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Diff Eqs for a block moving sideways down a plane

  1. Jul 2, 2014 #1
    1. The problem statement, all variables and given/known data
    I was looking at the problem discussed in the thread below.

    A block is initially kicked in the direction perpendicular to the downslope of the plane with initial speed V.
    The problem asks for the speed of the block down the plane after a long time if the coeff of friction is [itex] tan \theta [/itex].

    2. Relevant equations

    The thread I posted above mentions using the fact that [itex] \sqrt{v_x^2 + v_y^2} + v_y = C[/itex], where C is a constant, since the speed gained in the y-direction will be equal to the speed lost overall.

    C is initially V. Eventually [itex] v_x [/itex] goes to zero and [itex] v_y [/itex] goes to [itex] v_f [/itex]. So
    [itex] v_f + v_f = V [/itex] and the speed down the plane after a long period of time is [itex] \frac{V}{2} [/itex]

    However, I am interested if this problem can be solved in another way, using the equations of motion.

    3. The attempt at a solution

    If [itex] \gamma = arctan \frac{v_y}{v_x} [/itex], then the equations of motion for the x (perpendicular to the downslope) and y (down the slope) directions respectively will be

    [itex] -mg\,sin\theta\, cos \gamma = m \ddot x[/itex]
    [itex] mg\,sin\theta - mg\,sin\theta\, sin\gamma = m \ddot y[/itex]

    These simplify to

    [itex] \frac {-gsin\theta}{\sqrt{\frac{v_y^2}{v_x^2} + 1}} = \ddot x [/itex]
    [itex] gsin\theta \,(1-\frac{\frac{v_y}{v_x}}{\sqrt{\frac{v_y^2}{v_x^2} +1 }}) = \ddot y [/itex]

    Also, [itex] v_y = \dot y [/itex] and [itex] v_x = \dot x [/itex]. The total speed, V = [itex] v_y^2 + v_x^2 [/itex], is constant too. So we further can further simplify the equations to these:

    [tex] \frac{-gsin\theta}{V} \dot x = \ddot x [/tex]
    [tex] gsin\theta \,(1-\frac{\dot y}{V}) = \ddot y [/tex]

    I am wondering how to proceed in solving these equations. I am trying to find [itex] \dot y [/itex].
  2. jcsd
  3. Jul 2, 2014 #2


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    You can't take the next step algebraically and solve for [itex]\dot y[/itex]?

    Also, shouldn't The total speed be [itex]v^2 = v^2_y+v^2_x[/itex]?
  4. Jul 2, 2014 #3
    Right, the total speed should be [itex] v^2 [/itex], not [itex] v [/itex]. Then the square root makes it [itex] v [/itex].

    Seperating the equation for [itex] \dot y [/itex],
    [tex] \dot y \, = \, v \, (1 \, - \, \frac{\ddot y}{g\,sin\theta}) [/tex].

    As t -> [itex] \infty [/itex], [itex] \ddot y [/itex] -> 0 because the block will be moving straight down the incline. This can't be correct though, because that would mean [itex] \dot y [/itex] -> [itex]v[/itex].

    Finding [itex] y(t) [/itex] isn't as hard as I thought. I can integrate both sides of the motion equation to obtain [itex] v \,( t - \frac{\dot y}{g\,sin\theta}) = y + y_0 [/itex]. But we can say that [itex] y(0) = 0 [/itex] to get rid of the [itex] y_0 [/itex] term.

    Then I rearrange the terms to get [itex] \dot y + (\frac {g\,sin\theta}{v}) \, y = (g\,sin\theta)\,t [/itex].
    This is a first order linear diff eq that's solvable.

    [itex] y = Ce^{-\frac{g\,sin\theta}{v} t} + e^{-\frac{g\,sin\theta}{v} t} \int (g\,sin\theta) \,t \, e^{\frac{g\,sin\theta}{v} t} [/itex]

    I simplified this down to
    [itex] y = vt - \frac{v^2}{g\,sin\theta} + Ce^{\frac{-g\,sin\theta}{v}t} [/itex]


    [itex] \dot y = v - \frac{g\,sin\theta}{v} C e^{\frac{-g\,sin\theta}{v}t} [/itex].

    But, this also suggests that [itex] \dot y [/itex] goes to [itex] v [/itex] instead of [itex] \frac{v}{2} [/itex].
  5. Jul 3, 2014 #4


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    Why should be the total speed constant?

  6. Jul 3, 2014 #5
    I think the whole premiss that C as defined at the OP is a constant is just pain wrong. Why would that be the case?
  7. Jul 3, 2014 #6


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  8. Jul 5, 2014 #7
    That's a mistake. I realize now that total speed + [itex] v_y [/itex] is a constant, [itex] \sqrt{v_x^2 + v_y^2} + v_y = C [/itex]. I will check out the link you posted, which is the same problem, and try to redo my math.
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