Homework H2.J - Fa22

Any questions??

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.

HINTS:

• 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

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.

5 thoughts on “Homework H2.J - Fa22”

1. Katherine says:

In the very first instant of the animation, vA is going down and vD is to the left. How is this possible? I am pretty sure this contradicts the IC method. Shouldn't omega AD should start out counterclockwise, which means that vA goes down and vD goes to the right?

1. Joshua Charles Sander says:

Yes you are right. The video I think is inaccurate at that part... I was wondering this too. Just remember in the equations we can't assume the sign of anything, the only thing we know is that Vd is constrained to the i^ direction. The sign will work itself out in the math.

2. CMK says:

Katherine: I am very happy to you are challenging the animation results by applying things that you have learned in the course. In particular, you are considering the location of the IC for link AB to see if the directions of the velocity vectors for points A and D are correct.

If you carefully look at the animation at its start, when link AC is horizontal, you will see that D is actually moving to the RIGHT, as your IC analysis predicts. (I have included a freeze-frame capture of the animation above for that position.) This rightward direction of motion for D is short-lived, however, and D soon begins moving to the LEFT.

Again, you are thinking about things in a correct way, and it is consistent with the animation.

2. Aekum Kaur Mhajan says:

Does the contact point on the disk that the description notes as E, just point A?

1. Kenneth T Nakayama says:

I think point E is the contact point at the outer edge of the disk and the top of block B.