Exam I 2006 (Part 2)

# Exam I 2006 (Part 2) - ES 223 — Rigid Body Dynamics Exam...

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Unformatted text preview: ES 223 — Rigid Body Dynamics Exam II . March 1, 2006 NAME 3 ol M ‘i‘! W Closed book and closed notes. Show all work and enclose your ﬁnal answer with a box. Problems will be graded for nartial credit only if the work is neat and all intermediate steps are clearly shown and presented in an my to follow format. If you 8221! Newton 's Second Law or Imulse in a solution, you must show a FBD. All solutions must show a coordinate system. Equations: Uniformly accelerated rectilinear motion: Acceleration due to gravity: _ 2 v=vo+at g—9.81m/s 2 1 g = 32.2 ﬁ/sec s = so + vot + ——at2 2 v2 = v02 + 2a(s — so) Normal/tangential (path) coordinates: K = Vét v2 A O A (for circular motion) g = _ n + vet .0 v = r6 2 V ' 2 ' v2 . . an=—=pﬂ =vﬂ an:—=r02=v6 P r at=13=§ at=v=ré a 2 an 2 + a,2 Polar coordinates: Quadratic equation: ax2+bx+c=0 z—biVbZ—4ac g = fé, +ré ée g = (f—réz)é, + (r§+ 2ié)é9 x 2a (for circular motion) Vr = 0 Power: v0 = r0 P :El, a,=—rl92 e :Poutput a0 = r0 m Rnput Work-energy: Tl+Ul—>2=T2 T1+Vgl+Vﬁ+U,'_,2=T2+Vh+V¢1 U =jE-d5 T=-mv2 2 _1 2 2 . U192 “ -'2—k(x2 " x1) (Sprlng) Vg=mgh V¢=-2— F = kx (Spring) Impulse-momentum: kx2 Coefficient of restitution: e _ (vs), —(v;),. ' (v1), —(v2),, Selected integgals: j “5‘ =11n(a+bx) a+bx b Isin(a + bx)dx = —— —ll;cos(a + bx) j cos(a + bx)dx = —ll;sin(a + bx) 1. (25) The slotted arm revolves in the horizontal plane about the ﬁxed vertical axis through point 0. The 3-lb slider C is drawn toward 0 at the constant rate of 2 in./ sec by pulling the cord S. At the instant for whichr=9in.,thearmhasacounterclockwisean- m5; ;_ a, gular velocity to = 6 rad/sec and is slowing down at z. Z_ the rate of 2 rad/secz. For this instant, determine the tension T in the cord and the magnitude N of the force exerted on the slider by the sides of the smooth : ‘ :3; 7‘ —- 0" b 7 Se ‘* radial slot. Indicate which side, A or B, of the slot I7- ' 6 contacts the slider. (— /’7 r 9 ‘ It. an - m = m('r’-ré"> = auto-MW m => k |dex+aut {S on suﬁacc ZN: 0.3746le) 2. (5) State in words, Newton’s First Law. a Pm’i tole— re MAI n5 4+ hes-f 0t“ Cm+rﬂ M e: '1’“ move to!“ cum Firm felon ¢7 14 Ware \5 no Unbalanced Jone Qal'mg o“ ‘04, 3 . 30 . 3 ( ) Calculate the mgnitude of the velocity of the ball at Pomt B The 3-lb ball is given an initial velocity 11A = 8 ft/ sec in the vertical plane at position A, where the two horizontal attached springs are unstretched. ;- _ The ball follows the dashed path shown and crosses 1 a q point B, which is 5 in. directly below A. Calculate m 3 2.1—" O to 3 l S l the velocity 03 of the ball at B. Each spring has a stiffness of 10 lb/in. 3 o l Tﬁch-H/e' +11%}: T¢+V%1_+Vez \ I /"\ . UA / l .t / I L Dawn“ ed, A a; \{a‘zo '9 f‘x‘xf/éﬂ} 4 L ’ in‘ws‘“ = l3 33% ,L T.=~ i(0'0437-)..C8) = mm 4+. \b Ve‘=\$‘ll><z= 0 because x =0 1 T1; Mower-Mug)“: 0.0M 1175 L» m 7. Val: -Wau: — 3 (57,1): _\.15- 4M5 VeL‘QihYZ = (‘03 03413711): 0.8333 41H}: 30> 2.5182. +0 +5 = 0.04%U’31- [.15 +0.8333 “UB1”: 72H} 7&3 \$.54 4th“ 4. (5) The equation I; + Vg1 +Vel = T2 + V82 + Val is an expression of the Law of Conservation of 0‘%DQM\CQ ‘ Eh e g g a . (Two words) P ‘L 4 5. (30) The steel ball strikes the heavy steel plate with a velocity v0 = 24 m/s at an angle of 60° with the horizontal. If the coefﬁcient of restitution is e = 0.8, compute the velocity v and its direction 0 with which the ball rebounds from the plate. [Ill/1L: a L, C O b : M l s \ An 1; ' \ 1 / I -_ :_ M ®§ ‘ F // v7 1);”- an. smloo am? /s I)? A / “ “st E:(v'z’)n~(v7')q=o—éfl_a’) v CU—kaﬂn ——20.78-'0 0.8 ‘- 03') l - “J .- ’1): ‘ a Z’ amx 9 (‘ )n “Ob lot sphere. Q I mllvtlt; FA‘QHX; => (vilifkuﬁé : ‘7- “LS w ' U.‘:\l whisk? => 115.3 QO'b'Mlsl v: 9 , -\ nun. r2- 9~l15lh (-71“ \e=5L\.L°l 6. (5) Observe the Newton's Pendulum. The steel spheres all have the same mass and the coefficient of restitution of steel on steel is about 1. When one sphere comes in with a velocity v and impacts the remaining four, one comes out the other side with the same velocity v. We know that momentum of the system must be conserved [(mv)in = (mv)out] because there are no external impulsive forces. But momentum would still be conserved if two spheres came outwith half the velocity. Why does this not happen? For ea] enema? :5 01/50 Cam/159094. so Cmdl¢lm I: Cmvlmtwvlom“ J-"W-LBW =6 “\V-Lbou‘l'. Two Spheres canning 9 0u+ q'l- 3% does no?“ \$44419 energy Cmservof/m, ...
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## This note was uploaded on 04/25/2010 for the course ES 9111 taught by Professor Morrison during the Spring '07 term at Clarkson University .

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Exam I 2006 (Part 2) - ES 223 — Rigid Body Dynamics Exam...

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