1.

For the RRPP

mechanism shown
in figure 2 write
down the
displacement
equation for the
loop ABCDE in
terms of complex
numbers. Use
Figure 2 complex number
analysis to find the
absolute velocity and acceleration of the slider at E and the
point D on link CDE when the link AB is rotated counter
clockwise at 1 rad/s in the RRPP mechanism shown. At the
instant shown, AB and CD are horizontal, while BC and DE are
vertical. CDE is a single L shaped link. The slider rod EF is
fixed to the ground at F. AB=BC=CD=DE=1m at the current
position. Note that the lengths BC and FE change with time.
(3+3+3+3+3)
p
θ=
3 const , θ=
2 , ω=
2 1,=l 1
2
l exp ( iθ 2 ) + s3 − il + l = 2l + is4
⇒ iω2l exp ( iθ 2 ) + s3 =
is4
⇒ iω2l cos (θ 2 ) = s4 , −ω2l sin (θ 2 ) + s3 = 0
⇒ s3 = 1, s4 = 0
−ω2 2l exp ( iθ 2 ) +  is4
s3 =
⇒ −ω2 2l cos (θ 2 ) + 
s3= 0, −ω2 2l sin (θ 2 )= 
s4
⇒ 
s3 = 0, −1 = 
s4
2. In the RPPR mechanism shown in figure 3 the slider at A is
rotating counter clockwise with a constant angular velocity of 1
rad/sec. Based on the nature of the joints state what will be the
angular velocities and accelerations of the L shaped link ABC
and the slider at C. Using graphical vector analysis find the
angular velocity and acceleration of the point B with respect to
the ground. AB=BC=1 m. Drawing is to scale. At the instance
shown AB is vertical and BC is horizontal. Note that the
lengths AB and BC change with time. (1+1+1+1+5+6)
Figure 3
In the mechanism shown in figure 4, the link CE can slip
(i.e. both roll and slide) at the contact with the circular
hump that has its centre of curvature at D. Link AB rotates
counter clockwise at 1 rad/sec. Use the graphical method of
instantaneous centres to find the (a) IC of link CE with
respect to the ground (designated as 1), (b) IC of link AB
with respect to CE, (c) angular velocity of link CE and the
velocity of point E. AB = 1 m. Diagram is to scale.
(4+6+3+2).

Figure 4
1. A four bar RRRR mechanism is in a crossed dead center
configuration as shown in figure 3. OA (link 2) = 2 cm, CE (link 4)
= 1 cm. The coupler ABCD (link 3) is a square with diagonal
AC=1 cm. Given angular velocity of link 2 is constant and 1 rad/s
counter clockwise, find the acceleration of the points A, B, C and D
and the angular acceleration of the link 4 at the instance shown.
Use graphical vector method only. (3+3+3+3+3)
ω2 =
1rad / s, ω3 =
−2rad / s, a 2 =
0,
aC3/ A2 + a A2 = aC3 = aC4
⇒ aC3/ A2,t + aC3/ A2,n + a A2,t + a A2,n = aC3,t + aC3,n = aC4,t + aC4,n
⇒ α 3 × AC + ω 3 × ω 3 × AC + α 2 × OA + ω 2 × ω 2 × OA = α 4 × EC + ω 4 × ω 4 × EC
⇒ α 4 × EC − α 3 × AC = ω 2 × ω 2 × OA + ω 3 × ω 3 × AC − ω 4 × ω 4 × EC
a A 2 = a A 2,n = ω2 × ω2 × OA = 2cm / s 2 , aC 4,n = ω4 × ω4 × OA = 0, aC 3/ A 2,n = ω3 × ω3 × AC = 4cm / s 2
1.155
aC3,t= α 3 × AC
= 1.155cm / s 2 ⇒ α=
3 = 1.155rad / s 2
1
2.31
aC 4 = aC 4 ,t = α 4 × EC= 2.31cm / s 2 ⇒ α 4 = = 2.31rad / s 2
1
4 1.155
a D=
3, n ω 3 × ω 3 × CD = = 2.83rad / s 2 , a D= 3 ,t α 3 × CD
= = 0.817 rad / s 2 , a D 3 = 2.66cm / s 2
2 2
4 1.155
a B=
3, n ω 3 × ω 3 × CB = = 2.83rad / s 2 , a D= 3 ,t α 3 × CB
= = 0.817 rad / s 2 , a=B3 1.54cm / s 2
2 2
Link 2 in the mechanism
shown in figure 4 is
rotating at 1 rad/s in the
direction shown. OB (link
2) is the crank connected
to a hollow slider (link 3)
by a revolute joint at B.
Link 6 is a slider moving
along a vertical slot on the
ground (link 1). Link 4 is a
hollow slider connected to
the ground by a revolute
joint at A. Link 5 is a rod
connected to link 6 by a
revolute joint at C, while
passing through the
hollow sliders that form the links 3 and 4. OB=BC=AB=1 cm in the current configuration. OC is a vertical
line while OA is horizontal. Find the acceleration of the (a) slider 6 and (b) the point A on link 5, as well as (c)
the angular acceleration of link 4. Use complex number analysis only. (5+5+5)

Displacement : lOB exp ( iθ 2 ) =

lOA + l AB exp ( iθ5 ) , ilOC =
lOA + l AC exp ( iθ5 )

(
iω5l AB + lAB exp ( iθ5 ) , ilOC =
Velocity : iω2lOB exp ( iθ 2 ) = iω5l AC + lAC exp ( iθ5 ) ) ( )
ω2lOB exp {i (θ 2 − θ5 )} + ilAB ω2lOB cos (θ 2 − θ5 ) 
⇒ ω5
= ⇒ ω5
= = , l AB ω2lOB sin (θ5 − θ 2 )
l AB l AB
ω l sin (θ5 )  ω5l AC
ilOC = iω5l AC cos (θ5 ) + ilAC sin (θ5 ) , lAC = ) ⇒ lAC 5 AC =
cos (θ5 ) ω5l AC sin (θ5= , lOC
cos (θ5 ) cos (θ5 )

Acceleration (a 2 = 0 )
( ia − ω ) l exp ( iθ ) = ( ia l + 2iω l − ω l + l ) exp ( iθ ) ,
2 2
2
OB 2 5 AB 5 AB 5
2
AB AB 5

⇒ (a l + 2ω l + iω l − il ) exp {i (θ − θ )} − iω l =
5 AB 5 AB 5
2
AB 0 AB 2 5 2
2
OB

⇒ (a l + 2ω l ) cos (θ − θ ) − (ω l − 
5 AB 5 AB l ) sin {(θ − θ )} =
2 0 5 5
2
AB AB 2 5

(ω l − l ) cos (θ − θ ) + (a l + 2ω l ) sin {(θ − θ )} − ω l =

5
2
AB AB 2 5 0 5 AB 5 AB 2 5 2
2
OB

ω2 2lOB sin {(θ 2 − θ5 )} − 2ω5lAB 

⇒ a=
5 , l AB= ω5 2l AB − ω2 2lOB cos (θ 2 − θ5 )
l AB
(
ilOC = ia 5l AC + 2iω5lAC − ω5 2l AC + 
l AC exp ( iθ5 ) )
(
⇒ −ω5 2l AC +  )
l AC cos (θ5 ) − a 5l AC + 2ω5lAC sin (θ5 ) =0 ( )

=
lOC (a l 5 AC ) (
+ 2ω5lAC cos (θ5 ) + −ω5 2l AC + 
l AC sin (θ5 ) )

l AC =
(a l 5 AC + 2ω5lAC ) sin (θ ) + ω 5 2
l AC , 
a 5l AC + 2ω5lAC
lOC =
cos (θ5 ) cos (θ5 )
5
2. In the mechanism shown in figure 5 (next page) the center B of the wheel (link 2) is translating at a speed
of 1 cm/s in the direction shown. The wheel (2) has a radius of 1 cm. The wheel slips at the contact with the
ground (1) at A, but rolls at the contact with link 3 at C. Link 4 makes an angle of 30 degrees with the
horizontal. At the given instance the slider 5 is moving in a vertical line. Link 3 is connected to the ground by
a revolute joint at D. Link 4 is connected to wheel (2) and slider (5) by revolute joints at B and E respectively
and does not interfere with link 3. Find (a) the angular velocity of link 3 and (b) link 4 and the (c) velocity of
slider 5 with respect to the ground and (d) the relative velocity of sliding between the slider 5 and link 3 and
(e) wheel 2 and ground. Show all relevant ICs clearly. Use method of ICs only. (3+3+3+3+3)
 5

4 E
C
B
3 3
2
A D
1

9
 1,2

5,4
2,3
5,3
4 E
C 2,5
B
3 3
2,4
2

5,3 A 1,5
D1,3
1