Exam1 - EM311M - Dynamics gums-OHS Exam 1 Wednesday, Sep...

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Unformatted text preview: EM311M - Dynamics gums-OHS Exam 1 Wednesday, Sep 23, 2009, BEL 328, 6:00-8:00 pm 1. A particle is moving in a straight line with a an acceleration a(t) = 005 — 2)2, where c is a positive constant. At the time to = 2, the corresponding position and velocity are :30 and v0, respectively. Derive the formulas for position a:(t) and velocity v(t) at any time t (5 points) .5 ' 2, . Q=V: C(‘f—Z) / jagg— , i' c 21‘ 21 W“/ =‘ 70-3) v ‘ AI / + z 6 o C 0'3 / L w #va _ —§-/7’ 7;, 1‘ ~> 446/: 70+ 34—4] / v, re / .7 - t v 6) t c 6' Kfli r/Vof‘!” f i 2. J , _C_/ '-’/ x01~ x9 : \’o@'&j -' /2. '{—4/ 2. At point r = 2m, 0 = 30°, the cylindrical components of the acceleration vector of a particle at the point are a, = 2, a0 = —1 [m/sz]. Calculate the Cartesian components a1, (1,, of the acceleration vector. (5 points) ol ' ° r‘ //‘O f—(‘Cp ‘51:] A 30‘ ax=arcgg v ‘V,, .fll2nar ,.. m? =/g-/—§+/-é :4253[5—;/ Z L C O We” , a' 30"+ a Ca 30° J a .— ar Jl[( a g __.H__J ,7 r > (7 7 r. m 7 5.21% + [w] 13 *WMA‘ 54/ 3. Derive the Henét formulas for the velocity and acceleration vectors. (5 points) A A," 4:” A: > V: am ———- :AM -——- ~— - A . 4.5 At? 5 V/ I N who Alf 4“”) £5711) {$wa = 5 3?? = 5 Et- /' W“ :7 ’1‘] V. $2 + .5" 5f ’ 49'1- "‘6" 9e” -— «13' ‘3 -3 4465 f ‘2 on 5 c ‘53-!— +‘Tgn. W W 4. Derive the formulas for acceleration vector components of, a0, a; in the cylindrical system of coordinates (5 points) J 2\ 6% #16 e 3;: Ar' 1" {Lae- n r 4 de a ' 3: If: Ffr' + r+r6 7L Ede-8' “‘9 69 a ‘ ' szg/‘f' +£§f .19. v7” 9' V 04 a as . f e‘. a r g-0_0 6. 62,3 r rgr +r/§Ea+ rare +"6’5a rewefizqe w a. .1! no . 0 no = (r—f‘e Jér + W 4% figs §f=(m9/ flue/O) LL d6!“ ' .__ e 7 dfe- r ~ _ -—S’1:6 0 -= —e 06 s “76/ o J “P 5. A particle moves along a parabola y = 11:2 with a constant speed 1). Determine the Cartesian components of the velocity vector as a function of coordinate x. (5 points) 27-: Xz' Coasifircu‘nco/ I’mo 74'th , r0 6/Cét/g (6;: xxx Vz=;z+ 31, 2 5%,. 4X99 _— )<°<’(/+¢x"-) X: g i. __V__ ,, F + uv @ 6. If the pitching wedge the golfer is using gives the ball an initial angle 00 = 50", what range of velocities vo will cause the ball to land within 3 ft of the hole (Assume that the hole lies in the plane of the ball’s trajectory.) (25 points) CO H .' 6cm [LE/3 a, fan‘uf 3/; «air/(f 74; 44C I’m/c, . r 3 573‘? X= Vow 5-0 "I "- 53 => '1‘: foajf’o' Vfi‘ v v 2 W v067«50t_ 33.2 :— _ 5 34) {1 V0 V/VVO _ /6n 2 3 4Z?36,3 ‘ = 36. 326 V04 - /— Va = 34. I? A 1?- ‘7 QC tifVLA./)ln( SfQCC/ hutde [t ’(‘6‘ \/6C 7261174"; if 3/. 2 5 Va s 542/? ‘5 7. The collar A slides on the circular bar. The radial position of A (in meters) is given as a function of H by r = 2cos0. At the instant shown, 0 = 25°,0 = 4[rad/s] and 6 = 0. Determine the velocity and acceleration of A in terms of polar coordinates. (25 points) 7“=.ZC(739' = 4/— d"; =— ~29“? = wee/a— [7”] z __’a(_ 53;"; :—z;an.9¢—-/,?/s;-,—_/ Vr:/:.— figs; = —0.:?é5‘-4 =~3.3<?[T » a “4. V9: mg _— MWS-f = 2.252, [T .0 :6 I are r—rr-Q Z ,0 u all: 4 4r- « fl 4.: c g: - W’w-‘Jc?(fej" wfiba+o¢9 = 4.9/3. /6 — “M [57 ~~ w —~ @ are r-rQ = ~23 fl/gP/g /g = ~51? 5; 0 Kc. "‘ '71. . ' :r “3.33) 4‘ 46- ye +zre z [ J 8. The car increases its speed at a constant rate from 40 mi/h at A to 60 mi/h at B. Determine the magnitude of its acceleration when it has traveled along the road a distance 120 ft from A (1 mi = 5280 ft). (25 points) 57"” =52.67[3s£f7 *0 3goo : ax [:f-i/ ll _. -v—/ I : 2/50.? J5: a6 = 7.3/5. 52;] no V 4(1 j o 5‘. ,, fi' 1/: 7217‘6 5 357.? a me /O , a = 13’: Id: __ 44.5/5‘ [7?] O " f lice . 7 76/17 _ Z .Z, =- l a; quz+af7 _ ¢Z£f5 {-44.3/5‘ 45 [12' 5 ...
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Exam1 - EM311M - Dynamics gums-OHS Exam 1 Wednesday, Sep...

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