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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.)

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DISCUSSION
In some sense, this is a very standard kinematics of rigid bodies problem. A rigid link connects points A and B. To relate the motion of these two points, you will need the following kinematics equations:

v_A = v_B + ω x r_A/B 
a_A = a_B + α x r_A/B – ω² r_A/B

The nuance in this problem is the acceleration of point A. From what we learned earlier in the semester, the acceleration of a point can be written in terms of its path coordinates; that is, here the acceleration of A can be resolved into its tangential and normal components. As you proceed on this problem, you first need to recognize the tangential and normal unit vectors, e_t and e_n, for the motion of A. Then write down the acceleration of A, first in its path components, and then in its Cartesian component. In the end, you will have two scalar equations coming from the rigid body acceleration equation in terms of two unknowns.

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DISCUSSION and HINTS
This mechanism is made up of three links: AB, BD and DE. You are given the rotation rate of link AB, and are asked to find the angular velocities and angular accelerations of links BD and DE. At the position shown, it is known that ω_AB = constant.

From the animation below, we are reminded that B and D move on circular arc paths with centers at A and E, respectively. The velocities of B and D are always perpendicular to the lines connecting the points back to the centers of the paths. Can you visualize the location of the instant center (IC) of link BD as you watch this animation? For the position of interest in this problem, you see in the animation that the velocities for all points on BD are the same – is this consistent with the location of the IC for BD at that position?

Velocity analysis
Where is the instant center of link BD? What does this location say about the angular velocity of BD?

Acceleration analysis
Write a rigid body acceleration equation for each of the three links in the mechanism:

a_B = a_A + α_AB x r_B/A – ω_AB²r_B/A
a_D = a_E + α_DE x r_D/E – ω_DE²r_D/E
a_B = a_D + α_BD x r_B/D – ω_BD²r_B/D

Combining together these three vector equations in a single vector equation will produce two scalar equations in terms of α_BD and α_DE.

Shown below is a freeze-frame of the motion of the mechanism at a time corresponding to the instant on which this question is based. As can be seen the velocity for all three points on link BD are the same (in both magnitude and direction) This is consistent with the conclusion that you would make based on knowing that the IC for link BD is at infinity for this instant where AB and DE are parallel.

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Discussion and hints:

Note that since the top contact point of the disk rolls without slipping on the upper fixed surface, this contact point has zero velocity. In addition, the horizontal component of acceleration of that point is also zero (the vertical component of acceleration is NOT zero, however).

For the velocity problem, write down the velocity kinematics equation for the disk (where “C” is the contact point):

v_O = v_C + ω x r_O/C = ω x r_O/C

and solve for the angular velocity of the disk. For the acceleration problem, write down the acceleration kinematics equation for the disk:

a_O = a_C + α x r_O/C – ω²r_O/C = a_C j + α x r_O/C – ω²r_O/C

and solve for the angular acceleration of the disk.

Then, write down and use the acceleration rigid body kinematics equation relating points O and B.

Carefully study the velocity (BLUE) and acceleration (RED) information for the point on the circumference of the disk shown in the animation below. Recall that the velocity of the point on the disk in contact with the fixed upper surface is zero, and its acceleration has only a vertical component. Do you see this in the animation?

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DISCUSSION and HINTS

In this problem, end A of the bar is constrained to move along a straight horizontal path with a constant speed of v_A, whereas end B is constrained to move along a straight, angled path. As you can see in the animation below of the motion of the bar, the speed of B is NOT a constant (the acceleration of B is non-zero, and is, in fact, increasing as B moves along its path).

In your solution, it is recommended that you use the rigid body kinematics equations relating the motion of ends A and B:

v_B = v_A + ω x r_B/A
a_B = a_A + α x r_B/A – ω²r_B/A

For these equations, you know: i) the magnitude and direction for the velocity of A; ii) that the acceleration of A is zero (constant speed along a straight path); and, iii) the direction for the velocity and acceleration of B. These two vector equations produce four scalar equations that can be solved for four scalar unknowns: v_B, a_B, ω and α.