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DISCUSSION THREAD
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Discussion and hints
From the simulation results above, we see that point A travels on a cycloidal path, with the velocity vector for A being tangent to this path, as expected. In addition, the acceleration of A points inward to the path (again, expected). The angle between v and a is initially obtuse, implying that A is initially decreasing in speed. At some point, this angle because acute indicating that the speed of A begins to increase. In fact, this rate of change of speed becomes very large as A approaches the surface on which the disk rolls. Do you know why?
The velocity analysis is a straight-forward application of our rigid body kinematics equations where we write a velocity equation for each rotating member:
vA = vC + ωdisk x rA/C =ωdisk x rA/C
vA = vB + ωAB x rA/B = vB * i + ωAB x rA/B
From these, you can solve for ωdisk and ωAB.
Applying the same procedure to acceleration:
aA = aC + αdisk x rA/C - ωdisk2 rA/C
aA = aB + αAB x rA/B - ωAB2 rA/B
produces too few equations for the number of unknowns. It is recommended that you also use the following equation in your acceleration solution:
aO = aC + αdisk x rO/C - ωdisk2 rO/C
I didn't find a need for a third equation for my acceleration.. is the third equation really necessary?
I think you need it to find Ac
I believe that a_c != 0. Instead a_o = 0, which can be used to calculate a_c using equation 3, leaving 2 equations and 2 unknowns if the other two equations.
* (a_c != 0) in the x direction. (a_o = 0) in the y direction.
* (a_c != 0) in the y direction.
I believe that ao is zero in the j direction since it is moving on a horizontal surface and ac is zero in the i direction since it is the contact point. This would mean that the j component of aoj = 0 and ac = acj. You can then find acj in terms of omega disk and R. You can then plug this in for ac in a prior acceleration equation. This is at least how I thought to go about it.
Since v_c =0, why doesn't a_c also =0?
Emily: Please review pp. 91-92 in the lecture book. This gives a brief review on rolling without slipping. That provides an explanation regarding your question.
Isn't doing this, the same thing as using the derivation from the book for acy?
Are we assuming the v_a is in i and j direction? Or just one?
No assumptions needed. Write a velocity equation relating A and C . You will find that v_A has both x- and y-components.
When using the a_O/C equation, we can solve for both a_c and alpha_disk, right? So the j direction of this equation leaves us with alpha_disk = 0.
Why do you say that alpha_disk = 0? The acceleration equation that you mention relating the motion of O and C has three unknowns: a_C, a_O and alpha_disk. You cannot solve for these three unknowns with only the two scalar equations that come from your acceleration equation. As suggested in the discussion above, you need to also relate the acceleration of points C and A.
I think I am still a bit confused. Now we have four unknowns in three equations. Is there an assumption we are suppose to make about a_O? Is it only in one direction, not both?
Not sure of your confusion. You have three unknowns. And you have three equations. You can solve these for the three unknowns.
Point O moves along a straight path (rho = infinity, or 1/rho = 0) parallel to the surface on which the disk rolls. Therefore, it does not have a component perpendicular to the path. The acceleration of O is in only one direction. Please note that all of this is NOT an assumption, it is a consequence of the motion.
Do we need a 3rd equation for the velocities to find omega_ab and omega_disk? Because we have two unknowns and only 2 equations, or am I missing something?
Never mind, I found my mistake.
I am having the same issue, what was your mistake?
I had the same mistake originally but splitting the equations into i and j components allowed me to isolate one of the variables into all known or easily calculated values.
Once you have the equations relating A with B and C, you can set the i^ and j^ components of each equation together, leaving you with two unknowns (the angular velocities) and two equations to solve them with.
Is aO always equal to 0?
aO is zero only under special circumstances: when O is traveling at a constant speed along a straight-line path (which, in turn, corresponds to zero angular acceleration of the no-slip disk). Here, although O is traveling along a straight path, it is not given that O has a constant speed.
In general, one should avoid making any assumptions on motion for any problem. Assumptions (such as zero acceleration) are NOT needed since you can use the kinematics equations in order to determine the result.
When do we use a no slip value equal to R*omega^2 in the j?
For a disk rolling without slipping on a stationary surface, the acceleration of the no-slip point C is in the direction of the center of the disk O and has a value of r*omega^2. You do not need to memorize this - just work it out: write out the acceleration equation relating points O and C.
Is it just me or was anybody else's acceleration solution a super long equation? I think I got it right (positive alpha disk, negative alpha link), but I was just curious if that's how it was for other people.
Nevermind, I had some sign errors.
Yes, this is what I had. I found the acceleration for O in terms of C, then went from O to A, then from A to B, and using the knowledge that B has no acceleration, I worked my way backwards all the way to O. This process took a while (and I am not quite sure that I did it correctly.) but I understand the basic idea of it.
Would a_B be 0 in this case?
Yes. B is traveling along a straight-line path at a constant speed. This says that both v_dot = 0 and rho = infinity, making both the tangential and normal components of a_B = 0.
For some reason I keep getting a positive result for omega disk. Any idea why this might be?
It might help to use the 3x3 matix techique for cross products in order to avoid sign errors. In my experience it helps me avoid careless mistakes
Check that you have your position vectors correctly noted as it can get confusing sometimes.