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Curvilinear Motion

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A package is dropped from the plane which is flying with a constant horizontal velocity of va=150 ft/s. Determine the normal and tangential components of acceleration, and the radius of curvature of the path of motion (a) at the moment the package is released at A, (b) just before the package strikes the ground at B.

    317a340.png

    2. Relevant equations

    [tex]s = s_{o} + v_{o}t + (1/2)(a_{t})_{c}t^{2}[/tex]
    [tex]v = v_{o} + (a_{t})_{c} t[/tex]
    [tex]v^{2} = (vo)^{2} + 2(a_{t})_{c} (s - s_{o})[/tex]

    [tex]\rho = \frac{1 + ((\frac{dy}{dx})^{2})^{3/2}}{|d^{2}y/dx^{2}|}[/tex]

    3. The attempt at a solution

    [tex]s = s_{o} + v_{o}t + (1/2)(a_{t})_{c}t^{2}[/tex]

    [tex]x = 0 + (150ft/s)t + 0[/tex]

    [tex]x =150t[/tex]

    [tex]t = \frac{x}{150}[/tex]

    [tex]s = s_{o} + v_{o}t + (1/2)(a_{t})_{c}t^{2}[/tex]

    [tex]y = 0 + 0 + (1/2)(-32.2ft/s^{2})t^{2}[/tex]

    [tex]y = (-16.1ft/s^{2})t^{2}[/tex]

    [tex]y = (-16.1ft/s^{2})(\frac{x}{150})^{2}[/tex]

    [tex]y = -0.000716x^{2}[/tex] This is the equation of the path.

    [tex]dy/dx = -0.00143x[/tex]

    [tex]d^{2}y/dx^{2} = -0.00143[/tex]

    [tex]x = 1023.785ft[/tex]

    [tex]\rho = \frac{1 + ((\frac{dy}{dx})^{2})^{3/2}}{|d^{2}y/dx^{2}|}[/tex]

    [tex]\rho = \frac{1 + (((-0.00143)(1023.785 ft))^{2})^{3/2}}{|-0.00143|}[/tex]

    [tex]\rho = 3900.339ft[/tex] I am assuming this is the [tex]\rho[/tex] for (b).

    I am unsure how to get the [tex]\rho[/tex] for (a).

    The [tex]a_{t}[/tex] I think for (a) is 0 because the velocity in the horizontal is constant. I am unsure the [tex]a_{n}[/tex] because I need [tex]\rho[/tex].

    I don't understand how to get [tex]a_{t}[/tex] and [tex]a_{n}[/tex] just before it strikes the ground because I don't have a time, do I just assume 1 second before hitting the ground?
     
    Last edited: Oct 6, 2009
  2. jcsd
  3. Oct 6, 2009 #2

    kuruman

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    Nice work! :cool:

    As soon as the package is released, you know the speed and you know that the acceleration is directed towards the center of the circle and is perpendicular to the velocity, i.e. it is centripetal. So ...
     
  4. Oct 6, 2009 #3
    [tex]a_{n}[/tex] at the moment the package is released

    [tex]a_{n} = \frac{v^{2}}{\rho}[/tex]

    [tex]a_{n} = \frac{(150ft/s)^{2}}{3900.339ft}[/tex]

    [tex]a_{n} = 5.769 ft/s^{2}[/tex]
     
  5. Oct 6, 2009 #4

    kuruman

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    3900 ft/s is the radius of curvature for (b) not (a). When the package is released you know the centripetal acceleration, you don't need to calculate it. Use the known value to find the radius of curvature.
     
  6. Oct 6, 2009 #5
    Is it gravity?:

    [tex]a_{n} = \frac{v^{2}}{\rho}[/tex]

    [tex]32.2ft/s^{2} = \frac{(150ft/s)^{2}}{\rho}[/tex]

    [tex]32.2ft/s^{2} = \frac{(150ft/s)^{2}}{\rho}[/tex]

    [tex]\rho = \frac{(150ft/s)^{2}}{32.2ft/s^{2}}[/tex]

    [tex]\rho = 698.758 ft[/tex]
     
  7. Oct 6, 2009 #6

    kuruman

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    Yes it is. Instantaneously at the point of release the package is going around in a circle and the acceleration of gravity is perpendicular to the velocity. You are done.
     
  8. Oct 7, 2009 #7
    Ok thanks.
     
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