| Problem statement Solution video |
DISCUSSION THREAD
Discussion
The animation above shows the motion of the mechanism over a range of input angles of link OA. For a given position, envision the location of the instant centers (ICs) for links AB and CD. Do the directions and magnitudes for the velocities of points B, C and D agree with the location of these ICs?
Shown below if a freeze-frame of the mechanism motion at the position for which you are asked to do analysis. From this figure, where are the two ICs for links CD and BC? In particular, how does the position of the IC for AB relate to the relative sizes of the speeds of points A, C and B? What is the angular velocity of link AB at this position?
Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.
Is there meant to be a numeric value for Woa? I am unsure on how to calculate the speed of other points without.
Leave your answers in terms of omega_OA and L.
I'm confused about how to proceed if IC_ab is infinity. How would we find V_B?
I also got IC_ab as infinity, but that would mean that V_a = L*omega_OA = V_b. I would still be stuck on how to find V_c.
If the IC for a link is at infinity, then the angular velocity of the link is zero. If the angular velocity for link is zero, then all points on the link have the same velocity, in both magnitude and direction.
Based on the freeze-frame, v_D is perpendicular to v_C=v_A=v_B but how would that relate to determine v_D.
Construct perpendiculars to v_C and v_D. The intersection of these two perpendiculars locates the IC for CD. Since you know v_C, you can use that location of the IC to determine the angular speed of CD, as well as v_D.
If a link is pinned to a point that would make it automatically its instant center correct?
That is correct.
Is the velocity at each point on a rigid body equal when w=0?
When w=0 the body is in pure translation, meaning that all points along that rigid body are moving at the same speed in the same direction. This can be seen in the Lecture Book examples on page 107.
At first, I thought since there were non-zero velocities to the links omega couldn't be zero. I guess remembering that omega is rotational velocity and not just a velocity value is important. Hopefully this gives a bit of an intuitive 'spin' to what Niklas said.
Yes, that is true.
When looking at the distance from C to IC of CD, would that distance equal half of OA? The image looks proportional to me, and by eye it looks like C is roughly at the midpoint of OA in the vertical plane. If not, how do we find the distance from C or D to the IC?
Michael: Consider the right triangle AOB. Point C is at the midpoint of the hypotenuse of the triangle. Therefore, its vertical position is at half of 0.6L up from the base, and is half of 0.8L along the base.
Does this help?
Yes that helps a lot! I was able to solve for it from there. Just to confirm, in my new Vc and Vd equations, the distance I use is the distance relative to the IC correct? I used .3L from C to the IC, then .4L from D to the IC.
Michael: Yes, that is correct.
I don't understand how to find omega_CD so that you can in turn find v_D. How can I find this value?
I believe that Va, Vb, and Vc have the same magnitude and direction because their IC goes to infinity. This means that you would use the equation wCD=r*Vc
Mason: I think that you meant vC = r*wCD.
You can find omega_CD by using your known value of Vc. You can use it in the equation Vc = omega_CD*r(from C to your IC). Since we know the distance from C to the IC for CD, you can solve for omea_CD. Then you use that to find VD. Let me know if I messed that up at all. Thanks!
I am still a little confused, if wOA is zero and we are trying to calculate Va, are we supposed to still use the w=r*V equation?
Mason, be sure that you aren’t mixing up the rotational velocity of the rigid body AB and the rotational velocity of point A. These two values are distinct, and you can use the w = r * V equation for point A.
omega_OA is given. I do not think that omega_OA = 0. Furthermore, the equation goes as: v = r*w.
I am a little confused on what the instant center for beam BA is supposed to be as the perpendicular component to the velocity is just infinity.
Not exactly sure of what you are saying here. Since the IC for AB is at infinity, ALL points on AB have the same velocity; that is, v_A = v_B = v_C (in both magnitude and direction).
For part A, when it asks for the direction of pin C, can we leave it in terms of i,j,k vectors?
Yes, give the direction of motion for C in terms of i and j components. There will not be any components in the k direction.