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Unformatted text preview: ES 223  Rigid Body Dynamics
Exam 11 February 23, 2005 NAMEJSECTION S ‘3 lUL‘il M Put your name on the back of the exam also. Closed book and closed notes. Show it]! work and enclose your final answer with a
box. Problems will be graded for partial credit only if the: work is neat and an
intermediate steps are clearly shown and presented in an easy to follow format. If
free body diagrams do not include a coordinate system you will receive very little
partial credit. Equations:
Uniformly accelerated rectilinear motion: Acceleration due to gavity:
d 2
1 g = 32.2 ﬁlm
3 : 50 + vet +5412
v2 2 v02 + 2a(s — so)
Normal/tangential (path) coordinates:
1’ : v5.
2 A M (for circular motion)
'2. 2 We» + vet .
P v = r6
2
v   2 . .
an:_—:pﬁ2=vﬂ anzv—zrﬁzzvﬂ
P r
a,=1>='s' a,=v:ré
a : a"2 + 0,2
Polar coordinates: Quadratic gguation: vzrer+r6iea g=(f—réz)é,+(r§+2fé)ég _ abinz—tiac x:
2a (for circular motion)
v, = 0 Power:
v3 = r9 PEE— v
a, 2 462 Paw .. em  —_
00 = r9 'Piuput Workenergy:
Tl + UH: : T2
T1 + V3,I + V“I +U"'_,2 = T2 + ng + V‘r1 U=jgd_r_ T=lmv2
2
1
Um —§k(x§—x3) (Spnng)
V =mgh V‘=—kx2 Coefﬁcient of restitution:
e : (van—(17;),
(Vl)n _ (v2)n Selected integmls: j d” =11n(a+bx)
a+bx b Isin(a+bx)dx = —%cos(a+bx) Imam + bx)d5c 2 ésinw + bx) 1. (36) The 0.6 kg slider is released from rest at A and slides down the smooth parabolic guide under the inﬂuence of its own weight and the spring. The
spring constant is 120 Nlm, and the unstretched length of the spring is 200 mm. The guide lies in the vertical plane. (.3) Determine the speed of the slider as it passes point B. The“? 0+0+T73 = 0.37fBL—a.943+0.)5' 2. (25) The four 5 pound spheres are rigidly attached to the crossbar frame having
' ' ' ' ntM = (0.5: + 0.8) lbﬁ, where t is in . The ﬂame is inmaily at rest. (a) Determine the speed of each of the spheres after 4 seconds. eloeity of the ﬂame after 4 seconds. Use (b) Determine the angular v
this part of the problem. vector notation for your answer to La) “0‘ + £000 : o + {20.51; +0.2)At : Lier
@ swims) '— Magwyv) db) “UHu) 3. (5) Stateinwords, Newton’sFirstLaw. a Pennants a‘f
Tesi of continues +0 Nove— umL unrform Velocity {9 Mama 1.: no unbm’ancﬂl farce 00:"ng on If.
4. (5) The time rate of doing work is called EPOUJ g1: . 5. (35) Two smooth billiard balls, A amrB have equal masses of m = 200 g. A stokes B with a velocity of 15; = 2 m/s as shown. Ball B is originally at
rest, and the coefﬁcient of restitution e = 0. 75. Determine the ﬁnal velocities of the balls just after the collision. Your
answers should include the magnitudes of the velocities and arrows that show the directions. Be sure to include the angles that the an'ows make
with the horizontal axis. LoafMos 40 =  I332 Ms (an :qsm Ho '  ~185mI$ ’"U‘at 0 “when '
cmsermiumbi‘ai mameniulm 6'1"
36412»: m “n" dlré'ohm x (n) mm)n+m3(w3jo : meMOM 014307631 M's are :1,“ equal 3:: 44%} cancel Thaw. I,5'37. '‘ (mega +615er ® 6:0.75': @1490 “Offs/)0 _ LUAI)A“MO” I :LI‘Hw‘Pf’
(W); 44.537.) =>WL JO Sub m+o (p: 4.531: 1.149%)" +QJGQnemﬂa—t3405m Cindi Oral)”:0.IQS (“)5 n 5
Conﬁrmlam a? mammiMM 0‘9 €49!“ Paf+'°lc In “ ’5" tithe—chm: mam): me(vn')e 3305;): “1.23599
mm); me (ml 296%” Then 'Un ‘‘ \imms"? ms 1'= I75 an ‘ 1.7.8; __ z I D
9'9" W (ﬁne) 99“ 8"; ‘—‘ ‘ ’U’H: L7, m/s 31.5] U8: 036 6"“ ...
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 Spring '07
 MORRISON
 Velocity, Trigraph, Coordinate system, Polar coordinate system, Rigid Body Dynamics

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