Problem Set 2 Solution

Conservative forcedisplacementsiststa non 1 x10 and 2

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Unformatted text preview: Equation (2.105), and has the same c requirement2that m2 m1t, 2 for the0equilibrium to exist. When the system is in motion, the when v ! c , we have ! ! .55 year 10 ined from the force laws: 3 descriptive equations are obta t 99c m2 s when v = 99% c, we have t ! ! 6.67 year& b $ x1 ' m1 x (x 2 $ g) (6) !! 10 199( !!1 $ g ) ! 4 x2 2-53. To examine stability, let us expand the coordinates about their equilibrium values and look at 2 is abehavior for small when there ex Le 1 . xsingular potential function h ( calculations, tF-51.conservative forcedisplacements.iststa-non-1 $ x10 and -2 . x2 $ x20 . In tUex) satisfying heir F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations take terms in -1 2and -2 , and tb heir time derivatives, only up to first order. Equation (3) then mv0 dv dv a) m ! " bv # % ! " % dt # v(t) ! )terms of y becomes -2 " $(m1 m2v 2 1 . When written inbt/Fx$ m Fthese new coordinates, the equation of dt m v0 & / /y /x motion becomes 999m ! 138.7 hou3 s . r2 Nowso on. t) = v0/1000 , one finds t ! 2 2 and let v( v0 b g 4 m1 $ m2 !! ! $ -1 (7) a) In this case all relations above ar1e satisfied, &sm1F is indeed a conservative force. 4 m1m2 o / m2 ' d v & Fx & # ' bx 2 /U & ayz ) bx ) c % U & # ayzx # # cx ) f1 ( y , z) 2 /x (1) where f1 ( y , z) is a function of only y and z Fy & # /U & axz ) bz % U & # ayzx # byz ) f 2 ( x , z) /y t b) t x(t ) ! % vdt ! 0 m & btv0 $ m ' ln ( ) b* m+ We use the value of t found in question a) to find the corresponding distance (2) 77 NEWTONIAN MECHANICS—SINGLE PARTICLE where f 2 ( x , z) is a function of only x and z Fz " % !U " axy # by # c $ U " % ayzx % byz # f 3 ( y , z) !z (3) then from (1), (2), (3) we find that U " % axyz % byz % cx 2 % bx 2 #C 2 where C is a arbitrary constant. b) is Using the same method we find that F in this case is a conservative force, and its potential U " % z exp( % x) % y ln z # C c) Using the same method we find that F in this case is a conservative force, and its potential is ( using the result of problem 1-31b): U " % a ln r 2-54. a) Terminal velocity means final steady velocity (here we assume that the potato reaches this velocity before the impact with the Earth) when the total force acting on the potato is zero. mg = kmv and consequently v " g k " 1000 m/s . b) x 0 dv F dx dv vdv " " %( g # kv) $ " dt " % $ dx " % & $ dt m v g # kv & g # kv 0 v0 xmax " 2-55. g v0 g # 2 ln " 679.7 m kk g # kv0 where v0 is the initial velocity of the potato. Let’s denote vx 0 and vy 0 the initial horizontal and vertical velocity of the pumpkin. Evidently, vx 0 " vy 0 in this problem. m v x 0 % v xf dvx dv dx " Fx " % mkvx $ % " % dt " x $ x f " dt vx kvx k (1) where the suffix f always denote the final value. From the second equality of (1), we have % dt " dvx % kt $ vxf " vx 0 e f kvx (2) vx 0 ' % kt 1% e f ( k (3) Combining (1) and (2) we have xf "...
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