| Problem statement Solution video https://youtu.be/a0EJn-XhJLY |
DISCUSSION THREAD
NOTE: Please leave your final answer in terms of variables defined in the problem, such as: M, m, g and ι. Also, the bar moves in a VERTICAL plane.
Ask your questions here. Or, answer questions of others here. Either way, you can learn.
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.
Is "l" the full length of the bar? The instructions weren't completely explicit on that.
Yes, it is the full length.
Can we write the variable as "L" or does it have to be "l"
You may use either.
aGx and aGy become in terms of the angular acceleration after doing the kinematics, does that mean Na and Nb are also going to be in terms of the angular acceleration after we combine the equations?
True. Keep in mind that your final answers for the normal forces need to be written in terms of the given parameters of the problem, such as: M, m , L, g and theta.
Do we assume that the bar is released from rest?
Yes.
Is there gravity in this problem? The picture on the blog shows it, but the PDF does not.
Please see the note above regarding this.
Where is B, these instructions are not clear
B is at the other end of the bar that is against the ground - opposite from A.
B is on the opposite end of the bar from A. It appears that the figure picked up extraneous shading in the copy process, and this covered up the “B”.
since the bar is released from rest, does this mean the angular velocity is zero?
Yes. omega = 0. However, alpha Is not zero.
Is the center point of the bar a pin/ would it have reaction forces like at A?
The center point of the bar is the center of mass of the bar. It is not a pin to ground. If it were a pin to ground, the bar cannot move.
How do you find alpha for this problem?
nvm got it
You will have three kinetics equations (two Newton and one Euler). Alpha will be an unknown in the Euler equation. In Step 4 you will solve the thee kinetics equations along with kinematics to find alpha and the reactions at the ends of the bar.