Velocity and acceleration analysis analysis using vectors Problem 1 Link 2 rotates at a constant angular velocity. Determine the velocity and acceleration of points C and D.  

Solution 1 : Velocity analysis for C VC4=VC3=VC3A2+VA2 VA2 = 1 rad/s*2in =2in/s Normal to link 2 Hence direction and magnitude are both known for VA2 VC3A2 is normal to link 3. Only direction is known VC4=VC3 is horizontal. Only direction is known Absolute velocity vectors VA2 and VC3 start from Ov. Relative velocity vectors never start from Ov. A2 VA2 = 2in/s D, M VC3A2 =1.45 in/s (approx) D, M=? Ov VC4=VC3 =1in/s (approx) D, M=? C3 VC3A2 =1.45 in/s Hence w3= VD3A2 /AC =1.45/8=0.18 rad/s Scale: 1 in=1 in/s

Solution 1 : Acceleration analysis for C aC4=aC3=aC3A2+aA2=anC3A2+ atC3A2+anA2+atA2 atC3A2 Only Direction is known – normal to AC aC4=aC3 Direction is known - horizontal anC3A2 = w23* AC =0.182*8 in/s2 =0.26 in/s2 Direction- Radially inward along AC aA2 = w22* OA = (1 rad/s)2 *2 in=2in/s2 Direction – radially inward along O2A aC4=aC3 =1.48 in/s (approx) D, M=? Oa C The vector diagram aA2 = w22* OA = (1 rad/s)2 *2 in =2in/s2 D, M atC3A2 =1.38 in/s2 (approx) D, M=? atC3A2 =1.38 in/s2 Hence a3= atC3A2 /AC =1.38/8=0.17 rad/s2 anC3A2 = w23* AC =0.182*8 in/s2 = 0.26 in/s2 (approx) D, M

Solution 1 : Velocity analysis for D VB5=VB3=VB3A2+VA2 =VB5D6+VD6 VB3A2 = w3*AB= VD3A2 * (AB/AC) = 0.18 rad/s*4 in = 0.72 in/s Normal to link 6. VA2 = 1 rad/s*2in =2in/s Normal to link 2 VD6 is normal to link 6. Only direction is known VB5D6 is normal to link 5. Only direction is known Absolute velocity vectors VA2 and VC3 start from Ov. Relative velocity vectors never start from Ov. VB5D6 = 1.3 in/s (approx) D, M=? VB3A2 = 0.72 in/s (approx) D, M A VD6 =1.5 in/s Hence w5= VD6 /O6D =1.5/4=0.38 rad/s D VD6 = 1.5 in/s D, M=? VB5D6 =1.3 in/s Hence w5= VB5D6 /BD =1.3/4=0.33 rad/s VA2 = 2in/s D, M B Ov Scale: 1 in=1 in/s

Solution 1 : Acceleration analysis for D aB5=aB3=aB3A2+aA2= aB5D6+aD6 anB5D6+ atB5D6+anD6+atD6=anB3A2+ atB3A2+anA2+atA2 atD6 Only Direction is known – normal to O6D anD6 = w26* O6D = (0.38 rad/s)2 *4 in=0.57 in/s2 Direction – radially inward along O6D atB3A2 Only Direction is known – normal to AC anB5D6 = w52*BD = 0.332*4=0.44 in/s2 Direction – radially inward along BD atB5D6 = a3*AB = atC3A2*(AB/AC) =1.38*4/8=0.69 in/s2 Direction – normal to BD anB3A2 = w23* AB =0.182*4 in/s2 =0.13 in/s2 Direction- Radially inward along AC aA2 = w22* OA = (1 rad/s)2 *2 in=2in/s2 Direction – radially inward along O2A

Solution 1 : Acceleration analysis for D aB5=aB3=aB3A2+aA2= aB5D6+aD6 anD6+atD6+anB5D6 + atB5D6 =anA2+atA2 + anB3A2+ atB3A2 Since D cannot be located using the current order of vector addition, locating B is also difficult. atD6 D, M = ? Where is B3 ? atB3A2 D, M=? anD6 = 0.57 in/s2 D, M Where is D6? Where does anB5D6 begin? Oa anB5D6 = 0.44 in/s2 D, M atB5D6 D, M aA2 =2in/s2 D, M anB3A2 =0.213 in/s2 D, M Scale: 1 in =0.5 in/s2

Solution 1 : Acceleration analysis for D atD6 D, M = ? anB5D6 = 0.44 in/s2 D, M atB5D6 = 0.3 in/s2 D, M Oa Change order of addition of vectors anD6 = 0.57 in/s2 D, M aA2 =2in/s2 D, M atB3A2 D, M=? atD6 = 4.1 in/s2 D, M = ? anB3A2 = 0.213 in/s2 D, M aB5=aB3=aB3A2+aA2= aB5D6+aD6 anD6+atD6+anB5D6 + atB5D6 =anA2+ anB3A2+ atB3A2+atA2 aB5=aB3=aB3A2+aA2= aB5D6+aD6 anD6+atD6+anB5D6 + atB5D6 =anA2+atA2 + anB3A2+ atB3A2 atB3A2 = 2.8 in/s2 D, M=? aD = 4.05 in/s2 Scale: 1 in =0.5 in/s2 B D

Velocity and acceleration analysis analysis using vectors Problem 2   Determine the linear acceleration of AP4 if a2=0

Solution 2 : Acceleration analysis for P w2 Sense of rotation is found from direction of VP2 Analyze available information  VP2 = 10.0 in/sec. OPP=0.95. Hence w2= VP2/OPP= 10.53 rad/sec. Hence anP2 = w22 OPP=105.34 in/sec2 a2 = 0. Hence atP2 = 0 VP4P2 = 10.77 in/sec. Hence acP4P2 = w2VP4P2 = 10.53 *10.77=113.38 in/sec2. Vector equation aQ4=aP4=aP3= aP4P2+aP2=asP4P2+ acP4P2+anP2+atP2

Solution 2 : Acceleration analysis for P asP4P2 D, M=? aP4 D, M=? asP4P2= 55 aP4 =18 aP2=anP2 (Since atP2=0) = w22 OPP =105.34 D, M acP4P2 =2w2VP4P2 = 113.38 in/s2 aP4 Direction – Along OP (OpPO) acP4P2 =2w2VP4P2 = 113.38 in/s2 asP4P2 Direction -Along AP atP2= 0 anP2 = w22 OPP=105.34 Radially inwards along OpP (OpPO)

Velocity and acceleration analysis analysis using vectors Problem 3 Link 2 rotates at a constant w2 = 1 rad/sec. q2 = 60o. Find VA4 and aA4 if O2A = 2.4 cm

Solution 3 : Velocity analysis for A VA3A4 is along the slider in 4. Only direction is known VA3= w2*O2A = 1 rad/s*2.4cm =2.4 cm/s Normal to O2A VA2=VA3 +VA2A3 = VA3 = VA4+VA3A4 = VB4 +VB4A4 +VA3A4 = VB4 +VA3A4 VB4 is along the slider in 1. Only direction is known VA4 = VB4 (since 4 is translating only, all points on 4 have same linear velocity) A3 VA3= w2*O2A = 1 rad/s*2.4cm =2.4 cm/s VA3A4 = 1.2 cm/s D, M=? A4 VA4=2.08 cm/s D, M=? Ov

Solution 3 : Acceleration analysis for A aA2 = aA3 = anA2+atA2 = aA4+asA3A4 +acA3A4 Here we observe a link 3 which slides relative to a moving link 4. Hence there should be a Coriolis term atA2 = a2*O2A =0*2.4=0 Hence aA2 = anA2 aCoriolis=wcontaining link*Vrelcontained link aA4 = aB4 = ? Along O2B4 aCA3A4 = 2w4* VA3A4 But w4 = 0 as link 4 translates relative to the ground. Hence acA3A4 = 0 asA4A3 =? Along A2B4 aA2 = anA2 = w22*O2A =2.4 cm/s2 Radially inward from A to O2 aA4 = 1.2 cm/sec2 D, M =? A4 Oa asA3A4 = 2.08 cm/s2 D, M =? aA3 = aA2 = anA2 = 2.4 cm/sec2 A3

Problem 4 w2 = 1.000 1.0 1.631 3.56 4.41 Find the relative acceleration of the slider block with respect to the curved slot in the mechanism shown in figure. Assume w2=1 unit angular velocity clockwise, a2=0, O2A=1 unit length. Use graphical method of acceleration analysis using vectors.

Velocity diagram tangent OV A4 A2 , A3 VA4 = 0.835 w4=0.5 VA2=1 VA4 A3 =1.016 A2 , A3

Acceleration diagram OA w42O4A=0.41 a4O4A A2 , A3 A4 2w4VA4A3 +V2A4A3/4.41/ aA4A3,sliding tangent normal