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DISCUSSION THREAD
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DISCUSSION and HINTS
In this problem, we desire to relate the rotation rates of link AB and link OC. With the two rigid bodies connected by a pin-in-slot joint, we are not able to use the rigid body kinematics equations by themselves. Let's discuss that below.
Velocity analysis
Here, we can use the rigid body velocity equation to relate the motions of A and B:
vA = vB + ωAB x rA/B
However, we cannot use a rigid body velocity equation to relate the motion of points O and A (the reason for this is that O and A are not connected by a rigid body). In its place, we can use the moving reference frame velocity equation with an observer attached to link OC:
vA = vO + (vA/O)rel + ω x rA/O
where ω is the angular velocity of the observer, and (vA/O)rel is the velocity of A as seen by our observer on link OC. Note that with the observer being attached to link OC, this observer sees motion of A only along the slot.
Combine these two equations to produce two scalar equations.
Acceleration analysis
We will use the same procedure for acceleration as we did for velocity - use a rigid body equation for AB and a moving reference frame equation relating O and A.
aA = aB + αAB x rA/B + ωAB x (ωAB x rA/B)
aA = aO + (aA/O)rel + α x rA/O + 2ω x (vA/O)rel + ω x (ω x rA/O)
where α is the angular acceleration of the observer, and (aA/O)rel is the acceleration of A as seen by our observer on link OC. Again, note that with the observer being attached to link OC, this observer sees motion of A only along the slot.
Combine these two equations to produce two scalar equations.
This is not about this homework question but about the zoom recording of the meeting of Sunday's exam review session. The video is only a 12 second clip with no audio. Is it possible to post the video? Thanks.
We will work on getting that fixed. Thanks for letting us know.
Just to confirm, this homework will be due the day of our next class on Friday?
Correct. Since there is no class on Wednesday, the homework is due on the day of the next regularly-scheduled class on Friday.
Should the final answer be in terms of the axis of the observer or the standard xy axis?
I believe that it should be in terms of the observer (ie xy not XY).
Either set of coordinates is fine. All answers should be in terms of the same set of components.
O and B are fixed, meaning their velocities vO and vB are zero in this situation, right?
That is correct.
Are we supposed to use + or - 2 for w(ab) in our calculations? It shows clockwise but it says that the angular velocity is positive. Just want to make sure
I used -2 for w(ab) in my calculations. This gave me the correct signs for the velocity of a. You can always confirm using the directions. When drawing v_a the vector, it will point diagonally towards the bottom right. This implies both a negative i and j component of the velocity.
When you end up doing the calculations, v_a should equate to some negative value in both i and j as a result.
When I solve for w(oa), I am getting 0. Should we be using the equation w(oa) = L*w(ab) since they are perpendicular?
I also got 0, which i think makes sense for this particular instance since the velocity of A points towards pin O, so link OC has no velocity perpendicular to its axis acting on it. However I believe it still has an angular acceleration.
For this homework, we use the simplified motion equations for the 2D MRF rigid body equations described in the concluding remarks for Monday's lecture, correct?
3
Yes, that sounds correct
The given section says that Wab is in the direction given, which is clockwise, which would indicate a negative angular velocity. However, the given value is positive (2 rad/s) indicating counter-clockwise motion. Should we use 2 rad/s or -2 rad/s?
I used positive 2 and got the right signs for v_a and and v_a/o rel, as in the velocity is pointing toward O, in the negative i direction.
I used the negative sign with association to k, so that when doing the w_ab X r_ab, I got that my value would be in the negative i direction. Then when actually plugging in values I used 2.