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Drag force with differential equations, finding max speed

  1. Apr 14, 2017 #1
    1. The problem statement, all variables and given/known data

    A submarine engine provides maximum constant force ##F## to propel it through the
    water.

    Assume that the magnitude of the resistive drag force of the water experienced
    by the submarine is ##kv##, where ##k## is the drag coefficient and ##v## is the
    instantaneous speed of the boat.

    When the engine is switched on to full power, the submarine starts from rest and
    moves horizontally in a straight line.

    If the mass of the submarine is ##M##

    1) Write the differential equation of motion of the submarine in terms of its
    speed ##v##

    2) Solve the equation of motion from part (1) to find the speed ##v(t)## as a
    function of time ##t##

    3) What is the maximum possible speed ##v_{max}## the submarine can reach with
    this engine?

    2. Relevant equations

    Newtons second states ##F = ma##. Then given force ##F## I have ##F = ma##.


    3. The attempt at a solution

    For the drag force ##-kv## I can write this as ##-Kmv## where ##K## is some constant
    such that ##-Kmv = -kv##.

    This gives me

    $$
    ma = -Kmv
    $$

    Dividing by ##m## and noting that ##a = \frac{dv}{dt}## gives

    $$
    \frac{dv}{dt} = -Kv
    $$

    Which is a differential equation of motion in terms of the speed ##v##.

    Solving this differential equation as

    $$
    \frac{1}{v} dv = -K dt
    $$

    Then integrate both sides for

    $$
    \ln(v) = -K t + C_1
    $$

    Here I can take the exponential of each side and note that ##e^{x + b} = Ae^x##
    where ##A,b## are constants.

    So I have

    $$
    v = Ae^{-Kt}
    $$


    For the maximum speed I need to consider where the force of motion by the engine
    is balanced by the force of the drag.

    At this point I will have

    $$
    kv_{max} = ma
    $$

    Using the solution from part (2) we can sub for ##v## as

    $$
    k\left( A e^{-K t} \right) = ma
    $$


    I'm not sure how to obtain the maximum speed here though?

    The expression can be rewritten as

    $$
    \frac{kA}{ma} = e^{Kt}
    $$

    Given ##K## is some constant this is just (where ##D## is some constant)

    $$
    \frac{kA}{ma} = e^{t} e^{K} = De^{t}
    $$

    So

    $$
    \frac{kA}{Dma} =
    e^{t}
    $$

    I don't think that this is correct, or even know if it makes sense.

    It's suggesting that the constant ##k## multiplied by the constant ##A## divided by
    the product of mass, acceleration and ##D## is equal to ##e^t##.

    So I'm confused

    Thanks
     
    Last edited by a moderator: Apr 14, 2017
  2. jcsd
  3. Apr 14, 2017 #2

    Ygggdrasil

    User Avatar
    Science Advisor

    The submarine engine is propelling the submarine with force F and this force is being opposed by the drag -Kmv. Therefore the net force on the submarine will be F-Kmv.
     
  4. Apr 14, 2017 #3
    cheers -

    so I can use that at the equilibrium I will have

    $$
    F - kv = 0
    $$

    And this is the net force, so these are also equal to ##ma##, and can be expressed as

    $$
    F - kv = ma
    $$

    Note that ##a = \frac{dv}{dt}## which gives

    $$
    F - kv = m \frac{dv}{dt}
    $$

    From here I can divide through by mass, letting ##p = \frac{1}{m}## gives

    $$
    pF - pkv = \frac{dv}{dt}
    $$

    Then arranging the differential for integration gives

    $$
    \frac{1}{pF - pdk} dv = dt
    $$

    Integrate both sides

    $$
    \int \frac{1}{pF - pdk} dv = \int dt
    $$

    Here I can make the sub ##u = pF - pkv## such that ##du = -(pk) dv## and
    ##\frac{1}{-pk} du = dv##

    Then subbing these in gives

    $$
    -\frac{1}{pk}\int \frac{1}{u} du = \int dt
    $$

    Solving this gives

    $$
    -\frac{1}{pk} \ln(u) = t + C_1
    $$

    Subbing back for ##u## gives

    $$
    - \frac{1}{pk} \ln(pF - pkv) = t + C_1
    $$

    Then if I wanted to solve for ##v## I can multiply both sides by ##- \frac{1}{pk}##
    for (using a new constant ##C_2##)

    $$
    \ln(pF - pkv) = -pkt + C_2
    $$

    Then take exponentials as

    $$
    pF - pkv
    = e^{-pkt + C_2}
    $$


    Then note that ##e^{-pkt + C_2}## is equal to ##e^{-pkt} \times C_3##. Rearranging
    from this gives

    $$
    pkv = pF - C_3 e^{-pkt}
    $$

    Dividing through by ##pk## gives


    $$
    v = \frac{pF - C_3 e^{-pkt}}{pk}
    $$

    As the solution to the differential in terms of ##v##


    Is this approach right?
     
  5. Apr 16, 2017 #4

    Ygggdrasil

    User Avatar
    Science Advisor

    A good way to check the result, would be to plug it back into the original differential equation to see that m dv/dt and F - kv are indeed equal. But, the answer does look correct to me (provided you use the boundary conditions to find the value of C3 and use m instead of the placeholder variable p).
     
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