Question

A small spherical ball of density p and radius a is released from rest and falls under gravity in a viscous liquid of density pf and viscosity u. Its vertical position is denoted by z(t), and we choose the origin such that z(0) = 0. The ball is subjected to two forces: its weight corrected for buoyancy F, = (p-p Vg, where g is the acceleration of gravity pointing in the - z direction and V = 47a/3 is the ball volume, and a viscous drag force Fa = -6pav, where v is the velocity vector. (a) By applying Newton's second law F,+Fd = ma (where a = dv/dt), show that the vertical velocity component vz satisfies the ODE: m2 + 6 pav, = -map-pg. dit (b) Solve for the velocity vz(t), for a ball starting at rest (v (0) = 0). (c) What is the value of the terminal velocity as t oo? (d) How long does it take for ball to reach 50% of its terminal velocity? (e) Solve for the vertical position z(t) of the ball.

Answer 1

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Soline : Given, nosily of spherical ball - $ Radius of Spherical ball = a Density of viscous liquid = Bp Viscosity of liquid = -. 1 FG av Weight of -lhe spherical ball acting vertically downwarel, W = mass x acrelevation due to gravity acceleration due to gravity = (Densityx volume - svg Bouyancy force acting on the spherical ball vertically upward, Fe = weight of tbe liquid displaud by spherical ball = Density of liquid x volume of liquid displaced by bill ng = St Vg { vol of liquid diaplecach - vol. of ball} Drag -force acting on the spherical ball verilically upusavel, Fa = 6th a V, where vace vertical velocity of the topherical ball

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Then, by Newton's Ind Law, Fret = ma =) & W- FB - Fd = ma ... -> sug- &;vg - Grupa v.ce) = m dixit {:; a- dvies} > maxime + TMAV (4) = (3-S4) vg {v= vol. of spherical ball e ta} => mdva + 60 Mavy (t) = femme de Ta? (8-944. (DE) b) We have, mdvz + 617 Mava(t) = geta(8-8)9 => dret (sipa) 4(e) = 45a708-) – bet t o . Then the above out on be written as, m 3m Defa) 47 - 449%84) Equation (1) is of the form dy + px=Q , where P and Q are dx and p = 4Ta’ (8-3679. It's solution is : 4x 1F = S QxlF dx + c 3m fp.dat Now, Integrating Factor, 1.F= sorradt 671 Therefore, the solution of equation (1) is given by Vz(1)x1.5 = 5 Qx1.5 dt + C , -> vz(t)x e setove - | 4ta? (8-!)3 xe mat. dt +. GT 3m

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Зм Gпна - 40). е . 4та? (-4) » М. etc. әч,0)ve“га*(-1)сес — (0) . Given, V2C) = 0 Як аф .., Gn (1) =) он га* (*-*) с - С= - 24* (-?) як се, чn (2) =) ЧА Як о» * , ал' (-4). * эл" (-4), од гр* - 244. [cape - ) | -) Як c) Terminal velocity is the maximum, velocity attained by a spherical ball dropped in a viscous líquid. It occurs when the weight of the spherical ball (W =svg) is balanced by the sum of the buoyancy -очи ( Fға = feva) and u draa -ivu (Gтка 4 ) . t, w = Fe + F в а , => sva - frvз + вткаў, =(-) V3 = 6трavz - V; - (- $4)vg - (-4) 4та?3 - 2 (9-р ) a®y GTI да 6T Ma

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Thore fore, terminal velocity is, V = 2 (8-S4) as d) 50% of terminal velocity is, Vy = 0.5x 2(8-54 Dag que = (3-84) ág 9A Substituting 'n egn (3), we have (8=lf Das e S* = 241-999 le compt-1] 91 Get a Qe Santana - a both sides, Taking logarithm on both sides, 611M97 - en 2 min 2 This is the time taken for the ball to reach 50% of terminal velocity 8) From equation (3), vece) x e Supt = va’(8-8)9 be cukat -17 | >> V2C) = dá? (8-5429 [1-e cuma ) 9M - qr We know, Vz(t) = dz(t)

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Therefore, dzce) - 24°(8–84)9 [l-economat] de qu Integrating worlt t on both sides, GHO 1. qu za) = Ra°(0-4)9 [+ + **+] + c = 6). To find c, use the given initial condition z(o)=0. Therefore, ean (4) => 0 = Qa°(8=84)9 [0+ *.] +c. qu GTima =) 0 = dl ma(3-84)9 - 27TM² + C : 1. c = -ma (P-9f 9 27TM² oo, ean (4) =) -611 Mat 7t 24) = x2 (922 (e m e sentit ) - ma (-3). t + GIMA 27 M2.

Therefore, dzce) - 24°(8–84)9 [l-economat] de qu Integrating worlt t on both sides, GHO 1. qu za) = Ra°(0-4)9 [+ + **+] + c = 6). To find c, use the given initial condition z(o)=0. Therefore, ean (4) => 0 = Qa°(8=84)9 [0+ *.] +c. qu GTima =) 0 = dl ma(3-84)9 - 27TM² + C : 1. c = -ma (P-9f 9 27TM² oo, ean (4) =) -611 Mat 7t 24) = x2 (922 (e m e sentit ) - ma (-3). t + GIMA 27 M2.