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Please help w/ mass, distance, energy = ?

  1. Apr 18, 2007 #1
    For anyone who replies, tx in advance,

    I have a valve with a pin, being struck by a hammer which is energized by a spring over a distance of about 8 inches. What I am trying to find out is between the PSI of 400-800 in the valve, whether a lighter hammer will be more efficient in opening the valve more, with the same spring putting pressure on the hammer. In this case, what would be more important speed or momentum? Equations are what I would like, I can't seem to figure out the word I am trying to find the equation for. "weight" or "energy", "power" "work". None seem to fit the equation I am looking for.

    Again tx,
    Todd
     
  2. jcsd
  3. Apr 18, 2007 #2
    I'm not an expert, but shouldn't the spring give equal energy to both hammers? The light hammer will be moving faster at the end of the path of the 8 inches, but have the same kinetic energy as the slower, more massive hammer. So the force acting on the pin of the valve should be the same.

    E (kinetic) = 0.5 * m * v^2
     
  4. Apr 18, 2007 #3
    I sort of was assuming the same thing, but even if the force/energy (?) is equal the action on the valve will be different. If there is an equation for what I am looking for I can at least make an edjucated guess. What I do think is the more massive hammer will open the valve less for a longer period of time, and the lighter hammer will open the valve more for a shorter period of time. The problem is that I am also having trouble figuring the speed of either hammer. I have many constants so exact #'s aren't greatly important.
    1) spring (force) x (?)
    2) distance = (?) If Iknew the action the the hammer was doing at the end of its travel ?
     
  5. Apr 18, 2007 #4
    All the equatuions that I think could be useful have "time" involved. It is very hard to measure tthe speed of this hammer, I suppose if one could tell me the equation for finding time based on force, mass and distance?
     
  6. Apr 18, 2007 #5

    Hootenanny

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    Okay, here's my take. We know that the spring will impart an energy of 1/2kx2 on the hammer. We can write the kinetic energy (T) of the hammer as;

    [tex]T = \frac{1}{2}mv^2 = \frac{1}{2}\frac{p^2}{m}[/tex]

    Hence, combining the two equations we obtain;

    [tex]\frac{p^2}{m} = kx^2[/tex]

    [tex]p = \sqrt{mkx^2}[/tex]

    Now, the impulse (I = Ft) is equal to the change in momentum. If we assume the force the hammer exerts on the valve is constant and the hammer is at rest after the collision (both reasonable assumptions) then we can write;

    [tex]I = \sqrt{mkx^2} = x\sqrt{km}[/tex]

    So a larger mass will result in a larger impulse.
     
  7. Apr 18, 2007 #6
    In the first part you right k as mass. Then you rewrite m as mass, does that mean later k = spring constant. and what is x at the end? I have some physics but I think you lost me
     
  8. Apr 18, 2007 #7

    Hootenanny

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    k was never mass, k was always the spring constant. Recall that the potential energy of a spring is PE = 1/2kx2, where x is the extension/compression of the spring.
     
  9. Apr 18, 2007 #8
    In your first reply you have veloicty in an equation, and when you combined the two v2 disappeared (?). You're equation seems to be correct except it's not what I am looking for. I want the energy (joules)(?) of an object (mass) from rest when at a set distance with a specific force applied applied. Must I calculate the velocity of the object at the set distance, and if so it seems to me that I will need to know the time it takes 1 of 2 objects to reach said distance.
     
  10. Apr 18, 2007 #9
    Please help: force on mass = ? at a set distance

    I am trying to measure the force/energy an object(mass) will apply or have when it reaches a certain distance when a specific force is applied to it while at rest. 1)object(mass)(variable) 2) force applied (constant)(spring) under tension 3) at a set distance. Anything hopefully without velocity and time is greatly appreciated. Time is hard to measure as is the velocity. If many difficult equations will be required a key would be appreciated. Also looking to know if the object when at said distance is considered energy, impulse, momentum? TX in advance.
     
  11. Apr 18, 2007 #10
    I did not understand what you were trying to ask. Please clarify the question.
     
  12. Apr 18, 2007 #11
    The test is hardly a test, but an answer is needed for a different forum. When a hammer coiled by a spring strikes an object (valve pin) 8 inches away it strikes with what? (force) and how to calculate it, like calculating a crash but velocity time and accleration are unknown.

    Then a different hammer lighter or heavier coiled by the same spring strikes the valve pin at the same distance...Which one strikes harder longer? I believe that a heavier object, slower object will hold the valve open longer but open it less and the smaller hammer will open the valve more for a less amount of time. This has to do with airflow and dwell, BUT if I can decide which one strikes harder (with more energy?) while the spring and distance remain constant. (Efficiency and recoil)
     
  13. Apr 18, 2007 #12

    Doc Al

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    (I merged the two threads--please do not create multiple threads on the same topic!)
     
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