| Problem statement Solution video https://youtu.be/NTuAy-7KsLA |
DISCUSSION THREAD
Ask your questions here. Or, answer questions of others here. Either way, you can learn.
DISCUSSION and HINTS
A horizontal force is applied to the center of mass of the thin bar, with the bar attached to a block that is constrained to have horizontal motion. Here want to know the resulting angular acceleration of the bar.
Recall the following four-step plan outline in the lecture book and discussed in lecture:
Step 1: FBDs
Draw individual free body diagrams (FBDs) of the bar and block.
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 horizontal direction. 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 A to the acceleration of G:
aG = aA + α x rG/A - ω2rG/A
Combining the above kinematics equations will provide you with the relationships among aGx, aGy, aA and α.
Step 4: Solve
From your equations in Steps 2 and 3, solve for the angular acceleration of the bar.
We should draw reaction forces at A for the separate FBDs, correct?
Yes.
Parallel axis theorem would need to be utilized in this problem right?
Depends on your choice of point that you choose for your moment (Euler) equation.
If you choose the center of mass G, then no need for the PAT. This is recommended.
If you choose A, then, yes, you will need the PAT. A is NOT recommended since you cannot use the short form of the Euler equation with A.
So when you are solving for your I value, you would only take into account the beam then? Not the block?
As has been discussed in class, the Newton/Euler approach needs one to draw individual FBDs for each of the two bodies. From each FBD can be written the Newton/Euler equations for that body.
When you write down the Euler equation for the bar, you will use the mass moment of inertia of just the bar since that equation applies to just the bar.
At the instant shown in the problem, G has no acceleration in the y direction right?
I think you are correct since the Force is applied in the horizontal direction when the object is at rest, the only initial acceleration would be in the x direction. If you write a newton equation in the y direction, you only get normal forces and gravitation forces, which would have to cancel out to zero otherwise the system wouldn't be at rest.
If in doubt, write down the acceleration equation relating the motion of points A and G. This will give you the answer to your question. As Garrett points out, since the bar is at rest when the force is applied, its angular velocity is zero, and your kinematics equation will then show that G has zero acceleration in the vertical direction. If the system is NOT at rest, G WILL have a non-zero vertical component of acceleration.
can i make the assumption that omega would be zero?
Not an assumption. The system is at rest.
We usually take angular acceleration to be in the +k direction so I assume that applies here as well but is there reason why it is usually in the +k direction, how can we visualize that?
I think about in terms of the axis of rotation. Because angular acceleration is not a linear motion, it is described based on the axis the motion is around. In 2D planar problems, this axis is out of the page (the k direction). You could also accept the right-hand rule which states that rotational motion is in the direction of the cross product of the planar vectors (i and j in most cases).
Since the bar is rotating around the end, the moment of Inertia will be 1/3 M L^2, correct?
For a bar I of G should equal (1/12)mL^2 which you can then plug into I*alpha to solve the moment about G
Or is it 1/12 M L^2 because we are looking from the perspective of G?
Keep in mind that Euler's equation says that:
M_G = I_G*alpha
That is, if you sum moments about the center of mass (as you should) you need to use the mass moment of inertia about the same point G. Here, that would be I_G = (1/12)*m*L^2.
When relating the acceleration of G to A, is it assumed that at this position a_A = 0? Or do we have to treat it as an unknown variable?
The acceleration of A is not zero at this position. This acceleration should be treated as one of the unknowns in the problem. It will appear in the Newton equation for the block. This acceleration is related to the acceleration of the center of mass of the bar through the rigid body kinematics equation.