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Unformatted text preview: ES 2120 Name: ﬁouoﬂolq Cartesian Coordinates F:xf+ﬁ+zk
6 A. [A d}7
Vxl—lVyj'lVZC: d1 o A. A d? "5 no. "A
:axl+ayj+azk=—dZ—:xz +yj+zk =xf+yj+2k <l
ll Q! Normal & Tangential Motion Radius of curvature, p MW
m Exam 1 Equation Sheet Section: WC” Polar Coordinates (2D) rmr vrur +12qu 2 rer + r6»,
a = arur + czqu
. .. ‘2 dV,
w1th a, = r ~r6 =1), 7 ~rw2
r a9 :ré+2f9=ra+2fw Equivalent Acceleration Expressions dv dv dv
a 2— :V— :\)———
dt dx ds
da) da)
a :M : a)—
dt d0 Uniform Accelerated Motion Expressions ,3”: '0+vot+—~at ; v=vo+at; v2:v02+2a@—_§0) 1
6:60 +w0t+—2—at2; a)sz +051; 602 26002 +2a(c9—60) Relative Position, Velocity, Acceleration
Expressions for Planar Motion 973 = 97A + SEE/A ‘78 25A +178/A WlthGB/A :CTJXFB/A — .‘ “ ' ‘ _“ 7 2:
aB—aA+aB/Aw1thaB/AuaxIB/A—a)IB/A ES 2120 Exam 1 February 15, 2010 1) Short answer questions 5point each.
a) Determine the maximum speed the jeep can
travel over the crest of the hill and not lose
contact with the road (i.e., notjump). I memmijws it a, a
Wm remit (mm Wt» 3?" b) A ball is thrown vertically upward with a speed of 15 m/s.
Determine the time of flight when it returns to its original position. g “:3 36 «t» w; {5; «t» c) The boat is traveling along a circular path with a
a speed v: (0.0625?) m/s, where tis in M" seconds. Determine the magnitude of its acceleration when t = 10 seconds. \3
t 4 K "1:0 m » 3: 00 g +3». \\\ i
<9 4:: m to Cit tw f
, 22m t, “W?”
Va . {0:23 r’” @ij r W /€ 2: (O 0 ts M 49 E 3 i tie? "t: to a a ltl' * d) The car has a speed of 55 ft/s. Determine the
angular velocity 9 of the radial line CA at this instant. Z
N Kiwi "m W m w W I” ﬂ
0‘“ z/é ”” (" q%too M :rznf'JQQJE (5%;
m H! (I (in w (A r" ’3 m ‘(w “l” Tm ﬂ a 2* V m “ﬂ "”.Mi i
“team x Q “t we £9 t " t 20 points a _
t} v a: e ES 2120 Exam 1 February 15, 2010 2) After deploying its drag parachute, an airplane has a deceleration of a = O.004v2 (m/SQ).
a) Determine the time required for the velocity to decrease from 80 m/s to 10 m/s.
b) Determine the distance to plane covers during that time. Answers : t: Zlf’l sec S: Wt W
‘ ‘ mm?”
0t: mapped?“ "3 V“ “W”? [<33 2:: "MCDJV (W CW» lo 20 points ES 2120 Exam 1 February 15, 2010 3) ln slowpitch softball, the underhand pitch must reach a maximum height of between
1.8 m and 3.7 m above the ground. The pitch is made with an initial velocity v0 of
magnitude 13 m/s at an angle of 33° with the horizontal. Determine, a) the maximum height the ball reaches,
b) if the pitch meets the maximum height requirement, and
c) the height of the ball as it reaches the batter. Answers: hmax = 3.. it» m
(“Les or No
h = {34.33% m 6:53) (W. 1 éijh‘fwgggﬁ 5:3 a “it _ “a WW ” W k w gem
A M ah} 2 W egg WE 4%.. tg . 53% g a ago + “totftélﬁtﬁ ~« l/zt‘ttﬁ‘tttfiﬁf “ﬁes. hm"; QuQ§v%“ M m mmmmmﬂmm 20 points ES 2120 Exam 1 February 15, 2010 WW
4) Block A moves down withl constant velocity of 1 m/s Determine, a) the velocity of block C, wxrﬂwmuN‘WWW b) the velocity of collar 8 relative to block A, y c) the relative velocity of portion D of the cable with respect to block A, and d) the tension in the cable supporting block A; Aﬁseumwn Erma rt rttr‘tﬁt We:
(Note: Be sure to indicate the direction of velocity in your answeg‘) CONESWlMNT“ EQMEﬁ
warrants t
3% °%” {ﬁemgemé xi:
an M
2,5ﬁmsﬁm “W” mrﬁﬁ
flafwmtrw s m’”)
@ 2“an "m" ’4‘: if :2
tygmfggﬁ % m} eff Kr, w x “v MW {ﬁv 20 points Answers: vc= t/vt‘rn/g T
VB/A: \m/g «£4 t
7 W tttt New
0 coﬁﬁmm‘t‘”
\lemmrrj
t W M ﬁg feat get ﬁt”? mt: mt: WW 1 W65? #3? V"
k e it‘m/st are: ware)
‘ : Arm/{,Q/ ES 2120 Exam 1 _ February 15, 2010 5) The twodimensional motion of collar B is defined by the relations r = t3 — 2t2 and
6 = t3 — 4t, where r is expressed in m, t in seconds, and e in radians. Knowing that the collar has a mass of 0.25 kg and moves in a vertical plane (gravity is included),
determine: a) the radial acceleration, a, at 2.1 seconds,
b) the tangential acceleration, a9, at 2.1 seconds, and
c) the normal force exerted by rod CA on collar 8 att = 2.1 seconds. Answers:
“‘9 ar= mmwo m/s?”
ae= (Mu? : FOA= mos M
an
a W G? saws (52:? mm :5:
r”: l; m. Mt wrist m (3:: at as asst % sltwll: 4am m/t @« «f 619133
M W “‘1 9&0 "782 6:3; mm \Mo Wit/$2
a) an: \23 M We?» 3: 8,40% caA—tlﬂlm‘f “iﬁﬂwt 'm/st lo» dim w” «v ELQLCZB a: oaMlOWW) “‘“l‘ gmtgﬁéq‘aﬁ 0‘4“? Rho/62w
c) s Q 2,95 s swims 40133“ (f E44, W 01${ql80603 (T46???)fo a: ('9‘. 2%“; (£741.31) ﬁat} ﬁg 2 ~ MW" 1%”me
Es 3“ Zwﬁrwgg N 20 points ...
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 Spring '10
 Kobbe
 Dynamics

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