At the instant shown bar AB has a constant angular velocity

At the instant shown bar AB has a constant angular velocity. If w_ab = 4 rad/s (CW), w_bd 2.5 rad/s (Ccw)w_de = 6.67 rad/s (CW):Estimate the angular acceleration of bar BD. Estimate the angular acceleration of bar DE.

Solution

Relative position vectors

r _B/A = - 0.2 j

r _D/B = 0.16 i

r _DE= -0.06 i – 0.12 j

Angular velocity: _AB = 4 rad/s CW, _DE = 2.5 rad/s CCW,         _BD    = 6.67 rad/s CW

Angular acceleration: _AB = 0 (As AB has the constant velocity),    _BD = _BDk, _DE = _DEk

Where “” is the angular acceleration and “” is in vector form

Now for Bar AB ( Rotation about A) :

a_B = _AB×r _B/A- _AB ² × r _B/A

= 0 – (4²)(- 0.2 j )

= 3.2 j

Now for Bar BD ( Rotation about B + Translation with B)

a_D = a_B + _BD×r _D/B - _BD ² × r _D/B

= 3.2 j + _BDk × 0.16 i – (6.67)² (0.16 i)

= 3.2 j + 0.16 _BD j – 7.12 i

= – 7.12 i + (3.2 + 0.16 _BD) j _{-------------------------------------------- (1)}

Now for Bar DE ( Rotation about E)

a_{D         = DE}×r _D/E - _DE ² × r _D/E

= _DE k × (-0.06 i – 0.12 j) – (2.5)² (-0.06 i – 0.12 j)

= -0.06 _DE j + 0.12 _DE i + 0.375 i + 0.75 j

= (0.12 _DE + 0.375) i + (0.75 -0.06 _DE) j                      _{-------------------------------------------- (2)}

Now, equating like components of from equation 1 and 2

– 7.12 = 0.12 _DE + 0.375

Or _DE = - 62.46 rad/s² (Angular acceleration of bar DE)

And

3.2 + 0.16 _BD = 0.75 -0.06 _DE

Or _BD = - 38.75 rad/s² (Angular acceleration of bar BD)