Homework H2.J - Sp23

Problem statement
Solution video


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Before starting this problem, make note of the type of motion for each component in the mechanism:

  • Links OA and BC are in pure rotation about their centers of rotation O and C, respectively. From this, we know that the paths of points A, B and C are circular, as seen in the animation below.
  • Block E is in pure translation.
  • Links AB and DE have both translational and rotation components of motion.

Question: What are the locations of the instant centers (ICs) of AB and DE at this instant? Reflect back on the observations above in answering this. What do these locations say about the angular velocities of AB and DE at this position?

Once you have found the angular velocities for all of the links, you can then tackle the acceleration analysis.

  • For finding the angular acceleration of links AB and BC, use the following rigid body acceleration equations:
    aA = aO + αOA x rA/O - ωOA2rA/O
    aB = aC + αBC x rB/C - ωBC2rB/C
    aB = aA + αAB x rB/A - ωAB2rB/A
    This will give you the equations that you need to solve for the desired angular accelerations.
  • Repeat the above for link DE to determine its angular acceleration:
    aD = aC + αBC x rD/C - ωBC2rD/C
    aE = aD + αDE x rE/D - ωDE2rE/D

25 thoughts on “Homework H2.J - Sp23”

        1. omega_DE = 0 tells us that all points on DE have the same velocity. Therefore, vE = vD = L*omega_BC.

          The equation that you quote above, vE=wDE * rDE, is not valid because vD is not zero.

  1. In the first acceleration equation a_A = a_O + α_OA x r_A/O - ω_OA^2(r_A/O). Does a_O = 0, since its velocity (v_O) is constantly 0 because its pinned? And does α_OA equal 0, since w_OA is constant as stated in the problem?

  2. When looking at omega_OA compared to omega_BC, do we need to care about the directional sign since we are going to use it to just find the speed of E? I end up getting -Lomega_OA for V_E if I include the sign for direction of rotation while substituting, but in the animation V_E is going the same direction as link OA.

    1. I guess my question boils down to should I keep omega_BC equal to -omega_OA or do I disregard the different rotation directions and have the prior equal +omega_OA?

      1. What I did was use Va = Vb ---> Vc (in terms of Vb) then use the IC eqn (wR=V) for Vb which you can use to fin Wbc in terms of Woa. Also, my Wbc is positive.

      2. MIchael: The instant center approach is based on speeds (unsigned numbers). Deal with positive values. Get the direction of the rotations by looking at your figure.

  3. When looking at the link that includes B, D, and C, which link should we use when using the IC approach for v_D (in relation to the angular velocity of BC). Because D is in between B and C, would that equation be v_D = w_BC * BC or v_D = w_BC * DC. I know this has come up before I was just never sure of which one was correct

    1. Once you have found w_BC, then use v_D = w_BC * DC.

      The other option that you provide above, v_D = w_BC * BC, is not correct - that expression is for v_B, not v_D.

  4. When looking at the acceleration equations, I have found a value for the acceleration of A in terms of A and wOA, but now I am stuck on where to go from here. Looking at the two equations solving for a_B, there are three unknowns and two equations. Is there some inequality that I'm forgetting about or some way that I can write one variable in terms of another? Or should I set the two equations equal to one another and solve from there?

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