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Applied exam question watch

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    A particle moves under gravity in a medium whose resistance varies as the square of the speed. The terminal speed is V.

    a) If y is the coordinate of the particle in a vertical direction and x the coordinate in the horizontal direction, show that

    x(double dot) = -g[(xdot(xdot^2 + ydot^2)^(1/2))]/V^2

    y(double dot) = -g[1+ ((ydot(xdot^2+ydot^2)^(1/2))/V^2]

    b) If the direction of motion makes an angle theta with the horizontal, so that
    xdot = vcostheta
    ydot = vsintheta
    with v = (xdot^2+ydot^2)^(1/2) being the speed, show that

    dtheta/dt = (ydoubledot xdot - xdoubledot ydot)/(xdot^2 + ydot^2)

    c) Combine to show that d/dtheta (1/xdot^2) = -2/(V^2cos^3theta)

    The resisting force is -k(x'^2 + y'^2)w, where k is a positive constant (which we are going to work out) and w is a unit vector in the direction of the particle's movement. Since w = (x', y')/sqrt(x'^2 + y'^2), we can write the resisting force as

    -k sqrt(x'^2 + y'^2) * (x', y')


    m x'' = -k sqrt(x'^2 + y'^2) * x'
    m y'' = -mg - k sqrt(x'^2 + y'^2) * y'

    When the particle is travelling at terminal velocity, x'' = y'' = 0 and so x' = 0 and y' = -V. Hence -mg + kV^2 = 0 and k = mg/V^2. Putting k = mg/V^2 into the equations of motion,

    m x'' = -(mg/V^2)sqrt(x'^2 + y'^2) * x'
    m y'' = -mg - (mg/V^2)sqrt(x'^2 + y'^2) * y'

    x'' = -(g/V^2)sqrt(x'^2 + y'^2) * x'
    y'' = -g - (g/V^2)sqrt(x'^2 + y'^2) * y'

    x'' = -g[x' sqrt(x'^2 + y'^2)/V^2]
    y'' = -g[1 + y' sqrt(x'^2 + y'^2)/V^2]

    = v (-d(psi)/dt sin(psi)) + v' cos(psi)
    = -y' d(psi)/dt + v' cos(psi)

    = v (d(psi)/dt cos(psi)) + v' sin(psi)
    = x' d(psi)/dt + v' sin(psi)

    y'' x' - x'' y'
    = (x'^2 + y'^2) d(psi)/dt + v' x' sin(psi) - v' y' cos(psi)
    = (x'^2 + y'^2) d(psi)/dt + v' v sin(psi) cos(psi) - v' v sin(psi) cos(psi)
    = (x'^2 + y'^2) d(psi)/dt

    d(psi)/dt = (y'' x' - x'' y') / (x'^2 + y'^2)

    x'' = -g[x' sqrt(x'^2 + y'^2)/V^2]
    y'' = -g[1 + y' sqrt(x'^2 + y'^2)/V^2]

    y'' x' - x'' y' = -g x'

    d(psi)/dt = -g x'/v^2

    (d/dt) (1/x'^2)
    = (-2/x'^3)x''
    = (-2/x'^3)(-gvx'/V^2)
    = 2gv/(x'^2 V^2)

    (d/d(psi)) (1/x'^2)
    = [(d/dt) (1/x'^2)] dt/d(psi)
    = [(d/dt) (1/x'^2)] / [d(psi)/dt]
    = [2gv/(x'^2 V^2)] [-v^2/(gx')]
    = -2v^3/(x'^3 V^2)
    = -2/(cos^3(psi) V^2)
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Updated: February 5, 2006

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