which at any instant depends on the value of r* at that instant. Operator equation The relation (4.2.4) may be conveniently expressed by the following operator identity : d d c - tox dt (4.2.6) S S' and may usefully be applied to other similar vectors. Applying the operator equation to the velocity vector ū, in place of displacement vector i 'du (a dū dt tw.x u dt S S dυ oito Substituting for ū from (4.2.5), we obtain d (u' + xr)s +ū x (TP + w x 7), dt/s dt dū do 'dr + xr twx dt dt S S du dt + () +(xi") + w x (x1) + (W x ū”) + (w x ū) + w x (w x ) 24*2)+3x+ dw xr dt du' dt dt

or, ã = ã' + 2 (W x Ū') + W x W x ñ) + dū Хт dt (4.2.7) S' where ă' is the acceleration of the particle in the rotating S'-frame and ã the acceleration in the fixed inertial frame S. The term 2 (W x ū') is known as Coriolis acceleration, after its discoverer, which appears only when Ū' exists, i.e., the particle moves in the rotating S'-frame. The term ū x W x 7") is the centripetal acceleration, while the last term (dw/dt) x is the acceleration due to variation of ū with time. In a situation where the angular velocity w of the rotating frame is constant, dw/dt = 0 and the acceleration is also zero. The quantity w x W x ñ + (dw/dt).x is called the aceleration of transport at. For a uniform angular velocity, dw/dt = 0 and equation (4.2.7) becomes a =ā' + 2xy + x xr) müxņ)s Multiplying (4.2.7) by the mass m of the particle and rearranging, we obtain ma' mā – 2m(w x Ő) – mū < (w x 0) – m (ü x (4.2.8) or, F' = } – 2m(W x 7') – mū x (w x 7) – m ( x )s = + F where F' (= ma') is the effective force on the particle of mass m in the rotating fráme S' and F(= mā) is the true force on the particle in the inertial frame S and Fo = -2m(W x ū”) – mū x W x 7) – müx m)s stands for the fictitious force in the non-inertial frame. When, however, w = constant, we have, ü = 0 = m ( x P) = 0 F = -2m x U') – mū x W x r') i.e., Fictitious force = Coriolis force + Centrifugal force

we have formula, az a'+2 (WxD”)+ W8Wwx) 12 V nota 12 velocity in (a) so, whene, ã a acceleration in inertial froame. 해 tating froame :: Ter = r And position ññ ară (in both froame they came) so, Force body in inertial froam is ΣΕ Ôi a +2 (wx ūv) + xſ xn") =) EF -2mw xr nWx (w x ñ ame Force acting on bod =m a m m TEL w mo (Wxwo)

3> Let P be a point on the surface of the earth at a latitude 1 and (so) at a colatitude o = 90° - 1. Let a cartesian coordinate system XY Z have its origin rigidly fixed to the earth at P and the Z-axis be taken vertically upward at P, i.e., radially outward from the centre o of the earth (Fig. 4.4). The xy-plane is horizontal containing P, the positive X-axis pointing east and Y- axis towards north; the earth rotates from west to east. The angular velocity vector w is in yz-plane directed parallel to the polar axis NS about which the earth rotates. Thus, W has no component east-west, i.e., along the X-axis. : W = writwy i t w zh (: W x = 0) →X East = w cos lŷ + wsin tî Let a particle be projected from P horizontally Equator with a velocity ū. Then, we have I = Vai tvy i tuzh = ii týị tz Fig. 4.4 (: v= = = = 0) (º=0) Z N: kدند + ژوند : Р S Polar axis ci

(6) We have given a= 50° (Horth) earth angulars velocity, w = 21 rad/s 24 X 3600 => W - W cos Coriolis force, E-2 50° ſ + w sina air plane flying along se aseis (East) =) = 750 km/or î FC =- 2m (W xe) " Coriolis acceleration 2 ( W x D) -2 (wcosa ŷ to +wsina a) x +2vw (cosa în sinai) : Horizontal component, (āc), - -2vw sina ſ in m/s unit =) ă * ( 1 ) = -2x750X163 277 a х sin 50° 24x3600 3600 lâc) H =-0.023 Î m/s Vertical component, lãc), 2vw cosa = [(ac), 0.0195 ê m/s

(e) To compen sate for the horizontal component, we need fonce component along, ty aris. Let the tilt is o with any plane. Fe be the lift fonce in to its wing. So, now, Fr will make o angle with zazri's Component along +yanis is, - Fisino Fesino = 115ch H I SO Z & Y => Fe x 5 o = sin- (ac) H Fi =mw2R (2) centrifugal boce, (magnitude) Conio lis fonce (magnitude) = m (ac) + (ae), i. Ratio mlacl Mac) H2 + (ae) H mw2 R W2R H V 243C00)+6370x303 = 1.12 M (0.023)2+(0.0195) :{Ratio = 1.12