1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Magitude of accelaration
Loading...