| Problem statement Solution video |
DISCUSSION THREAD
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DISCUSSION
As always, we should follow the four-step plan:
STEP 1: FBD
Draw an FBD of the bar. Since the support at B is smooth, the reaction on the bar at B will be perpendicular to the bar.
STEP 2: Kinetics
You should write down the two Newton equations, and the Euler equation. As always, take care in choosing your reference point for Euler's equation. Since we have no fixed points for the bar, you should choose the center of mass G as your reference point.
STEP 3: Kinematics
Note that the path of the center of mass G is tangent to the surface of the bar (that is, the bar can move only along the support, not into the support). With the bar being released from rest, the acceleration of G is tangent to the path. Specifically, the acceleration of G is directed along the direction of the bar. Write down the rigid body kinematics equation that relates the accelerations of A and G. This vector equation represents two scalar equations (x- and y-components).
STEP 4: Solve
At this point, you will have three kinetics equations and two kinematics equations, for a total of five equations. You will have five unknowns. Solve!
I found it useful to start by using the moment about G, the forces in the x and y directions, and acceleration kinematics to find alpha.
Using this method I was still short an equation. I used sum of forces for x and y, sum of moments about G, and used acceleration kinematics to relate the acceleration of G to A. Is there some assumption I am missing? obvious A is constrained to only move in the x direction but is there something else?
I am also in the same boat, I have acceleration of g in terms of acceleration of a and alpha but no way to solve for acceleration of a.
Please re-read the four-step plan discussion above. In particular, focus on the STEP 3 discussion. This discussion leads to the conclusion that the acceleration of G is along the bar; that is, the acceleration of G does not have two unknown components - it has a magnitude of acceleration directed along the line of the bar.
Should there be Bx and By reaction forces?
Yes. When you combine them (as in B=sqrt(Bx^2+By^2)), the reaction force B will be perpendicular to the bar. Using the given length of the bar and the height of the guide, you should be able to use trig to figure out the exact values of Bx and By in terms of B.
Additionally, can we say that acceleration of a is purely in the x-direction?
Acceleration of a should be solely in the x, I believe.
One important note that confused me for a second is that the bar has no thickness, and point b is touching the center of mass of the bar. This means a contact force at B goes through the center of mess of the bar, and has no affect on the moment equation.
I seem to have 5 equations and 5 unknowns (assuming that the acceleration of a in the y is 0) but the system is still unsolvable. Where might I be going wrong? I've got the kinematics in X and Y, the sum of forces in X and Y, and the moment about G.
It may be hard to tell without seeing your equations, but have you related the x and y components of the acceleration of G? Since G travels tangential to the horizontal support B, you know how the x component and the y component of G relate to the total acceleration of G. This may give you another equation for solving this problem.
Will w be zero since we are determining the angular acceleration upon release?
Yes.
Yes. w will be zero since it starts at rest.
Given that the normal force on the bar from the guide B is perpendicular to the bar, shouldn't we need the angle that the bar is initially resting at to express the normal force from B in its x and y components in the Newton equations?
You can solve for angle the bar sits at theta with the given height to the center of gravity, b, and the length of the bar to the center of gravity, L/2.
Should I assume the acceraleration of point A and G to be positive or negative when doing the kinematics for this problem?
I found it helpful when solving to write and assume them in the direction they were going in my coordinate system based off of thinking about how the bar would move and the discussion. This way I didn't have to correct +/- later on in my solving. I could just plug things in as I went.
Something that helped a lot was putting acceleration of G in the x and y directions in terms of just acceleration of G using trig. This eliminates the need to solve for the acceleration of G in both directions.
I found it useful to use the equation sum of moments about A as well as the equation for sum of moments about G. Using this equation along with the sum of forces and the acceleration equation relating A and G allowed me to solve the system for the angular acceleration.
I could be wrong but I found it interesting how I did not use the i hat component of the kinematics equation as the j hat component relates the acceleration of G and alpha.
The way that I found best to approach this problem was by taking the moment about G and the sum of the forces in both the x and y directions. It important to note that the acceleration of G is along the bar as talked about in the hints. Taking this into consideration, we should be able to use the equations to solve for alpha.
While the hint says you will end up with 5 equations and 5 unknowns, it's more reasonable to say it's 6 equations and 6 unknowns. The final equation you should come up with will be based on how the acceleration of point G is tangent to the force applied by B onto G. This way you can relate the acceleration of point G in both x and y direction to each other.
The trick to this question is two fold, firstly the acceleration of G occurs along the bar down to the left and the acceleration of A exists only the x direction. Knowing this, if you drew your normal at B as two reaction forces you can equate these to a single normal using trig functions.
Is it safe to assume that the angular velocity is zero when we are relating the angular accelerations for A and G?
It is given that the bar is at rest. So, it is not actually an assumption.
Does the acceleration of G being tangent to the force of B indicate that Nb=MAg? I'm not sure I fully understand how to form an equation from this relationship.
You do not need to know that a priori. N_B will be one of your unknowns that appear in your Newton/Euler equations. Solve for that just as you would for the other unknowns in the problem.
Is it fair to say that we can rewrite the coordinates for x to be along the bar and thus Agy is zero? I am in the kinematics and because of switching the coordinates I am having trouble with writing acceleration of A in terms of x and y.
You are free to define your coordinate axes in whatever way that you feel is best. I do not see a real advantage of one orientation over another. The simplicity gained from one orientation is sometimes diminished by other factors.
It is useful to remember that the angular velocity is equal to zero. The reason being is that the system starts at rest, this should help to simplify the equations when solving.
It is important to remember that the acceleration of G is along the path of the rod tangential to B so you can split it up into components using sin and cos. This helps simplify 5 equations and 6 unknowns to 5 equations and 5 unknowns.
Is the initial angular velocity zero since it starts from rest
I found this problem to be similar to the first problem on the most recent exam. Since we know that aG is along the bar, we can set the magnitude of aG equal to a function of alpha. Then, we use the Newton equations to find the normal force at A and Euler's equation to solve for alpha.
Applying the normal force at point A. When taking the moment Rember to use the force component perpendicular to the Bar.
For people doing this again for further clarification:
I initially solved but did not include the normal force of B. There are two components both in the x and y directions. If you account for the bar alone and solve for these forces it makes a lot more sense. Keep in mind that you can also check your sanity for signs as the final answer should be negative due to the spinning going into the page based upon right hand rule.