Any questions??

Discussion:

Let’s first take a look at the motion of point D. This motion of D is shown in the simulation results below.

The motion of D is circular, with the center of the path located at point A. This is expected since link AD is pinned to ground at point A.

Now, let’s attach an observer to link BE. Keep in mind that this observer is unaware that they are moving. The motion that this observer sees is straight, with this straight path aligned with the slot cut into link BE. You can see this is the following animation shown from the perspective of the observer attached to link BE.

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Discussion

An animation from the simulation of the motion of the robotic arm system is shown above for a fixed observer. The motion of end-effector B is quite complicated, as can be seen. However, if we attach an observer to section AD of the end link of the arm, the observed motion of B is greatly simplified. In fact, the observer simply sees a back-and-forth motion of B in the x-direction, as shown in the animation below.

HINTS: The velocity of point B can be found from the following moving reference frame velocity equation:

v_B = v_A + (v_B/A)_rel + ω × r_B/A

With the observer on link AD, we know the following:
v_A = (L θ_dot) i
ω  = φ_dot k
(v_B/A)_rel = b_dot i

Use similar logic for writing down the acceleration of B. Be careful in writing down the expression for a_A ; in particular, a_A ≠ 0. (A is traveling on a circular path with a constant speed.)