The mechanism produces intermittent motion of link AB. If the sprocket S is
turning with an angular velocity of Ws = 6 rad/s, determine the angular velocity
of link AB at this instant. The sprocket S is mounted on a shaft which is
separated from a collinear shaft attached to AB at A. The pin at C is attached to
one of the chain links.
Solution:
Kinematic Diagram: Since link AB is rotating about the fixed point A,
then VB is always directed perpendicular to link AB its magnitude is
VB =
ω AB rAB =0.2 ω AB . At the instant shown . Vs is directed at an angle 60°
with the horizontal. Since point C is attached to chain , at the instant shown ,it
moves vertically with a speed of Vc = WSrS = 6(0.175) = 1.05m/s
Instantaneous center : The instantaneous center of zero velocity of link BC at the
instant shown is located at the intersection point of extended lines drawn
perpendicular from vB and vC . Using law of sines , we have
rB/rC
Sin105°
=
0.15
sin 30°
rB/rC = 0.2898 m
rC/rC
Sin45°
=
0.15
sin 30°
rC/rC = 0.2121 m
The angular velocity of bar BC is given by
ω BC =
vC
1.05
=
rC / rC
0.2121
= 4.950 rad/s
Thus ,the angular velocity of link AB is give by
vB =
0.2
ω BC rB/rC
ω AB =4.950(0.2898)
ω AB =7.17 rad/s