hw2 - PROBLEM 31 1.153 A‘satellite will travel-...

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Unformatted text preview: PROBLEM 31 1.153 A‘satellite will travel- indefinitely in a circular orbit around a planet if the normal componentxof thexacce’leration of the satellite is equal to g(R / r’)2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the) planet, and r iswthe distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity atvits surface is 274 m/sz, determine the radius of the orbit of the indicatedflamt assuming that the orbit is circular. Earth: (anew)orbit = 107 Mm/h. SOLUTION , For the sun, ; _ L g 274 I'D/82, / and ' R = —1—-D =‘[l]r(1.39 x 109) = 0.695 x109 iii I 2 j 2 gRZ 2 Given that an = T and that for a circular orbitta-"s ; r v2 Eliminating an and solving for r, U ‘1 / ; r r = For the planet Earth, ~ ‘ , v = 107 x 106mlh = 29.72 x103 m/s 4 : ‘- 1 y . i _ 2 , ‘ , / 5 (274)! 0.695% x: 109! Then, r - = 149.8 x 109 m r = 4 » - (29.72)2 PROBLEM 1 1 :1 62 The oscillation of rod 0A about 0 is defined by the relation 0 = (4/ 7r)(sin m‘), where 0 and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from 0 is r = 10/ (t '+ 6), where r and t are expressed in mm and seconds, respectively. When t'='1 s, determine (a) the velocity of the collar, (b) the total acceleration of the collar, (c) the acceleration of the collar relative to the rod. SOLUTION Differentiate the expressions for r and 0 with respect to time. 10 . . 10 20 ’ =——mm, r=- 2mm/s, f: 3mm/52 1+6 0+6) 0+6) 9 = iSinflll‘ad, 9 = 4cos ntrad/s {9 = 47min grad/52 7r Att=1s, , r=flmm; Him, ham/.2 7 49 343» 0:0, 9=—4rad/s, 9:0 (a) VBIOCity 0f the collar. /v:= r' = 0.204 mm/s, v, = ré: —5.7lmm/sr ( \ x - (5.71 mm/s)e,9 4 (b) Acceleration of the collar. ‘ 343 7 09 = ré + 23.9 = (0) + (2)[—%) (-4) = 1.633 mm/s2 a3 = —(22.8 mm/s2)e, + (1.633 mm/s2)e9 4 (c) Acceleration of the collar relative to the rod. aB/OA = Fe, = ———e , a3,0A = (0.0583 mm/s)e, 4 PROBLEM 11.165 The motion of particle P on the parabolic path shown is defined by the equations r = 2t\/1+ 4t2 and 0 = tan—12!, where r is expressed in meters, 0‘ in_ radians, and t in seconds. Determine the velocity and acceleration of the particle when (a) i = 0, (b) t = 0.5 s. SOLUTION Differentiate the expressions for r and 0 with respect to time. r = 2t\/1+ 4t2, r' = 2 1+ 4:2 + 8t2(1+ 4t2)_1/2 —3/2 'r' = 24t(l + 4?)”2 — 32:3(1+ 41‘2) , 0 = arctan2t 9 = 2(1+ 4t2)_1, 0 = ‘—-16t(1+ 4:2)—2 r = 0, r‘ = 2m/s, 0 = 0, 9 = 2rad/s, v =f=2m/s, vg=r6l=0, v=(2m/s)e,4 a,=F—r92=0, a9=ré+2fé=8m/sz, a=(8m/sz)e94 (b) Att=0.55, r=J§m, i=3J§m/s, i=5JEm/s2 0 = grad, t9 = lrad/s, 3? = —2 rad/s2 v, = r“ = 4.24 m/s, v6, = ré: 1.414 m/s -v = (4.24 m/s)e, + (1.414 m/s)e9 4 a, = 'r' — r65? = 5J5 — \/§(1)2 = 5.66m/s2 a0 = r'é+ 2r'é= JE(—2) + (2)(3J§)(1) =' 5.66 m/s2 a = (5.66 m/sz)e, + (5.66 m/sz)eg 4 PROBLEM 12.6 A 0.1'-kg1model rocket is launched vertically from rest at time t = 0 with a constant thrust of 10 N for one second and no thrust for t > 1 s. Neglecting air resistance and the decrease in mass of the rocket, determine (a) the maximum height h reached by the rocket, (b) the time required tb'reacliefliis height. SOLUTION Forthethrustphase, +12F = ma: F} — w = F; — mg = may 01:51—g=—(1)31~9.81=90.19m/s2 m Att=ls, v =y at =~(90.19)(1~) = 90.19m/s y = %at2 = —;—(90.19)(1)2 = 45.095 m For the free flight phase, t > Is. a = —g = —9.81 m/s2 v =~v1 + a(t — 1) = 90.19 + (—9.81)(t — 1) Atv=0, t—l=-90'—19=9.194s, t=10.19s 9.81 v2 - V12 = My - y1)= -2g(y - yl) _v2 — v12 _0 — (90.19)2 2g _ (2)(9.81) (a) ymax = h = 414.5’89‘+ 45.095 h = 460 m 4 y _ yl = = 414.589 m (b) As already determined, t = 10.19 s 4 SOLUTION PROBLEM 12.18 The systemishown is initially at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys, determine (a) the acceleration of each block, (b) the tension in each cable. Let y be positive downward position for all blocks. Constraint of cable attached to mass A: y A + 3 yB = constant aA+3aB =0 or aA =-3aB Constraint of cable attached to mass C: yC + yB = constant '1’. ma For each block +l 2F = ma: BlockA: mAg —TA = mAaA, __ ’3 aC+aB=0 or aC=—aB 3TA T: Te. l We man, mg} w:‘ or TA = mg — mAaA := mg —3mAaB Block C: mcg — TC = mCaC, or TC = ng — mCaC = mcg - mCaB mag — 3(mAg — 3mAaB) - (meg _ mCaB) = mBaB (a) Accelerations. (b) Tensions. TA = (10)(9.81)—(10)(2.26) TC 01‘ a3: a3 = —0.75462 m/s2 aA = —(3)(—0.75462) = 2.26 m/s2 ac = —(—o.75462) = 0.755 m/s2 (10)(9.81) — (10)(0.755) mB —3mA —ng = m3 +.9mA+mC 30+90+10. a3 = 0.755 m/SZT 4 aA = 2.26 xxx/5214 ac = 0.755 m/s214 TA = 75.5 N 4 TC = 90.6N 4 PROBLEM 12.43 ’,,;-+__+—. An airline pilot climbs to a new flight level along the path shown. The , motion of the airplane between A‘ and B is defined by the relation ‘ a '8" . , ' s = t(180 —‘ t) , where s is the arc length in meters, t is the time in seconds, “ r? and t = 0 when the airplane is at point A. Determine the force exerted by his seat on a 75-kg passenger (a) just after the airplane passes point A, (b) just before the airplane reaches point B. SOLUTION 8°7r ' Length ofarcAB. W = p0,“, = (6000)1 = 837.76m 00 Ejvdv = EABa, ds 2 2 v—3 — v?! = 2a,sAB or v; = v}, + ZatsAB = 1802 + (2)(—2)(837.76) = 29049 mZ/SZ, VB = 170.44 m/s may ' v V / For passenger, m : _75 kg, W ; mg = (75)(9.81) = 735.75 N "11’2 man +\)3F = Man: Woosg — N = 7 mv2 N = WcosB — — (1) p +/EF = mat: P— Wsin0 = ma, 1 P = Wsin0 + ma, (2) (a) Just after pointA, t = 0, v = 180 m/s, 6 = 8° 2 From Eq. (1), N = 735.750058" _ (Jim = 323.6 N 000 From Eq. (2), P = 735.75sin8° + (75)(—2) = —47.6 NA ' F = N2 +102 = 327 N, tanfl = % = 6.798, ,6 = 81.6° fl — 8° = 73.6° F = 327 N 3A73.6° 4 PROBLEM 12.46 During a high-speed chase, an 1100-kg sports car traveling at a speed of 160 km/h just loses contact with the road as it reaches the crest A of a hill. (a) Determine the radius of curvature p of the vertical profile of the road at A. (b) Usiné the value of p found in part a, determine the force exerted on a 70-kg driver by the seat of his 1400-kg car as the car, traveling at a constant speed of 80 km/h, passes through A. SOLUTION (a) v = 160km/h = 44.44 m/s Wheels do not touch the road. _ _ , Mat g i ZFy = —man: —mg = —mv2/p "‘3' ma” 2 2 ' p = V— : —(44'44) = 201.4 g 9.81 p=201m4 (b) v = 80km/h = 22.22 m/s r m = 70 kg for passenger ...
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hw2 - PROBLEM 31 1.153 A‘satellite will travel-...

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