1. 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.

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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 = 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

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

9. Solution 2 : Acceleration analysis for Pw2
Sense of rotation is found from direction of VP2
Analyze available information
 VP2 = 10.0 in/sec. OPP=0.95.
Hence w2= VP2/OPP= rad/sec.
Hence anP2 = w22 OPP= in/sec2
a2 = 0.
Hence atP2 = 0
VP4P2 = in/sec.
Hence acP4P2 = w2VP4P2 = *10.77= in/sec2.
Vector equation
aQ4=aP4=aP3= aP4P2+aP2=asP4P2+ acP4P2+anP2+atP2

10. 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
= in/s2
aP4
Direction – Along OP (OpPO)
acP4P2
=2w2VP4P2
= in/s2
asP4P2
Direction -Along AP
atP2= 0
anP2 = w22 OPP=105.34
Radially inwards along OpP (OpPO)

11. 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

12. 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

13. 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

14. 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.

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

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