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Math Help - Classical mechanics problem

  1. #1
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    Classical mechanics problem

    I need to find the slope \frac{dy}{dx} at the point P. The only thing I know is that:

    y = f(x)
    \frac{dx}{dt} = v_0 (constant)
    The acceleration at P is: \vec{a} = \frac{a}{\sqrt{2}}(\hat{t}+\hat{n}) (a constant and > 0, natural coordinates)
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  2. #2
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    Quote Originally Posted by Mondreus View Post
    I need to find the slope \frac{dy}{dx} at the point P. The only thing I know is that:

    y = f(x)
    \frac{dx}{dt} = v_0 (constant)
    The acceleration at P is: \vec{a} = \frac{a}{\sqrt{2}}(\hat{t}+\hat{n}) (a constant and > 0, natural coordinates)
    Post the actual question (or try defining all the symbols), that is incomprehensible.

    CB
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  3. #3
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    A ball moves along the curve y=f(x) with the constant velocity \dot{x}=v_0. At a certain position P, the ball has the acceleration \vec{a} = \frac{a}{\sqrt{2}}(\hat{t}+\hat{n}), a > 0

    What is the slope \frac{dy}{dx} at P.

    The acceleration is given in normal-tangential coordinates, i.e. \vec{a} = \ddot{s}\hat{t}+\frac{\dot{s}^2}{\rho}\hat{n}
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  4. #4
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    Well, isn't it true that

    \dfrac{\dot{y}}{\dot{x}}=\dfrac{dy/dt}{dx/dt}=\dfrac{dy}{dt}\,\dfrac{dt}{dx}=\dfrac{dy}{dx}?

    Since you're given \dot{x}, it remains to find \dot{y}. How do you suppose you could do that?
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  5. #5
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    It's probably really easy, but I can't figure it out. It's actually a two-part question where I also need to find the radius of curvature, but that one I can solve once I know dy/dx by using the arc length and identifying \rho with \dot{s} in the acceleration that we know.

    ds = \sqrt{1+\left(dy/dx \right) ^2}dx \implies \dot{s}=\dot{x}\sqrt{1+\left(dy/dx \right) ^2}
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  6. #6
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    In your post # 3, what is \rho? Is it known? Is f(x) known? If so, what is it?
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  7. #7
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    \rho is the radius of curvature, and it's unknown (I can find it once I know dy/dx). f(x) is also unknown.
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  8. #8
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    So, what is the representation of the velocity vector in terms of tangential and normal components? How does taking the derivative of that compare with the acceleration in tangential and normal components? Could you infer anything from that?
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  9. #9
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    \vec{v} = \dot{s}\hat{t} where \hat{t} is defined as \frac{d\vec{r}}{ds} and where s is the arc length and \vec{r} is the distance vector from the origin.

    \frac{\dot{s}}{\rho} = \frac{a}{\sqrt{2}} is given, but \rho is unknown.

    The answer is \frac{dy}{dx}\pm 1, but I don't know how to get there.
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  10. #10
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    So what happens when you differentiate both sides of

    \vec{v}=\dot{s}\,\hat{t}

    w.r.t. time?
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  11. #11
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    You get the acceleration I posted above: \vec{a} = \ddot{s}\hat{t}+\frac{\dot{s}^2}{\rho}\hat{n}

    \rho is defined as \frac{1}{\left|\frac{d\hat{t}}{ds} \right|}
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  12. #12
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    So what happens when you compare the two accelerations? Say, just the tangential component?
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  13. #13
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    \ddot{s} = \frac{a}{\sqrt{2}}

    Don't know how that helps me since the acceleration is only given for a single point, so I can't integrate it to get the velocity.

    Comparing the normal components gets us \frac{\dot{s}^2}{\rho} = \frac{a}{\sqrt{2}}
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  14. #14
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    I can't integrate it to get the velocity.
    Perhaps not exactly. But you could get it correct to first order, perhaps. That might be good enough.

    Question: how do you interpret the phrase "constant velocity \dot{x}=v_{0}"? Do you think that \dot{x}=v_{0} for all time? Do you also think that initially, the ball is moving only in the x direction with speed v_{0}?
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  15. #15
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    It is moving with constant velocity \dot{x}=v_0 all the time.
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