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Homework Help: Magitude of accelaration

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle moves along a circular path having a radius of 2.0 m. At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s2, what is the magnitude of the total acceleration of the particle?



    2. Relevant equations
    magitude= V^2/R




    3. The attempt at a solution

    But i dont know how to slove with this equation
     
  2. jcsd
  3. Feb 4, 2007 #2

    Dick

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    Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?
     
  4. Feb 4, 2007 #3
    Dick,
    v=3m/s
    a=5 m/s^2

    I guess I can find the magnitude when v=3 m/s . is that
    right?


     
  5. Feb 4, 2007 #4

    Dick

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    Ok, so what are the two components of the acceleration.
     
  6. Feb 4, 2007 #5
    the two components are ax and ay
     
  7. Feb 4, 2007 #6

    Dick

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    The components are normal to the circle and tangent to the circle, so x,y coordinates are probably not appropriate. I would say the tangent component is 5 m/s^2. Why would I say that?
     
  8. Feb 4, 2007 #7
    coz that is a constant acceleration?
     
  9. Feb 4, 2007 #8

    Dick

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    No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?
     
  10. Feb 4, 2007 #9
    this is my calculation

    a=Sq root of (at)^2+(ar)^2

    at= 5m/s^2
    ar=v^2/r= 9/2=4.5
    a=sq root(5^2+4.5^2)
    =6.7 m/s^2

    is this is right??
     
  11. Feb 4, 2007 #10
    towards its center
     
  12. Feb 4, 2007 #11

    Dick

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    Your calculation is fine. What does "towards its center" mean? You labelled it as at in your calculation?
     
  13. Feb 4, 2007 #12
    u mean my answer is right??
     
  14. Feb 4, 2007 #13

    Dick

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    Yeah. Does that surprise you? If you're clear on the two components of the acceleration then I think you can go on to the next problem...
     
  15. Feb 4, 2007 #14
    yaa i got it now dick thanks for you help
     
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