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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 75kg 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 highspeed chase, an 1100kg 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 70kg driver by the seat of his 1400kg 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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 Summer '06
 
 Acceleration, Velocity, m/s, normal componentxof thexacce’leration

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