Homework H2.H - Sp23

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?


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28 thoughts on “Homework H2.H - Sp23”

    1. 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.

        1. 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.

    1. 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.

      1. 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.

  1. 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?

    1. 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?

      1. 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.

    1. 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

    2. 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!

    1. 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.

  2. 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.

    1. 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).

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