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How do i find the radius of this path

  1. Nov 14, 2008 #1
    given
    [tex]\vec{R}[/tex]=(30t-t[tex]^{3}[/tex])[tex]\hat{x}[/tex]+(22t-4t[tex]^{2}[/tex])[tex]\hat{y}[/tex]

    find, for time t=2s
    1)the acceleration (a)
    2)the angular acceleration([tex]\alpha[/tex])
    3)the radius of the curve of the path it takes, at t=2s

    as far as i can see, i can integrate [tex]\vec{R}[/tex] twice to get the acceleration, but i need the radius to find out ([tex]\alpha[/tex]),...
    the only thing i could think of to find the radius is using the equation of a circle
    (X-Xo)[tex]^{2}[/tex]+(Y-Yo)[tex]^{2}[/tex]=r[tex]^{2}[/tex]
    then plugging in the values of [tex]\vec{R}[/tex](t=2s) in the X and Y, problem is i dont know the Xo, Yo (centre point) of the circle. thought maybe to plug in X,Y at t=0 for Xo Yo but that just seems wrong.
    any ideas
     
  2. jcsd
  3. Nov 14, 2008 #2

    Redbelly98

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    Well, actually you don't integrate R. Hopefully you know that, and just typed in the wrong thing.

    True, to get angular information there has to be a reference point for the rotation. Since none was given, I'm guessing we use the origin.
     
  4. Nov 14, 2008 #3
    sorry, meant take the derivative of r twice, but thats not really my problem, how do i find the radius?
     
  5. Nov 14, 2008 #4

    Redbelly98

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    Do you have a calculus book lying around? Curvature of a path or function is usually covered in the 1st semester.
     
  6. Nov 14, 2008 #5
    like i said, only thing i can think of is r^2=(x-x0)^2+(y-y0)^2
    dont know where the centre(x0,y0) is?
    any ideas
     
  7. Nov 14, 2008 #6

    alphysicist

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    Redbelly98,

    If I'm reading the question correctly I don't believe that is correct. The path itself defines the center of rotation at any particular time.

    As an alternative approach to find the radius of curvature, you could examine the tangent and normal components of the acceleration.
     
  8. Nov 14, 2008 #7

    Redbelly98

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    I'm not convinced. I'd like to know if this is a basic Phys 101 class that is studying introductory rotational dynamics. If so, I'd expect a straightforward use of the origin as reference:
    (x0, y0) = (0,0)
    r is simply the vector ( x(t), y(t) )
    Angle value is taken between r and +x axis.

    You're interpretation could be right, but I would only expect that in the case of an advanced honors track course -- which is of course possible. Presumably there is an example or discussion in devanlevin's text book that would clear this up.
     
  9. Nov 14, 2008 #8

    alphysicist

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    You mean find [tex]\frac{d^2\theta}{dt^2}[/tex] for the coordinate [itex]\theta[/itex]? That's possible, but to me doesn't seem to fit the rest of the question.


    If they are expected to find the radius of curvature, I think finding the tangential and normal components is something they would be expected to do. Since the velocity, speed, and acceleration were already found in part a, a dot product and using the pythagorean theorem is all that is required to find the two acceleration components. Then one component gives the radius of curvature, and the other gives (alpha) = aT/R.

    (So my point is that these ideas all go together, and the above work takes only a few lines so I think it is well within reason.)

    Please correct me if I misunderstand what you were saying, but I think this is much more straightforward than calculating [itex]d^2 \theta/dt^2[/itex] (which seems to involve taking the second derivative of the arctangent of y(t)/x(t) which is a lot of tedium).


    I definitely agree with that; just asking for the "angular acceleration" is rather ambiguous since it could go either way.
     
  10. Nov 15, 2008 #9
    sorry about the terminology, im not taking it from an english textbook and english isnt the language im studying in,,, what im looking for when i say "angular acceleration" is the amount of radians/s^2
    how do i find the tangent and normal components of the acceleration,
     
  11. Nov 16, 2008 #10

    alphysicist

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    Unfortunately that does not clear up the ambiguity. The question is whether the word "angular" in angular acceleration here refers to the angle that the position vector makes with the x axis, or to the angle it makes with the center of rotation at that particular time.


    As I was using the terms, the tangential component is parallel to the velocity, and the normal component is perpendicular to the velocity. (Tangential and normal to the path.)

    Since you already have the forms for the velocity, speed, and acceleration, you can use the dot product to get the component of the acceleration parallel to the velocity.

    How can you then use that to find the normal component?
     
  12. Nov 16, 2008 #11
    the angle it makes with the centre of the circle
     
  13. Nov 16, 2008 #12
    but how would i get the radius, or the "angular acceleration" from the component parallel or normal to the acceleration and velocity?
     
  14. Nov 16, 2008 #13

    alphysicist

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    The angular acceleration relative to the center of the path is given by:

    [tex]
    a_T= \alpha\ R
    [/tex]
    (similar to simple circular motion) where aT is the tangential component (component parallel to the velocity) and R is the radius of curvature, and alpha is related to the angle of the instantaneous center of rotation.

    The normal component is similar to the formula for the centripetal acceleration:

    [tex]
    a_N = \frac{v^2}{R}
    [/tex]
    where aN is the normal component (perpendicular to the velocity).

    Since you have the speed, once you find aN and aT you can find alpha and R.
     
    Last edited: Nov 16, 2008
  15. Nov 16, 2008 #14
  16. Nov 16, 2008 #15

    alphysicist

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