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Angular motion

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    Point P rotates around the center O at a constant rate.The magnitude of the velocity of point P relative to O is 3.7 m/s and the distance between the two points is 90 mm. What are the magnitudes of (a) normal and (b) tangential components of the acceleration of P relative to O?

    2. Relevant equations
    an = v2/r



    3. The attempt at a solution
    So I was able to find the normal component of acceleration using the formula an = v2/r and got value of 152.11 m/s/s. But for tangential component, I'm confused, the point doesn't actually accelerate (it travels at a constant speed). So does that mean that at would equal to zero? Or am I misunderstanding something here?

    Thank you!
     
  2. jcsd
  3. Dec 3, 2008 #2

    fluidistic

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    Gold Member

    I think you are right, the tangential acceleration of point P must be [tex]\frac{0m}{s^2}[/tex] otherwise it's SPEED (and not velocity as written) wouldn't be a constant with respect to point 0.
     
  4. Dec 3, 2008 #3
    Sweet, thanks!

    This is for another question, but I don't want to open another topic. So I'm given equation of a path through which the particle travels y=0.2sin(pi*x) and it has a uniform speed in the x direction of 2 m/s. We're required to find velocity at x=0.25. So can I take the derivative of the path equation in order to find vertical velocity of the particle? And then combine the two (x&y direction) to find resultant?
     
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