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Dynamics, ball in motion

  1. Apr 30, 2006 #1

    In a test of resistance to motion in an oil bath, a small steel ball of mass m is released from rest at the surface. If the resistance to motion is given by R = kv where k is a constant, derive an expression for the depth h required for the ball to reach a velocity v.

    [tex]mg - kv = ma = mv\frac{dv}{ds}[/tex]

    [tex]\int_0^hds = \int_0^{v_t}\frac{mv}{mg-kv}dv[/tex]

    [tex]h = -\frac mk\int_0^{v_t}\frac{-kv}{mg-kv}dv[/tex]

    [tex]h = -\frac mk\int_0^{v_t}\frac{mg-kv-mg}{mg-kv}dv[/tex]

    [tex]h = -\frac mk\left[v_t - \int_0^{v_t}\frac{mg}{mg-kv}dv\right][/tex]

    [tex]h = -\frac mk\left[v + \frac{mg}{k}\int_0^{v_t}\frac{-k}{mg-kv}dv\right][/tex]

    [tex]h = -\frac mk\left[v + \frac{mg}{k}\ln(mg-kv_t)\right][/tex]

    [tex]h = -\frac {m^2g}{k^2}}\ln(mg-kv_t)-\frac{mv_t}{k}[/tex]

    [tex]h = \frac {m^2g}{k^2}}\ln\left(\frac{1}{mg-kv_t}\right)-\frac{mv_t}{k}[/tex]

    But the answer sheet says
    [tex]h = \frac {m^2g}{k^2}}\ln\left(\frac{1}{1-(kv_t)/(mg)}\right)-\frac{mv_t}{k}[/tex]

    I've calculated this two more times and got the same result.

    Why is this so?
    Last edited: Apr 30, 2006
  2. jcsd
  3. May 1, 2006 #2
    What happened to your constant of integration?
    Once you have substituted the initial conditions and solved for the constant of integration, you should get the correct answer.
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