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Homework Help: Resistive force of air

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    At major league baseball games it is commonplace to flash on the scoreboard a speed for each pitch. This speed is determined with a radar gun aimed by an operator positioned behind home plate. The gun uses the Doppler shift of microwaves reflected from the baseball, as we will study in a later chapter. The gun determines the speed at some particular point on the baseball's path, depending on when the operator pulls the trigger. Because the ball is subject to a drag force due to air, it slows as it travels 18.44 m toward the plate. Use the equation below to find how much its speed decreases.
    Suppose the ball, with a mass of 0.145 kg and cross-sectional area of 4.20 10-3 m2, leaves the pitcher's hand at 98.0 mi/h = 43.8 m/s. Ignoring its vertical motion, determine the speed of the pitch when it crosses the plate. (Assume a drag coefficient, D, of 0.305 and a resistive force, R, of 1.2 N.)

    3. The attempt at a solution
    How do i know what the k stands for in V=V0e-kx?
    so i assume it is the drag coefficient, D, of 0.305

    43.8e(-0.305x18.44) = 0.158m/s
    the answer looks very wrong, and it is wrong indeed.
    but how do i know what is the k in the equation stands for? or perhaps i shouldn't just sub in the value like that?

    may i know what is the proper way of doing this kind of question?
  2. jcsd
  3. Sep 20, 2009 #2
    Did you use all of the information provided in the question?

    Think of your every day experience, what happens when you throw a sheet of paper, and what happens when you throw a crumpled up ball of paper? What's responsible for the difference?

    I seriously have no idea what that resistive force is supposed to be, though.
  4. Sep 20, 2009 #3
    seriously, i did not use all the information provided since the question ask me to use V=V0e^(-kx) to find the velocity. i have no idea at all =(
  5. Sep 20, 2009 #4
    Drag forces through a fluid (Gas or liquid) take the form:

    [tex]F=b_1S\cdot V[/tex] or [tex]F=b_2 S\cdot V^2[/tex]

    Where [tex]b[/tex] is a constant that depends on the properties of the fluid itself, and [tex]S[/tex] is the cross-sectional area of the object in question.

    Now the tricky bit here is to use dimensional analysis to find what your k is, exactly. But seeing how the dimensions of D are not given, it gets a bit tricky. :\

    Solving the differential equation for motion of the ball not affected by gravity provided me with the following equation:

    (The resistive force through air is dominated by the force that's proportional to the square of the velocity.)


    [tex]p\equiv \frac{bS}{m}[/tex]

    But again, I haven't the slightest on how you were supposed to realize this without knowing the form of the resistive force, or what it's proportional to.

    The [tex]R[/tex] resistive force is still confusing the heck out of me.
    Last edited: Sep 20, 2009
  6. Sep 20, 2009 #5
    will the k be the resistive force(without v2), since it is constant for the whole journey.
    resistive force(without v2), = 0.5D[tex]\rho[/tex]A
    so k = 0.5D[tex]\rho[/tex]A
    does this assumption make sense?
  7. Sep 20, 2009 #6
    The thing is, that if the resistive force were constant, the relation [tex]v(x)=v_0e^{-px}[/tex] wouldn't hold true, so that's an assumption you aren't allowed to make.

    What you are allowed to do, though, and what I, personally, would do, is to assume that:

    [tex]k\equiv \frac{D\cdot A}{m}[/tex]

    I assume that you have a textbook answer to compare it with, right?
  8. Sep 20, 2009 #7
    No, i don't have the answer :/

    so the drag coefficient,D=0.305, , cross-sec area,A=4.20x10-3, mass of the ball, m=0.145kg

    k=(0.305)(4.20x10-3) / 0.145


    but the answer seems to be not correct, after compared with my friend
    any mistake in my calculation?
  9. Sep 20, 2009 #8
    That answer looks very sensible. What did you friend do to reach a different answer?
  10. Sep 20, 2009 #9
    argh! she don't wish to share, so i've no idea. only know the answer is pretty close :/

    anyway, i've one more similar question.

    [PLAIN]http://img225.imageshack.us/img225/8897/symimage.gif [Broken]

    how do i define the s in the equation above?

    by using dimension analysis, R should be newton,N
    so the right hand side should be equal to newton,N

    N x [tex]\frac{s}{kg}[/tex] x m/s

    so i assume that s is equal to kg/(m/s), mass/speed. am i right?

    the full question for your reference:
    An undercover police agent pulls a rubber squeegee down a very tall vertical window. The squeegee has mass 160 g and is mounted on the end of a light rod. The coefficient of kinetic friction between the squeegee and the dry glass is 0.800. The agent presses it against the window with a force having a horizontal component of 6.00 N. (Downward motion is in the negative direction.)
    (a) If she pulls the squeegee down the window at constant velocity, what vertical force component must she exert?(-3.2304N)
    (b) The agent increases the downward force component by 36.0%, but all other forces remain the same. Find the acceleration of the squeegee in this situation.(-7.2681)
    (c) The squeegee then moves into a wet portion of the window, where its motion is now resisted by a fluid drag force proportional to its velocity according to the equation below.
    [PLAIN]http://img225.imageshack.us/img225/8897/symimage.gif [Broken]
    Find the terminal velocity that the squeegee approaches, assuming the agent exerts the same force described in part (b).
    v =____m/s
    Last edited by a moderator: May 4, 2017
  11. Sep 20, 2009 #10
    You don't define the [tex]s[/tex], it just stands for 'second' this time around. :)
    Those are just the units of the drag constant.

    The question isn't phrased all that well (Your book sounds horrendous. :()
    The drag force in water is dominated by the form [tex]F_{drag}=-kV[/tex]

    Dimensional analysis shows that the units of [tex]k[/tex] must be [tex]N\cdot m^{-1} s[/tex]
    Which are the units of force divided by velocity.

    The form of the drag force is: [tex]\vec F_{drag}=-k\vec v[/tex]
    And you have [tex]k=22 N\cdot \tfrac{s}{m}[/tex] as a given this time (Thank goodness!)
    So this question should be a bit more simple.
  12. Sep 20, 2009 #11
    argh.. finally i got the whole picture =D
    the R is -22v
    and the s is second, is not a symbol or coefficient.
  13. Sep 20, 2009 #12
    Oops, nevermind the comment in the spoiler. I was the one who got mixed up this time!

    You're still mixing some things up. ^^;

    The relationship you were given was:

    [tex]R=-(22 N\cdot \tfrac{s}{m})\cdot V[/tex]

    Substituting [tex]k\equiv 22 N\cdot \tfrac{s}{m}[/tex]

    We get:

    [tex]R=-k\cdot V[/tex]

    R is the force, and the 22 in whatever units it has, is the coefficient of the Resisting force.
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