Question

Rithal End - ENGR 330: Dynamics 3. The mechanism shown below consists of a homogeneous disk of radius R and a link of length L. Link BC is pinned to the disk at B and a slider at C. The disk is constrained at point A in its center and point C is constrained to move in one dimension as shown. Determine the angular velocity and acceleration of link BC, the velocity and acceleration of the piston C, and the velocity and acceleration of point B at the instant shown. 0.8 m -- 0.2 m w = 6 rad/s a = 3 rad/s?

Answer 1

Velocity of point B=(

Acceleration of point B = (0.61

Velocity of point C = (1.29 + 0.693

Acceleration of point C = 0

Angular velocity of link BC = 0.866 rad/s

Angular acceleration of link BC = 9 rad/s²

.0.8m Do2m xx w = 6 rod Positive z-axis is x=3 had pointing out of hage for right hand coordinate system • Write the important vector in cartesian form: 60 - 02 + (a) Here Bh is vector from point A to hoint B. -. 0.8î -* (b) Here CỔ is vector from point B to point c. ūé = C830'î + sim 3ooh D g ū . Hore ut is unit vector along 1300 the direction of slide W =-68 - Here w is angular velocity vector of dick ã =-3K > Here Ž is angular acceleration vellor of dishe Lac -- WBC - X = ABC - Here TBC and Woc are angular acceleration and velocity vector of link BC.

ac = acão Hoke ãe is acceleration vector of hoint C and ùe is unit vector along. the direction ae. The slides is constrained to move along We. Substitute Ue = 608 30° ê t sin 3o in above equation, ão = ac (cm 30cê + 81m 30 °9) ã = Che Cos 30° + Qe sin 300 — J = Ve Uc Here Te is velocity vector of hoint C and Uc is its magnitude, Substitute te = cos30° êtsin 300 Ve= . Veloszor îl + BimBoj) - L2, L8301 + 0.80 - - • Calculation of velocity and aceleration of hoint B: va = WX.BA Here To is velocity vector of hoine B Sulstilule W = -6K and BA = 0.2 US = -66k)x(@24) - î j Coass - Product To 02 ol Of 2. vector Do = 12 h. Jan Velouty as hont B. AB= X X BĄ - W²BÃO Here To is acceleration vector of hoint B, Substitute X=-3Ř BR = 0.29 and W = 6rad D = (-36) x(+0-22) - (6) (02)) = 1 i j 2 - 7.29 Cross - Product o 0:2 0 Co=40.69-79) 9. - Rokeration of beint 8.

• Calculation of velocity vector of facing Cuseing relative analysis of link BC: Ve = 0 + (Wbc X CB) Tuloslilile = 12ê , Wec = Wocſ and. CB = 0.8îl from equalion ®, ® and b) Ve = p2î + [[Weck) x (0-82)7 = lozî + l ßl To 8 - Cross-Product 10.8 o ol Je = 1022 +0=8Wbc Subbhtute Vo = U L830 îl + V. Bingooj from equation o in aliove equation 02C8830 + le simso 9 = pozî +0•8 Wech [Transfer barms and simplify? (VCCM 30° – 1.2)+ (W.Sim 300 – 0.8 Wee) = 0 -> Equate a comhoreno Torm of equation 1o lo gero, Ule (8630°–1.2=0 1.2 CO830° Ve= . 1.399 - 0 Equale ſ component form of equation o lo zero • 8080° – O-800 =0 0.8Wgc = lle 80M30° Wbc = Ue singoº Subslilule Ue = 1.39 m from equation ® in above equation Ue = 0.8 Wbc = .6439 310 30° Wbx = 0·866 god. Angular velocity of link BC. Subslilule ce = 1:39 m from equation ® in equalion Ⓡ to find the Vo = 1-39182ocê + 1.398in 309 V = 1.2 + 0-635 velocity vector of fant c

O 0 Ker • Calculation of acceleration vector of hoint C using selalive analysis of link BC? a = 9Bt. Rocx CB - WRC . Substitute Ta = 06Ê – 72, Foc = Bec , = 0.8î from equation , , (b) espectiwely and wer = 0·866 grad from equation 12, de = (orî – 7:2j) +[[X®CR)x(0-8)] - (0-866)° [0-82)] a = (6.61-7.27) + 1 2 7 1 2 1 - 0062 = .0.60 -7 2] + 0-896] -006? 10.8 OOT ac = . 0°8KBC] - 7.29 Substitute D = Accosocê tac singojo from equalios in above equation, A 8.983° + Qe 30 j = 0-8027.2ỹ [Tranfor tumas to @L45]. (Qc CB300) @ + (ac sim 30 – 0.88pc +7.2)ğ -0. -> Equate. î component of equation 13 to zro, . ACCOS 30° = 0 Ac=0 Equale ſ component of equation @ lo gero, Qe sin 30º – 058Mgc +7.2=0 Subelilula ac = o from equation in above equation 0 - 0.8%8c + 7.2=0 [simplify Equalion] XBC = 7.2 Plec = 9 god e Angelar acceleration of link Substitute Ae=o from equation ® in equation , Taco acceleration vector of point c. 0.8