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Discussion
The animation shown above demonstrates the kinematics of the motion of the bar. (Note that in the animation velocities are shown in BLUE and accelerations are shown in RED.) In particular, we are reminded from this that:
- End A moves along a straight line path dictated by the angled wall. Since the path of A is straight, the acceleration of A is always aligned with the wall surface.
- End B moves on a circular path centered on point O. Here, since the path of B is curved, the acceleration of B will have, in general, two components of acceleration: one component that is tangent to the path of B and the second component that is normal to the path of B. This second component is the centripetal component of acceleration, and is given by aBn = vB2/R, where R is the length of rod OB. Since the bar is released from rest, vB = 0; therefore, the normal component of the acceleration for B is zero. As a result, the acceleration of B is tangent to the path of B( that is, perpendicular to OB) on release.
The above kinematics observations are seen if we freeze-frame the above animation at the time that it is released from rest. See below. Here we see the acceleration of A is down along the slot, and the acceleration of B is horizontal.
HINT: As always, we should follow the four-step plan for solving this problem.
STEP 1: FBD
STEP 2: Kinetics (here, Newton-Euler)
STEP 3: Kinematics. See discussion of this above. You will need to use the known directions of the accelerations of A and B in order to write down the acceleration of G needed for Newton's equation. Click on this link to see how this is done. See Example 5.A.13 of the lecture book as an example on how to approach the kinematics of a problem such as this with constrained motion for two points on a rigid body.
STEP 4: Solve
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