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Linearization and energy equation

  1. May 18, 2013 #1
    Hey guys.

    We were given a problem, which was to consider bv^2 as the force acting upwards on a body falling in water. We were asked to find the depth at which a body entering the water at 5 * ve (terminal speed) would end up with a speed of 1.1 * ve.

    Starting from -bv^2 + mg = ma, the first step is to linearize the non-linear expression bv^2 at the equilibrium point (ma = 0).

    The equation at the chosen equilibrium point becomes b(ve)^2 = mg

    The linearization gives:

    b(ve)^2 + 2b(ve)(Δv).


    Substitution into the original equation gives -b(ve)^2 - 2b(ve)(Δv) + mg = ma.

    We know that at the chosen equilibrium point, b(ve)^2 = mg. Thus, mg - b(ve)^2 = 0. I can remove these 2 expressions in the above equation to get the linearized equation at the equilibrium point ve:

    Linearized upward force at terminal speed = 2b(ve)(Δv) = ma

    Also our teacher showed us that the 2b(ve) expression must be used to find the value of the linearized ve point. Thus, using the equilibrium equation above but substituting b(ve)^2 by the new constant coefficient 2b(ve) which multiplies ve, we get 2b(ve)*ve = mg -> ve = sqrt(mg/2b)

    Given the above equation in bold, I was able to solve for height h by using the following substitution:

    a = v * dv / dh

    Using this substitution I could integrate on both sides of the above equation and solve for h pretty easily. I confirmed with my teacher that our answer was correct.

    Here's my question:

    I would like to solve this problem by using the conservation of energy equation instead of solving the F = ma equation. Before my teacher confirmed that we had the right answer, I wanted to solve the problem using this other method because I could've verified my answer by myself. However, I don't know how to solve this. I put the initial energy + the work on the left hand side, and the final energy on the right hand side.

    Assuming the object falls into the water at a height h at an initial speed of 5ve,

    V1 = mgh
    T1 = m/2 * (5ve)^2
    Work = 2b(ve)(Δv) * h

    V2 = 0
    T2 = m/2 * (1.1ve)^2

    V1 + T1 - Work = V2 + T2

    I don't know what to do with the expression in bold, for the work of the linearized force. Obviously an integral is needed, but if I integrate as I specified above, I have to divide both sides by Δv since dv is on the right side of the equation. I end up with 2b(ve)h = stuff. The left side of this equation is NOT equal to the Force times acceleration because the force has a Δv on the left side.

    So, how can I find the "correct" work to put into the equation?
     
  2. jcsd
  3. May 19, 2013 #2
    I understand linearisation is mostly used in engineering, since real-life systems must use these approximations. I'm going to assume this is why physicists aren't really comfortable answering my question. Does anyone know a forum that could be better suited for this engineering question?

    Thanks again.
     
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