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DISCUSSION THREAD
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DISCUSSION and HINTS
After block B strikes and sticks to block A, the two blocks move together as a single rigid body. In deriving your equation of motion here, focus on a system with a single block (of mass 3m) connected to ground with a spring and dashpot.
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 A+B.
Step 2: Kinetics (Newton/Euler)
Use Newton's 2nd law to write down the dynamical equation for the system in terms of the coordinate x.
Step 3: Kinematics
None needed here.
Step 4: EOM
Your EOM was found at Step 2.
Once you have determined the EOM for the system, identify the natural frequency, damping ratio and the damped natural frequency from the EOM. Also, from the EOM we know that the response of the system in terms of x is given by: x(t) = e-ζωnt [C*cos(ωdt)+ S*sin(ωdt)]. How do you find the response coefficients C and S? What will you use for the initial conditions of the problem? (Consider linear momentum of A+B during impact.)
To find the response coefficient S, don't we need a numeric value for Vo?
Leave your answer in terms of v_0.
For the damped natural frequency, w_d, I seem to be getting an imaginary number. Is this something that should be possible?
I do not believe this to be possible, I would double check your damping ratio value as that should be under 1. This would allow you to solve for Wd with a reasonable value.
that is not correct, please check your solution. the damping ratio should be from 0<x<1 which is not the case for you
I don't think that would be possible since the ratio should be between 0 and 1.
This should not be happening since the damping ratio should be between 0 and 1. With this, when you square it and subtract it from 1 you should end up with a positive number. Take a look at the concluding remarks PDF for April 17th on the daily schedule on the blog and ensure that you are using the correct formulas.
It isn't. Check your calculation of damping frequency (which should not higher than 1)
I would double-check that your damping ratio is positive and less than 1. The only time when you'd get an imaginary number for w_d would be if your damping ratio is negative or greater than 1.
This problem is nearly identical to example 6.B.13 done in class. However, this time the lighter block collides with the heavier stationary block. This will change the equation for speed after collision.
Since we are finding the speed of A+B right after the impact, would it be fair to assume we must utilize LIM to solve for that?
That makes sense. Both objects begin stationary, and have velocities in relation to each other due to the impact. I'm assuming they both have the same speed after the collision because of the sticking nature.
I did the same as once they collided I treated it as one system with the same velocity.
Yes it's okay to use LIM because there are no external forces acting on the system when the system is defined as the two blocks. So you can find the mass and velocity in the initial condition and set it equal to the masses and velocity in the final condition.
Yes. I did that too.
Because A and B stick together upon impact, linear momentum is conserved, so you can use conservation of linear momentum to solve for part a.
After linear impulse momentum is used, the two masses are stuck together. The problem can then be treated like a normal spring problem with a mass of 3m having an initial velocity as determined by LIM.
we need to use the linear impulse momentum , in order to find the velocity at time 0 . then we only need to write our kinetics and kinematics to derive the necessary equations . good luck !
I kind of see what you're saying here, but I'm still confused as to what lead you to even think of that (it's an effective shortcut by the looks of it). Could you explain more on that?
When solving for x(t) later in this problem, it is important to know the initial conditions. Because our general form of x(t) has two unknowns, we need two known initial conditions. We already know the initial displacement condition, and then the LIM relationship for the impact of A and B can be used to find another initial condition.
Would the direction of kx be pointing to in the positive x direction since the motion is to the left? What is the theory behind this?
If the positive x-direction is defined as leftward, and we are assuming that the two blocks are moving in the positive x direction, then the forces of the spring and the dashpot would point rightward, as these forces always oppose the direction of motion (to return back to x=0, unstretched position).
From the collision, block AB has a velocity to the left when the spring is uncompressed. This means there will be a displacement to the left (positive as defined by the problem).
By assuming positive displacement, the spring is compressed. The compressed spring will apply a force on block AB in the right direction (negative).
What's keeping block B stuck to block A after the first oscillation? If the ground is smooth and B isn't physically bonded to A, wouldn't the spring and damper system send B outwards?
Because momentum is conserved during impact, you can use conservation of momentum for part a, and then treat both particles as one after impact for the rest of the problem. Parts b, c, and d are very similar to what we had to do for the last homework set that was due on Wednesday.
For this problem, it would be easier to not use the work energy theorem and instead use the linear impulse theorem.
What is the actual meaning behind the x(t) function or the theta(t) functions? What are they actually defining in the premise of the problem?
In this problem, x(t) represents the displacements of the particles with positive x being to the left. There is no theta(t) in this problem.
Those functions define the position or angle of the body as a function of time. In oscillation problems, they are usually a sinusoid. Taking the first derivative of the function gives you the velocity, and second derivative is acceleration.
For part D remember velocity at time 0 is the same as the velocity from part A
Refer to example 6.B.13 that was completed in class for a very very similar example on how to solve this problem. First, use conservation of linear momentum to find the initial conditions of the combined block and then go about solving your EOM.
Remember to use the masses of both A and B in the EOM because you are describing the motion after B sticks to A.
Would you be able to use the energy equation in this example, or is only linear momentum conserved?
When solving this problem the first step is to solve for the velocity after impact. This will become an initial condition and be equal to x_dot at 0 for part d of the problem.
Should we leave the answers in terms of the velocity of A+B right after impact, or in terms of the initial velocity of B?
I solved the problem in terms of Vo as it was a given variable in the problem statement.
I found it was most helpful to solve this problem by applying the linear impulse momentum equation to find the velocity right after the collision. For the remaining steps, I treated it as a normal spring/dashpot problem with a combined mass and worked through it that way.
Just for a bit of clarification, when considering the two blocks as the system, is the force of the spring and dashpot an external force, but just conservative force? I know this does not have an effect on this system but just wanted clarification, thanks!
This is a lot like the example done in class or in the lecture book 6.B.13. The velocity A and B will be equal to each other after they stick. Remember the equations change based off the type of damping, in this case it should be underdamped.
This question is almost identical to example 6.B.13. For this problem, it is important to consider an FBD for both upon impact and just after impact. You'll find that upon impact, the sum of forces in x is conserved, thus linear momentum in x is conserved. This will help you solve for the velocity after the collision. An FBD for just after the collision will help you find your EOM. Secondly, when solving for the coefficients of your equations, remember that the velocity value from part A is your initial condition for xdot(0).
This problem is very similar to problem 6.B.13, which was done in class. This example can be used as a framework of the steps needed to solve this homework problem.
Although it doesn't look that similar to the question that we did in class, it is. I would recommend setting it into two distinct parts to get the external force on the system.
Make sure to take the derivative of x(t) as it equals v0.
On an exam, if the problem defines a positive x direction I assume we will be required to use that definition?
That would be recommended, as the grader will be using a solution with that sign convention.
Is this a perfectly inelastic collision?
B sticks to A.
To find the speed after A sticks to B you know that linear momentum is conserved so you can use the linear momentum equation. To find the EOM I use the sum of the forces of the blocks after the collision.
In the case of the underdamped system, if we see that x(0)=0, are we allowed to say that C term is 0, or do we have to justify the assumption? On an exam, not having to derive the entire equation would save quite a bit of time, so I wanted to know how much work is required to be considered complete?
I think you would have to derive the whole equation in order to find the damping ratio. However I could be misunderstanding your question.
You should always derive the result, and not write down a result as that from memory. The problem with writing it down from memory is that you get no points if what you write down is incorrect. If you show your work and make an error, you can get partial credit.
Finding C first based on the x(0) condition helps to simplify the x(t) equation so that taking the derivative is simpler and then can use the x'(t) equation and the x'(0) condition to find S.
Assuming energy is conserved, I assumed that the kinetic energy or block A compared to the combined blocks were equal, which helped solve the velocity of v(a+b)
It is helpful to remember that the answer for part A is the velocity after impact in terms of the initial velocity. This means that the answer to A can be rearranged to solve for the initial velocity, which gives a condition needed to solve for the coefficients in part d.
Sorry I think I am incorrect. The answer to A is the initial condition and no rearranging is necessary.
Treat this as a conservation of momentum and only analyze the system immediately after impact right when they stick together.
This problem should first be separated into more distinct problems, a linear impulse momentum to find the initial velocity of the spring/damper problem. After finding that initial velocity of the spring/damper, techniques from the past few homeworks will be used, but remember to include the mass of both blocks as they stick together, and the entire 3m mass oscillates back and forth.
Thank you this along with following the format of 6.B.13 helped a lot.
When finding speed after A collides with B I noted that momentum is conserved and I used the linear momentum equation, I also found the EOM by sum of forces after the blocks collide.
It is useful to keep in mind that the velocity of block A found in part a) is used as the initial condition in the EOM.
Are we supposed to leave the answer in terms of v_0 as we are not given a numerical value for that quantity?
For this problem it is important to note that because we are lookcing at an instance right after block A and B contact, there is change in x and thus linear momentum is conserved. This is should simplify the problem greatly.
I found Example 6.B.13 to be especially helpful when solving this problem, since in both situations the Blocks A and B stick together after collision.
For the final part of the question, keep in mind that x is zero when A and B initially stick together (at time 0). This should help simplify solving for the derivative of the equation to find the other constant S.
On top of this, keep in mind that the other initial condition needed for this part was solved for earlier in part A.
Remember that the initial velocity for x(t) is not equal to the initial velocity given in the problem. To find xdot(0), use the linear impulse/momentum equation for the system consisting of A and B with no external forces in x
This problem goes back to using the conservation of momentum principles. The results you get of velocity based on the momentum, comes in handy when solving for the final equation of this problem. This problem is not particularly difficult, as long as you keep track of your signs and forces of motions.
We solved for the velocity term in part a so remember that when solving for term S in the response equation
In this problem, it is important to use conservation of linear momentum to your advantage. You must take into account that the two blocks will stick together when calculating the momentum. After the collision, the two blocks can simply be treated as one mass of 3m moving at one velocity. This velocity value is crucial later on in the problem when determining the coefficient values C and S.
It's important to check that your damping ratio for this problem comes out as underdamped to be sure you're proceeding correctly.
I found it useful to use work-energy equations for part one. Because there are no external forces on the system horizontally, there is no non-conservative work and linear momentum in the x direction is conserved before and right after impact.
I used momentum equations as I was under the understanding that energy is not conserved in an inelastic collision, which this is since the two blocks stick together. When I solve for step one using work energy, I get an answer that is a factor of sqrt(3) off from mine
Please recall from our work in Chapter 4 that energy is NOT conserved during impacts. Because of this, you should not use the work/energy equation.
For help with this question, problem 6.B.13 is extremely useful to reference. It is also important to note that the blocks stick together so for the FBD they can be treated as one object and linear momentum is conserved.
For help with this question, problem 6.B.13 is extremely useful to reference. It is also important to note that the blocks stick together so for the FBD they can be treated as one object and linear momentum is conserved.
I thought that it was very important to clearly define the forces and torques and their relationships with the system's components (like the cables, drum, and connectors). Knowing all of those can help us derive the equations of motion (EOM). This process not only helps in solving the problem but also enhances our understanding of the physical interactions at play.
For this question, I started with linear impulse momentum to solve for the final velocity of the two block system in terms of the initial velocity of block B. Then I used the sum of forces in the x direction (newton) equation to find the equation on motion. Once I found the equation of motion I solved the rest of the question following similar procedures as example 6.B.13 in the book.
You will have to use your answer from part A as an initial condition when solving for the response coefficients.
This problem is fairly straight forward. However, you will need to utilize linear impulse momentum to determine the final speed of block A after both blocks have collided and have stuck together. To complete this pay attention to the mass of block A before and after the collision.
Remember that all the variables that need to be solved for in part B are simple formulas, using parameters that have already been found. These same parameters are used to find the EOM in part C.
In analyzing the collision between blocks A and B, the conservation of momentum principle helps determine the final speed of block A after the collision.
you can use conservation of linear momentum to solve the xdot(0) value of A+B right after impact
This question can be approached by using the LIM to find velocity after impact. From there, this velocity can be used as an initial condition while solving for x(t).
If you notice when drawing the FBD during impact, there are no forces in the x-direction, meaning the sum of forces in the x is zero. Consider using LIM to help you find the velocity of the system after impact. Next, It's helpful to draw a separate FBD to analyze the system after impact. This FBD and the equation(s) that go along with it can help find the EOM.
The illustration seems to be a bit confusing. How does the block B stay attached to block A. Is in an inelastic collision where it stays stuck or is there an external force?
Yes, the collision is inelastic. There could be a physical latch on block A that captures block B on impact. Or, there could be some sticky material on A that keeps B stuck to A after impact. Or, ... There are a number of ways that this could occur.
When dealing with this inelastic collision (assuming thats what it is) 6B.13 is quite useful when it comes for help solving this problem.
It is important to remember that the this problem is right after the blocks collide. So when finding the velocity right after collision we can just use conservation. In addition, after we look at the two blocks as just one 3m block.
Using the LIM equation, the linear momentum of the system of blocks A and B is conserved, which helps to find the speed for Part A.
Remember that the conservation of linear momentum can be used to find the velocity in part A of the problem. From there, you can use EOM analysis to solve.
In order to solve this problem I found it helpful to first solve for velocity after impact. Once you have this the rest of the problem can be completed because you can set it equal to x_dot which is at 0.
The wording “of A immediately after B sticks to it" is important. Understand that both blocks act as one rigid body once they are stuck together. This means they have a total mass of both blocks added together and one velocity. This should help to solve part A using the LIM equation.
To solve this problem it is best to not overthink it. We want to solve using our knowns where we use linear momentum update to get the velocity at state 2. We then solve normally by now stating we are at state 2 progressing forwards which allows us to deduce the equation of motion. Then we must solve for our Acos and Bsin equation on the understanding the initial position is 0 (resulting in the cosine term disappearing) and then solve by taking the derivative and setting it equal to V(0)/3 (the velocity at state two). After this, it is quite easy to solve.
Hope this helps.
This problem is fairly simple using LIM and then following the regular steps for finding EOM.
For this homework assignment, I referenced example 6.B.13, as it followed a similar level of analysis and setup with two different FBDs that allow you to obtain the initial condition velocity using linear momentum equations (excludes the dashpot and spring to lead to a sum of forces in x of 0) and the other when including the spring and dashpot to obtain the components for the EOM. Since this situation is also underdamped, a similar derivation for the x(t) and x_dot(t) equations is very similar in form to the ones I found for the homework, with a similar setup in enforcing initial conditions.
If x is defined as positive to the left, and the forces of the Spring and dashpot act to the right, then would they be listed as negative on the Newton equations?
LUKE: Yes, that is correct.
Do not be afraid to leave your answers in terms of v_0, as we are never given a value for it.
Remember to sub in the initial velocity value we found at the beginning of the problem, for part d, to find your C and S coefficients.
equations pertaining to this and similar problems are:
w_n = sqrt(k/m)
c=2sqrt(km)
YT=c_coef/c
w_d=w_n*sqrt(1-(YT)^2)
Remember that the velocity of A after the collision is the initial velocity for the characteristic equation.