Before starting this problem, make note of the type of motion for each component in the mechanism:
- Block B is in pure translation, and moves to the left with a constant speed of vB.
- The disk is pinned to ground at its center C and moves in pure rotation; that is, C is the center of rotation of the disk. Because of this, the velocity of A is in the vertical direction.
- Slider D is in pure translation, and, at the instant of interest, D moves in the same direction as does Block B.
- Link AB has both translational and rotation components of motion.
Question: Where is the location of the instant center (IC) of AD at this instant? Reflect back on the observations above in answering this. You can use the location of the IC for AB to either find the angular velocity of AD, or to check your answer found from vector analysis.
- Use the rigid body kinematics equations relating the motion of the contact point on the disk (call that point E) to the center C of the disk to find the angular velocity and angular acceleration of the disk:
vE = vC + ωdisk x rE/C
aE = aC + αdisk x rE/C - ωdisk2 rE/C
- Use the rigid body kinematics equations relating the motion of points A and B to find the angular velocity and angular acceleration of the link AD:
vD = vA + ωAD x rD/A
aD = aA + αAD x rE/C - ωAD2 rD/A
Shown below is a freeze-frame capture of the animation of the motion for this mechanism at the position of interest (when link AC is horizontal). Note that the velocity of A is to the RIGHT at this position. The direction of motion quickly changes to the LEFT at the point where D becomes the IC of AD, as it should.