Dynamics Old Test 1

Dynamics Old Test 1 - Samur- L00 5 EGM 3400/3401 Exam 1...

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Unformatted text preview: Samur- L00 5 EGM 3400/3401 Exam 1 Name KG, ‘ ' Work any four of the five problems. Only four problems will be graded. Indicate whifitiproblem you do not want graded. If you do not indicate a problem, you will be graded on the first four prob ms. Each problem is ‘7‘ Worth 25 points. 1. BlockA moves down with a constant velocity of 3 m/s. Determine the velocity of Block C. ( M SAMYRJlA l“ DRvaw‘nj ‘ ) detain is fflsllwe) EGM 3400/3401 Exam 1 Name lg 2. A projectile is launched from point A with an initial velocity v0 of 40 m/s at an ang of 30° with the vertical. ‘ADetermine the radius of curvature of the trajectory described by the projectile at the point on the trajectory where the velocity is parallel to the incline. A 1 Vx: ngBOO \/ : V 5600306 l ML- M4230" V3 M14300 V} ’w \/x M1300 0030” Vxlwkfiivi: VOW3OO ’L Vt : Vvfiflf’ @3300 '2’- I 7\ ‘L '1 p Z. r't b V‘vx “’VE :VoM 30 ‘lVg 641150 4360 Z 1. , a ’2 c; '7‘? o sw‘z—/ ': ‘, 24/ A V 2 Voiwzw (1+ Wgflo = 40%“ *0 (H @530” 5333331 a 1 M — v2 - 53.5 Mt/Z an- L > fl- i:_ 5 5 f 9'4957 /57- EGM 3400/3401 Exam 1 Name K41“? 3. A spring AB of constant k is attached to a support A and to a collar of mass m. The unstretched length of the V‘spring is 1. Knowing that the collar is released from rest at x = x0 and neglecting friction between the collar and the horizontal rod, determine the magnitude of the velocity of the collar as it passes through point C. xo=l.2m m=5kg k=lON/m l=2m EGM 3400/3401 Exam 1 Name % 4. A SO-lb block is attached to spring A and connected to spring B by a cord and pulley. The block is held in "the position shown with both springs unstretched when the support is removed and the block is released with no initial. Knowing that the constant of each spring is 20 lb/in., determine the maximum velocity achieved by the block. X3 = Xisplmmw‘l 'th 245/ H U KAI 7x a». H A X5 T (Ky—KAI) - WS‘lfiwlY’ X3 C lxa/h Xfi( : WS‘l‘GA/L+ Cl\ KA’ 1(Xg/+)(B) (KAN) W9lml' (2) it u k =- 201b/in T _ x A X5 - A/l _ ,. o Lisp-7L \Z" l '1 Z V = ~1<‘E%>(32,2fi/jz> n 2 2‘ r ‘ \UXA ‘ i; KM ’JiKng "' 1i “V Magg’~ 5(“3‘3’31 “'X 374;”? K m .. 1 , 4+ 30 V2; «waggmzfi’l‘éCxt/l) 1 9f K m m gV1:!0.7655‘°§f9*3"6742—: L v l\2X‘ ‘E’ 2’ x42 :l m 0‘4 +17 'vai 5—5664” {lg/5 l—_______ 1 ' K Z ’ v :‘Jx’é’fig‘g‘ [N1 2'5'7‘0l/5 3‘ V! itoi~26v5mt£i l xt m ~r ‘ ; IK 5 _ 7a?“ ‘ a _ ~9lwx A e A X- ‘ \1 i IV; .— “‘2' %~ 573 5‘ my «Luv/aw: EGM 3400/3401 Exam 1 Name % 5. The acceleration of point A is defined by the relation a = -1.8 sin kt, where a and t are expressed in m/s2 and Aseconds, respectively, and k = 3 rad/s. Knowing that x = O and v = 0.6 m/s when t = 0, determine the velocity of pointA when t = 0.5 s. duel; dt alv: 0.let M 9/ A” 5/ A IQ? = 'G FQ C20 V2 W caflit Wat K, aft“); ,(rgsxigqo __________ _. 3/5 V- .041?! “is ...
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This note was uploaded on 05/17/2011 for the course EGM 3400 taught by Professor Matthews during the Summer '08 term at University of Florida.

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Dynamics Old Test 1 - Samur- L00 5 EGM 3400/3401 Exam 1...

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