| Problem statement Summary sheet for 3D MRF kinematics-1 Solution video |
DISCUSSION THREAD
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DISCUSSION and HINTS
The disk has two components of rotation:
- One component of ω1 about the fixed J-axis.
- The second component of ω2 about the moving k-axis.
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?
Would all of the axes be lined up in this case because the Y and y axes are aligned?
At this moment, yes, all axes are aligned. In the problem diagrams you can actually see all axes lined up, not just Y and y. However, as the disk rotates and we leave this moment, this will not always be the case.
I'm assuming the way to do part b is to first write an equation that relates O to C to find the velocity and acceleration of C and then write an equation that relates P to C to find the velocity and acceleration of P. However, since O and C are on the same rigid body, and P and C are on the same rigid body, would their respective relative velocities and accelerations be equal to 0 when computing the final answer?
Yes, I believe that the respective relative velocities and accelerations for V_p/c and a_p/c should both be zero given the observer's location as they share the same rigid body. This greatly simplifies the equations and should make the calculations for this part fairly straightforward. A more full explanation of why can be seen on page 162 of the lecturebook, with the equations necessary there. Just like on Earth we don't observe our movement through the universe and only relative motion on the planet, it is just the same for the observer on the disk.
Can the angular acceleration of the disk be 0?
I suppose that it could. But in this case, it is not.
It won't be because alpha would be equal to the derivative of omega with respect to time and we have a moving reference frame that changes with time.
If I wanted to write my velocity and acceleration of P with respect to O, is it accurate to say that the relative velocity of P with respect to O is equal to the relative velocity of P with respect to C plus the relative velocity of C with respect to O?
Nathaniel: What you say is true; however, I would recommend something else, one that cuts out the middle man. I would go straight from O to P. Since O and P are on the same rigid body ("rigid body" meaning that the distance between O and P is constant), an observer on the disk will not see any motion for P. That is, both the relative velocity and relative acceleration terms in the kinematic equations are zero.
Does that make sense?
Actually, that really helped. Thank you!
I'm having a hard time seeing how the distance between O and P is constant. Is O also rotating at the same rate as P?
Nevermind, I see now. The vector between O and P is not constant, but the distance between them is.
This problem is very similar to 3.B.6 from the lecturebook. Once you identify that the observer on the disk sees that p has no relative velocity and acceleration the rest of the problem is a straightforward solve.
For this question, when finding the velocity and acceleration of P, it is very helpful to remember that (Vp/o)rel and (ap/o)rel are 0. Secondly, make sure to take note that J dot isn't changing and is 0 also. This will help you find your angular acceleration.
When solving for the velocity and acceleration of the disk is O moving? I assumed it was not as it is at the center of the motor.
To clarify I mean the velocity and acceleration of point P on the disk
Yes. Point O is on the centerline of the shaft that is rotating about a fixed axis aligned with the shaft. Therefore, O is stationary.
I believe O is a fixed point.
Should our answers be in terms of i, j, & k or I, J, & K?
That is your call - either way works. Just be sure to not mix coordinates. For this particular problem, the ijk unit vectors are aligned with the IJK unit vectors, so there should be no confusion here.
Since the observer is attached to the disk, would the angular velocity feel the affects of both Omega1 and Omega2 in their respective directions?
Yes, the disk has both components of angular velocity. Since the observer is attached to the disk, then so does the observer.
Keeping track of the axes is important in this problem. For example, since the Z axis points downward on the second picture, the distance between O and P is actually positive on that axis instead of negative as it may first appear.