Homework H3.B - Fa22

Problem statement
Solution video

DISCUSSION THREAD

Any questions??


Discussion

Suppose that you decide to use the following velocity and acceleration equations:
vA = vO + (vA/O)rel + ω x rA/O
 aA = aO + (aA/O)rel + α x rA/O + 2ω x (vA/O)rel + ω x (ω x rA/O)

using an observer that is attached to link OB. With this choice, the angular velocity and acceleration of the observer are those of link OB:
ω  = ωOB k
α = α
OBk

Now to the question of what motion does the observer see for point A. The observer see A moving back and forth along the x-axis, where the x-axis is attached to OB:
(vA/O)rel = b_dot i
(aA/O)rel = b_ddot i

Note that A is traveling on a circular path of radius R at a constant speed vA. What does this say about the velocity and acceleration vectors for A and B,  vand aAat this instant? Watch the animation above to confirm your response.


 

6 thoughts on “Homework H3.B - Fa22”

  1. Since the problem states that end A of the arm is moving with a constant speed v_A, does that make the acceleration of the arm at A equal to zero? Because the animation shows an acceleration vector at A throughout the entirety of its motion, even at the instant shown in the problem. Or since the entire arm is moving about the slot with radius R there is the acceleration at end A in the positive x direction and it would just be the acceleration in the y direction that equals zero?

    1. Ethan: Think back to the second day of class when you were studying the path description for kinematics. On that day, you learned that the acceleration of a point moving on a curved path has two distinct components: one tangent to the path (rate of change of speed) and one normal to the path (centripetal component). The problem states that A is moving with a constant speed; therefore, the rate of change of speed of A is zero, and A has ONLY a centripetal component of acceleration. The component that you see in the animation is always perpendicular to the path, and is the centripetal component: (v^2/rho)*e_n.

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