Problem statement Summary sheet for Newton's Laws - 1 Solution video |
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Problem statement Summary sheet for Newton's Laws - 1 Solution video |
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Problem statement Summary sheet for 3D MRF kinematics-2 Solution video |
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Discussion and hints:
Your first decision on this problem is to choose your observer. Since an observer on the plate will have the simplest view of the motion of the insect, attaching the observer to the plate is recommended. Also, attach your xyz-axes to the plate.
Next write down the angular velocity and angular acceleration of the plate. Based on what we have been doing up to this point in Chapter 3, hopefully it is clear that the plate (and observer) has two components of angular velocity: Ω about the fixed X-axis and θ_dot about the moving z-axis. Take a time derivative of the angular velocity vector to find the angular acceleration of the plate (observer).
Following that, determine the motion of the insect as seen by the observer on the plate.
Use these results with the moving reference frame kinematics equation to determine the velocity and acceleration of the insect.
Problem statement Summary sheet for 3D MRF kinematics-2 Solution video |
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Discussion and hints:
It is recommended that you use an observer attached to the wheel. As we have discussed in class, your choice of observer directly affects four terms in the acceleration equation: ω and α (how the observer moves), and the relative velocity and relative acceleration terms (what the observer sees). Note that the remainder of the discussion here is based on having the observer attached to the wheel.
The wheel shown above has TWO components of rotation:
(Be sure to make a clear distinction between the lower case and upper case symbols.)
Therefore, the angular velocity of the wheel is given by:
ω = ω1J + ω2 k
The angular acceleration vector α is simply the time derivative of the angular velocity vector ω : α = dω/dt. In taking this time derivative,
With the observer attached to the wheel, what motion does the observer see for points A and B? That is, what are (vA/O)rel and (aA/O)rel, and (vB/O)rel and (aB/O)rel?
NOTE: Pay particular attention to the motion of the reference point O. What path does O follow? And, based on that, how do you write down the acceleration vector of O, aO?
Problem statement Summary sheet for 3D MRF kinematics-1 Solution video
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DISCUSSION and HINTS
Bar OA has two components of rotation:
Write out the angular velocity vector ω in terms of the two components described above.
Take a time derivative of ω to get the angular acceleration α of the bar. When taking this derivative, you will need to find the time derivative of the unit vector k. How do you do this? Read back over Section 3.2 of the lecture book. There you will see: k_dot = ω x k, where ω is the total angular velocity vector of bar OA that you found above.
Problem statement Summary sheet for 3D MRF kinematics-1 Solution video |
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DISCUSSION and HINTS
The disk has two components of rotation:
Part (a)
Write out the angular velocity vector ω in terms of the two components described above.
Take a time derivative of ω to get the angular acceleration α of the bar. When taking this derivative, you will need to find the time derivative of the unit vector k. How do you do this? Read back over Section 3.2 of the lecture book. There you will see: k_dot = ω x k, where ω is the total angular velocity vector of bar OA that you found above.
Part (b)
Here you will use the MRF kinematics equations of:
vP = vO +(vP/O)rel + ω x rP/O
aP = aO + (aP/O)rel + α x rP/O + 2ω x (vP/O)rel + ω x (ω x rP/O)
For these equations, employ an observer on the disk. The angular velocity and angular acceleration of the observer are the same as the ω and α of the disk found in Part (a). What are the relative velocity and relative acceleration terms ((vP/O)rel and (aP/O)rel ) in these equations? These represent the velocity and acceleration of P as seen by our observer. As our observer moves with the disk and with P being on the disk, what motion does the observer see for P?
Problem statement Summary sheet for 2D MRF kinematics-2 Solution video |
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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.
Problem statement Summary sheet for 2D MRF kinematics-2 Solution video |
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Discussion and hints
In order to create the animated GIFs shown below, a problem slightly different that that described in the homework statement has been solved. Here, the rotation rate of omega_OB is not constant, but instead, link OB is oscillating back and forth.
What you should sees from these animations is that the velocity of P as seen by a stationary observer is NOT tangent to the circular guide, whereas the velocity as seen by the observer on OB is tangent to the circular guide. Also, the observed acceleration of P always points to the inside of the circular guide, whereas the acceleration seen by a fixed observer does not always do so.
We need to treat this problem just as we have done for all planar mechanism kinematics in this course. For velocity analysis:
For acceleration analysis, repeat the above steps using the rigid body acceleration and MRF acceleration equations. Note that for the MRF equation, the acceleration of P as seen by the observer will have TWO components: one tangent to the circular guide (the observed rate of change of speed, vdot_rel) and one perpendicular to the circular guide (the observed centripetal component of acceleration, v_rel^2/R). In the end, you will have two equations in terms of vdot_rel and alpha_AP.
Problem statement Summary sheet for 2D MRF kinematics-1 Solution video |
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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:
vB = vA + (vB/A)rel + ω × rB/A
With the observer on link AD, we know the following:
vA = (L θ_dot) i
ω = φ_dot k
(vB/A)rel = b_dot i
Use similar logic for writing down the acceleration of B. Be careful in writing down the expression for aA ; in particular, aA ≠ 0. (A is traveling on a circular path with a constant speed.)