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DISCUSSION and HINTS
The motion of the bar is constrained by the two horizontal and vertical guides for the ends of the bar. These constraints add to the complexity of the kinematics (Step 3) of your solution, as we will discuss below. Take particular note of the direction of the acceleration of the center of mass G of the bar as it moves.
Recall the following four-step plan outline in the lecture book and discussed in lecture:
Step 1: FBDs
Draw a free body diagram (FBD) of the bar.
Step 2: Kinetics (Newton/Euler)
Take care with your Euler (moment) equation. The point about which you write down the moment must be either the center of mass or a fixed point. For example, you can NOT write ΣMA = IAαAB because A is neither a fixed point nor is it the center of mass.
Step 3: Kinematics
End A of the bar is constrained to move only in the vertical direction. The other end of the bar (let’s call it B) is constrained to move only in the horizontal direction. The first part of your kinematics should be directed to relating the motion of A and B through the rigid body acceleration equation:
aB = aA + α x rB/A – ω2rB/A
Since the Newton equation for the bar requires the usage of the acceleration of the center of mass G, you will also need to use a kinematics equation relating the acceleration of either A or B to the acceleration of G. For example, you could use:
aG = aA + α x rG/A – ω2rG/A
Combining the above kinematics equations will provide you with the relationships among aGx, aGy and α.
Step 4: Solve
From your equations in Steps 2 and 3, solve for the reactions NA and NB acting at ends A and B of the bar.