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Spring max compression

  1. Dec 7, 2013 #1
    Hey guys


    If I have a cylinder of mass 5 kg and release it along the vertical axis onto a spring with stiffness coefficient (k = 1.8 kN), how would I need to set up the work/energy equation(s) to find the maximum compression and deflection of the spring. And what would the maximum velocity of the cylinder be when the cylinder is located 0.1m above the spring.

    Work/energy relation
    U'A-B=ΔT + ΔVg+ΔVe
    ΔT = 1/2*m*v^2​
    ΔVg = m*g*h​
    ΔVe = 1/2*k*x^2​

    Have the first part for compression calculated by

    -m*g*x = m*g*0.1 - 1/2*k*x^2
    -5*9.81*x = 5*9.81*0.1 - 1/2*1800x^2
    Computed to give: 0.1059m (matches my answer sheet)

    But I'm struggeling with how to set up a formula for the max velocity and deflection for that speed. Have tried the following for velocity

    0 = 1/2*m*v^2 + m*g*compression - 1/2*k*compression^2
    0 = 1/2*5v^2 + 5*9.81*0.1059 - 1/2*1800*0.1059^2
    Computed velocity to be: 1.39 m/s (but this value doesn't match with the answer sheets - should be 1.493m/s)

    Not sure with calculating the deflection


    Thanks and keep on rocking...
     
  2. jcsd
  3. Dec 7, 2013 #2
    Hello,

    What happens when system comes to rest?
     
  4. Dec 7, 2013 #3
    Hi

    The assignment doesn't mention anything further... My assumption is that it stays at rest :) so the spring remains compressed
     
  5. Dec 7, 2013 #4

    Doc Al

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    OK, that's the point of maximum compression. (What's the speed of the cylinder at that point?)

    The point of maximum speed is completely different than the point of maximum compression. Hint: At what point will the falling cylinder stop accelerating and begin to slow down?

    Not sure what is meant by that question. The max compression is the max deflection. (Not sure what they have in mind if they mean something else. What does the answer sheet say?)
     
  6. Dec 7, 2013 #5
    Hi

    Thanks for reply... Well to my understanding of physics, the cylinder will stop accelerating when it hits the spring and slow down, so I'm going for this now

    0 = 1/2*m*v^2 - m*g*gap size
    0 = 1/2*5*v^2 - 5*9.81*0.1
    Computed velocity to be: 1.4 m/s (It's nearer the given value but still a slight difference)

    The deflection for the last question is given to be 27.2 mm. In the diagram drawing it shows it as the distance for how much the spring will compress (presumably at the calculated speed)

    I have calculations that gives me that value but, it is indicated as an imaginary number.
     
  7. Dec 7, 2013 #6

    Doc Al

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    Hint: Something stops accelerating when the net force on it is what? What forces act on the cylinder?
     
  8. Dec 7, 2013 #7
    When net force is at 0 theres no acceleration, hence a = Fnet / m.
    Forces acting on the cylinder would be gravity downwards and the spring impact/normal force upwards. Air resistance not included.
    But, am I missing a point (probably) in how to calculate the values correctly?

    I know of Hooke's law (F = k*x), so the deflection can be found as x = F/k <=> x = (m*a) / k, but the idea of the assignment is to the use the work and energy equation(s) to calculate the different values. Using Hooke's equation I get the 27.2 mm given in the answer sheets, but would rather be able to calculate it the other way around for the sake of practice and understanding.
     
  9. Dec 7, 2013 #8

    Doc Al

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    That 27.2 mm is the answer to what question?

    Did you find the maximum velocity?

    To find the point where velocity is maximum using energy methods, start by writing an expression for the kinetic energy of the cylinder as a function of x (the spring compression). Then find the value of x that maximizes that KE.
     
  10. Dec 7, 2013 #9
    This is also what I meant (F=k*x) when asking you what happens when system comes to rest.
    It's a situation when system is resting, why would you use energy equations?
    When spring is under influence of [itex]E_{k}[/itex] and [itex]E_{p}[/itex] there is greater compression than when they vanish.
     
  11. Dec 8, 2013 #10

    Doc Al

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    Ah, I missed this clarification. So that's the deflection at max speed, not the maximum deflection of the spring. Anyway, that value is correct. Do you understand why and how to get it? Show how you got an imaginary number.

    As I said before, you can determine the point of max velocity using dynamics (finding the point of zero acceleration and thus max speed), and then use energy methods to calculate that max speed. Or you can determine that point purely using energy methods (see my earlier post).
     
  12. Dec 8, 2013 #11
    Have drawn a picture of the situation here, if I have not been clear in my explanations.
    http://postimg.org/image/9uq1te8m7/

    I have found the maximum compression to be 0.1059m - matches the answer sheet... yaey

    I have found the maximum velocity to be 1.40 m/s - matches not quite the answer sheet (1.493 m/s) by converting the potential energy stored in the cylinder to kinectic energy using
    1/2*m*v^2 = m*g*h
    1/2*5*v^2 = 5*9.81*0.1
    Computes the maximum velocity to: 1.40 m/s

    It is given to be the amount of corresponding deflection from the maximum velocity in the answer sheets.

    Just to know and be able to understand how to calculate it different ways... The assignment is all about work and energy.
     
  13. Dec 8, 2013 #12
    Hey Doc Al

    I get the imaginary value by saying

    m*g*Δx = -m*g*h - 1/2*k*x^2
    5*9.81*(0.1-x) = -5*9.81*0.1 - 1/2*1800*x^2
    Computes to: ±0.02725+0.10078411333141746701i

    I have tried to fiddle around with the signs infront each, if that was the case but with no luck... :(
     
  14. Dec 8, 2013 #13

    Doc Al

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    You just found the speed of the cylinder as it first hits the spring (after falling 0.1 m). That's not the maximum speed.
     
  15. Dec 8, 2013 #14

    Doc Al

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    Why do you have a minus sign in front of the spring PE term? That's your error.

    Done correctly, this will give you the point of maximum compression--which I thought you already found. You are setting the initial energy (all gravitational PE) to the final energy at max compression (all spring PE).
     
  16. Dec 8, 2013 #15
    Hmm, I don't get the physics there then. How would the velocity still be increasing when it hits the spring?

    If I remove the minus infront, I get a value of 0.0545... Which in fact is the double value of what I'm seaking for. But then I'm not sure how to go further with that.
    If I keep the minus sign, then I get the value 27.2 mm seeked for (sort of - still imaginary). In other words, I don't quite see how to combine the work and energy equations to give me correct result
     
  17. Dec 8, 2013 #16

    Curious3141

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    Homework Helper

    The restoring force of the spring acts vertically upward on the cylinder from the time it begins to compress. What constant force acts vertically downward on the cylinder throughout? In the initial phase of the spring's compression, which force is the greater? Hence what is the direction of the net acceleration on the cylinder in this initial phase? What can you therefore conclude about how the velocity of the cylinder is changing in this phase?
     
  18. Dec 8, 2013 #17

    Doc Al

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    The velocity will keep increasing as long as there is an net downward force on the cylinder. At some point (what point?) the net force goes to zero and after that point the net force is upward, finally slowing things down.

    Not sure what you are doing. When you use (correctly) that equation you are finding the maximum compression, not the compression where the speed is maximum.

    Show your reasoning for using that equation with a minus sign.
     
  19. Dec 8, 2013 #18
    Mass * gravity is exerted on the cylinder downward, so (I'm a bit guessing here) this is then greater in the initial phase of the springs compression. So the net acceleration on the cylinder is still downwards at initial phase. Thereby the cylinder won't go fully stop, but gradually decrease its velocity which reflects in more travel distance so the velocity will be a bit higher than presumed before...

    Hmmm, physics is fun when you (hopefully) understand it :)
     
  20. Dec 8, 2013 #19
    I don't have a real reason for that equation other than it (by trying around) gave the value 27.2 which is the only way it got, at some extend - except for Hooke's equation. I'm aware the equation is wrongfully set up, but haven't been able to formulate the right equation with work and energy equations to calculate that "exac"t value
     
  21. Dec 8, 2013 #20

    Curious3141

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    Right. All of it so far.

    Actually, in the initial phase (until when?), the cylinder will continue to increase in downward velocity, albeit at a progressively lower acceleration. In other words, the cylinder still speeds up despite the spring being compressed, but the rate of its speeding up goes lower and lower. At one point (when?) this rate becomes zero momentarily. What can you say about the cylinder's velocity at this point?

    In Physics, it's good to always talk yourself through a qualitative understanding of a problem before you start writing equations (let alone numbers) down. If you cannot accurately state what's actually happening, you don't actually understand the problem, and the rest becomes meaningless.
     
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