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Acceleration and radius of curvature of a particle

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    The motion of a particle is defined as:

    x=[(t-4)3/6]+t2

    y=t3/6-(t-1)2/4

    Find the acceleration and radius of curvature at t=2

    2. Relevant equations

    a=(dv/dt)(et)+(v2/[tex]\rho[/tex])(en)

    where et and en are the tangential and normal unit vecotrs to the curve and [tex]\rho[/tex] is the radius of curvature.


    3. The attempt at a solution

    So I was able to get the acceleration by differentiating both of the equations twice and plugging in 2. The x component becomes 0 and the y is +1.5m/s. I'm stumped on the radius though. I differentiated to find dv/dt and v and tried solving for rho but it didn't work. That would have been essentially ignoring et and en since they are unit vectors, so the length is on it they shouldn't affect the outcome, right? Any ideas on how to solve the problem. Also, I think the way the equation is written with the unit vectors in there somewhat confuses me too so if there's anything that could help clear up any confusion with the equation that'd be great too.
    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 6, 2009 #2

    Redbelly98

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    You do need to account for those unit vectors, at least the en one since that is where the radius of curvature comes into play.

    Think of a as having a normal and tangential component (instead of x and y components). The normal component may be equated to v2/ρ, according to your equation.

    If you can come up with an expression for en, that may help get you going.
     
  4. Oct 6, 2009 #3
    the equations I have in my notes are

    en = -sin[tex]\vartheta[/tex]i + cos[tex]\vartheta[/tex]j

    et = cos[tex]\vartheta[/tex]i + sin[tex]\vartheta[/tex]j

    where [tex]\vartheta[/tex] would be the angle between the tangential vector and the x axis

    I worked out [tex]\vartheta[/tex] = 14 by taking the inverse tan (dy/dt)/(dx/dt).
    And then in the original equation, v is the length of <dx/dt, dy/dt> at t=2, and then I'm not exactly sure what dv/dt is. I was thinking that would be the final acceleration, but that would be what is on the left side of the equation so that doesn't make sense. Either way, I have feeling there is an easier way to do all this, any thoughts?
     
  5. Oct 7, 2009 #4

    Redbelly98

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    I'm not sure of an easier way here.

    I agree with the θ = 14° result, which can be used to determine en. Can you now find the component of a in the direction of en? That result will be equivalent to v2/ρ, so you could then solve for ρ.

    dv/dt will not be needed.
     
  6. Oct 7, 2009 #5
    ok, I was able to talk with my classmates today and figure out the problem without having to use theta. en can be found by taking the unit vector of the vector perpendicular to et. et is found by differentiating the equation and finding vx and vy. Then a = <0,1.5> is dotted with en to get an. Finally, set an equal to v2/rho and solve.
    Thanks for your help!
     
  7. Oct 7, 2009 #6

    Redbelly98

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    You're welcome, glad it worked out. :smile:
     
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