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Time of Motion with Negative Feedback

  1. Nov 30, 2009 #1
    A block of mass m slides down an incline plane of length xL. The friction factor b is constant, where the friction force increases as a linear function of forward velocity v.

    This produces exponential motion. My problem in building an educational simulator is this, I want to compute the time down the ramp of length xL rather than make educated guesses in the simulator time setting.

    I've tried solving for time in the DE, with no result. I've also tried to apply Conservation of Energy but I'm still struggling with an answer. Anyone recognize a solution to this apparently simple challenge?

    I've posed the problem in this thread:

  2. jcsd
  3. Dec 1, 2009 #2
    [tex]\frac{dv}{dt} = \frac{mg sin\theta - bv}{m}[/tex]

    [tex]\frac{d^2 x}{dt^2} = g\sin{\theta} - \frac{ b}{m} \frac{dx}{dt}[/tex]

    This is a nonhomogeneous linear 2nd order DE. The solution is (I think)
    [tex] x(t)=A + Be^{-bt/m} + \frac{mg\sin{\theta}}{b} t [/tex]

    where A and B are arbitrary constants.
    It is difficult to express t in term of x. But it can be solve numerically for a given x, say using Newton-Raphson method.
  4. Dec 2, 2009 #3
    This is the equation for differential work, where each force in the DE is multiplied by dx. Is this perhaps a DE with an exact solution of t(xL) where xL is the ramp lenth? Starting position is x = 0. Again any help is appreciated.

    [tex] m \frac{d^2 x}{dt^2}dx = mg\sin{\theta}dx - b \frac{dx}{dt}dx[/tex]

    I don't recall how to treat dx*dx so I'm checking my old Calculus textbook. While I'm at it, here's the power:

    [tex] m \frac{d^2 x}{dt^2} \frac{dx}{dt} = mg\sin{\theta} \frac{dx}{dt} - b \frac{dx}{dt} \frac{dx}{dt}[/tex]

    Again I don't recall the rules for distributing differential terms at this time.
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