Since the distance BC is not constant, this problem is well suited for using two frames of reference. Fix the local frame of reference, x-y, on bar BCD so that the motion of C can be easily modeled relative to B, (**r**_{C/B})_{rel}. Then the velocity of C is,
** v**_{C} = **v**_{B} + **Ω** × **r**_{C/B} + (**v**_{C/B})_{rel}
This equation includes the unknown angular velocity of bar BD, **Ω** = **Ω**_{BC} = **Ω**_{BD}, that is required, and can be determined. The global frame of reference is fixed to the background and is not moving.
The velocity of B is zero (it is pinned).
**v**_{B} = 0
The velocity of C with respect to B, **v**_{C/B}, is
** ** **v**_{C/B} = -v_{C/B}**i** = -v_{C/B}cos45**I** - v_{C/B}sin45**J** m/s
The velocity of point C, based on the motion of bar AC, in the global coordinate system is
** v**_{C} = -ω_{AC}r_{AC} **J** = -3(0.15)**J** = -0.45**J** m/s
The relative position of C with respect to B is
**r**_{C/B} = 0.15/cos45**i** = 0.15**I** + 0.15**J** m
Putting all the terms together, gives
-0.45**J** = -v_{C/B}cos45**I** - v_{C/B}sin45**J** + 0 +
+ Ω_{BD}**K** × (0.15**I** + 0.15**J**)
-0.45**J** = -0.7071v_{C/B}**I** - 0.7071v_{C/B}**J** +
+
0.15Ω_{BD}**J** -
0.15Ω_{BD}**I**
Summing the **I** terms gives,
0.7071v_{C/B}= -0.15 Ω_{BD}
v_{C/B}= -0.2121Ω_{BD}
Summing **J** terms:
-0.45 = -0.7071 (-0.2121Ω_{BD}) + 0.15Ω_{BD}
**Ω**_{BD} = -1.50**K** rad/s
Ω_{BD} = 1.50 rad/s |