Homework H2.B - Sp23

Problem statement
Solution video

DISCUSSION THREAD

 

Discussion and hints:

The solution for the velocity and acceleration of end B is a straight-forward application of the rigid body velocity and acceleration equations for member AB:

vB = vA + omega x rB/A
aB = aA + α x rB/A - ω2*rB/A

where vB = vB*j, vA = vA*(cos(θ)*i + sin(θ)*j), aB = aB*and aA = aA*(cos(θ)*+ sin(θ)*j)Each of the two vector equations above represents two scalar equations, providing us with the necessary equations to solve for vB, ω, aB and α.

HINT: The solution for this problem follows very closely that of Example 2.A.7 of the lecture book. If you get stuck on this problem, it is recommended that you review the solution video for Example 2.A.7 on the course website.


For the inclination angle used in the above simulation, we see that point B moves DOWNWARD along the vertical wall as A moves up along the incline. As B moves onto the same horizontal plane as A, the acceleration of B becomes very large (although A continues to move with a constant speed). Can you provide a physical explanation for this?

If we now consider a steeper inclination angle for A, as used above, we see that end B initially moves UPWARD along the wall; however, at some point B reverses its direction and begins to move DOWNWARD along the wall. Can you provide a physical explanation for this? Note also that the acceleration of B becomes very large as B moves onto the same horizontal plane as A, as it was for the initial value of inclination angle.

What is the value of the incline angle theta that defines the boundary between the types of initial motions for bar AB shown in the above two simulations? For the numerical value of the angle theta provided in the problem statement, which of the two simulations above agree with your results?


Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link above.

33 thoughts on “Homework H2.B - Sp23”

      1. Picky point, but this is not an "assumption"; instead, it is a consequence of given information. Since A moves with a constant speed along a straight path, A has zero acceleration. No need to assume.

    1. Recall the end B of the bar is constrained to move in y only, then can it have any velocity and acceleration in x? This problem is also similar to HW 2.A part b in terms of approach, so you may find some comments there helpful.

  1. I cannot figure out how to find angular velocity, but I know it should be in the k direction. Isn't angular velocity and velocity related somehow through R? Additionally, since the acceleration of A is 0, would the angular acceleration be 0 as well?

      1. I don't think that's right for this question. I found the angular velocity using the velocity equation. Since there's no velocity in the x direction for B, any components that affect the i component of B would cancel out, and this gave me the angular velocity in end.

            1. v_B is not relative to A.

              In the equation, v_B = v_A + omega x r_B/A, v_B and v_A are the velocities of A and B as seen by a stationary observer.

              omega x r_B/A is the velocity of B relative to A, as stated in the lecture book.

  2. Just to make sure, when calculating the position vector rb/a it should be the vector pointing from a to b correct? I feel that I did everything correct but my answers for velocity and acceleration came out in the positive j direction, which does not make sense.

    1. Yes, basically r_b/a is the position vector b with respect to a (I believe that is how it's said); this means that B is the final position and A is the starting position.

    1. I didn't use the arc length formula. It can be seen since 0 is directly below B and to the left of A. I did R_b = Rj and R_a=Ri. From here it can be seen that Rb/a = R_b-R_a. This gives all our equations to be in i and j

      1. Can we use this since it is no longer Rbi-Raj once it starts moving? In the example in class the vector was a function of the angle between Ra/b and the horizontal, but I'm not sure how we would get that angle in this problem.

  3. I looked at the example video 2.a.7 and I am confused - if a point is moving only in the j direction why does it follow that acceleration is in terms of j only? Does acceleration have to be parallel to the path of the point?

    1. Trisha,

      Think back to the second day of class. Then, we said that acceleration has two orthogonal components: one tangent to the path and one normal to the path. Here, "j" is tangent to the path, and "i" is normal to the path. The path is straight (rho = infinity), so the normal component of acceleration is zero. All that remains is the tangential component, and that is in the "j" direction. Hence, the acceleration must be in only the "j" direction.

      Does this help?

  4. I'm supposing omega and alpha are meant to be found using the velocity and acceleration of B. But what are we supposed to find v_B and a_B relative to? The origin or point A?

      1. But the equation for velocity is vB = vA + omega x rB/A. The position vector rB/A is vecotr rB relative to a point A. So, when I find the velocities in this problem, are the position vectors relative to the opposite ends of the rigid body or the still origin O? Does 'absolute' velocity & acceleration always mean relative to the coordinate origin?

        1. In that equation, v_B and v_A are absolute velocities (velocities seen from a stationary observer). In that equation, however, you need to use the position of B measured from the position of A, r_B/A.

          The term omega x rB/A is the velocity of B relative to A.

  5. Do we need to use the arc length? It would seem that we just need to multiply R by cos and sin to get the i and j components. Then once we get these components we plug them into the equation Vb bar = Va bar + omega x rb/a. From here break up the equation and solve for the i component since we know its zero. This would get us the omega value which we can then use to find j. Is this the correct thought process with this problem?

    1. Phillip: You need only the position vector from A to B: r_B/A = -R*i + R*j. No arc length is needed, nor does the angle theta come into play on that position vector.

    2. Nah, you don't need to use arc length. You pretty much just draw a line from point A to point B and then you'll realize that you have a right triangle. And then from there you just do the normal trig stuff to get the i and j components for r_B/A.

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