I get the feeling that we have to solve this using the observer on other link, which should give the same answer, just makes the math different.
As we discussed in class, you are free to choose whomever that you want as the observer. That choice, however, dictates the various terms in the MRF kinematics equations. Consider the following:
* If you place the observer on the turntable, then: omega = Omega*J^hat alpha = 0
and the relative velocity and acceleration terms are NOT zero.
* If you place the observer on OP, then: omega = Omega*J^hat +theta_dot*k^hat alpha ≠ 0
and the relative velocity and acceleration terms ARE zero.
You choose.
Will Vp be the derivative of the theta function? And then you are supposed to split it into X and Y? Or am I thinking of this the wrong way?
Aditya: I recommend using the standard MRF kinematics equation to find vP:
vP = vO + (vP/O)rel + omega x rP/O
with an observer on OP. With that choice of observer, the omega vector has two components: Omega about Jhat, and theta_dot about khat.
Does that help?
I personally think the method with the observer on OP was an easy way to solve the problem, but like they said it can be done correctly both ways.
When comparing XYZ to xyz, keep in mind that at t=0s they are lined up, but oriented differently.
0
When t=0, theta is also zero, so then by the given schematic, both coordinate systems are in perfect alignment, are they not?
This problem is very similar to H.3.I. Make sure to properly do your derivatives for the second rotation and keep in mind that the axes all line up when t = 0.
Hi guys! As a helpful tip for this problem if you have your observer on the link OP you will need to find theta_dot and theta_ddot using the chain/product rule. I got theta_dot = theta_o*b*cosbt and theta_ddot = -theta_o*b^2*sinbt which theta_ddot ends up equaling 0.
2
Since O is on the centerline of both rotating axis, is it safe to assume that V_o = A_o = 0?
Yeah, O doesn't move. I like to Imagine the problem in my head to double check with the math of being on centerlines. You see the ball spins around O and that the the end of the turntable as a point isn't moving when being spun.
Yes, the only time we can't assume V_0 and A_0 = 0 is if the point also rotate with lamda or omega. Hope that helps!
If you are confused for how to get your theta_dot and theta_ddot, you just need to take the first and second derivative of theta(t)
This problem is very similar to those previously done in Chapter 3. You are not given a suggestion for an observer, so you can either choose to place one on the turntable or on OP. Using the OP method, the main difference is that you have to manually compute the derivatives for theta(t) and substitute into the equation for alpha. Theta_ddot ends up being 0 (since sin(0) is 0) but theta_dot ≠ 0, since cos(0) ≠ 0.
Why is the pdf for H3.A?
I get the feeling that we have to solve this using the observer on other link, which should give the same answer, just makes the math different.
As we discussed in class, you are free to choose whomever that you want as the observer. That choice, however, dictates the various terms in the MRF kinematics equations. Consider the following:
* If you place the observer on the turntable, then:
omega = Omega*J^hat
alpha = 0
and the relative velocity and acceleration terms are NOT zero.
* If you place the observer on OP, then:
omega = Omega*J^hat +theta_dot*k^hat
alpha ≠ 0
and the relative velocity and acceleration terms ARE zero.
You choose.
Will Vp be the derivative of the theta function? And then you are supposed to split it into X and Y? Or am I thinking of this the wrong way?
Aditya: I recommend using the standard MRF kinematics equation to find vP:
vP = vO + (vP/O)rel + omega x rP/O
with an observer on OP. With that choice of observer, the omega vector has two components: Omega about Jhat, and theta_dot about khat.
Does that help?
I personally think the method with the observer on OP was an easy way to solve the problem, but like they said it can be done correctly both ways.
When comparing XYZ to xyz, keep in mind that at t=0s they are lined up, but oriented differently.
When t=0, theta is also zero, so then by the given schematic, both coordinate systems are in perfect alignment, are they not?
This problem is very similar to H.3.I. Make sure to properly do your derivatives for the second rotation and keep in mind that the axes all line up when t = 0.
Hi guys! As a helpful tip for this problem if you have your observer on the link OP you will need to find theta_dot and theta_ddot using the chain/product rule. I got theta_dot = theta_o*b*cosbt and theta_ddot = -theta_o*b^2*sinbt which theta_ddot ends up equaling 0.
Since O is on the centerline of both rotating axis, is it safe to assume that V_o = A_o = 0?
Yeah, O doesn't move. I like to Imagine the problem in my head to double check with the math of being on centerlines. You see the ball spins around O and that the the end of the turntable as a point isn't moving when being spun.
Yes, the only time we can't assume V_0 and A_0 = 0 is if the point also rotate with lamda or omega. Hope that helps!
If you are confused for how to get your theta_dot and theta_ddot, you just need to take the first and second derivative of theta(t)
This problem is very similar to those previously done in Chapter 3. You are not given a suggestion for an observer, so you can either choose to place one on the turntable or on OP. Using the OP method, the main difference is that you have to manually compute the derivatives for theta(t) and substitute into the equation for alpha. Theta_ddot ends up being 0 (since sin(0) is 0) but theta_dot ≠ 0, since cos(0) ≠ 0.