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Circular motion.

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A small bob is suspended from a fixed point by a string 0.50m long. It is made to rotate in a horizontal circle of radius 0.40m, the center of this circle being vertically below the point of support. Find the angle of inclination of the string, with respect to the vertical.

    2. Relevant equations



    3. The attempt at a solution
    What I did was to draw the diagram out and i used tan θ = 0.4/0.5. θ=38.7(1d.p.) Am I correct??
     
  2. jcsd
  3. Jun 24, 2012 #2

    tiny-tim

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    welcome to pf!

    hi jinhuit95! welcome to pf! :smile:
    no, tan = opp/adj (opposite over adjacent), but .5 isn't the adjacent side, is it? :wink:
     
  4. Jun 24, 2012 #3
    Yes yes I solved it already but now how do I find angular velocity?? Do I use 2pi/t?? The thing is I don't know how to find t.
     
  5. Jun 24, 2012 #4

    tiny-tim

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  6. Jun 24, 2012 #5
    Okay!! I have another question. Angular velocity is 2pi/t?? T is revolution per sec or radian per sec??
     
  7. Jun 24, 2012 #6

    tiny-tim

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    angular velocity is always radians per second

    so if the period is T seconds, that means it goes 2π radians in T seconds …

    ie 2π/T radians per second :biggrin:

    (but you won't need to find T … the centripetal acceleration formula already has the angular velocity, ω = v/r, in it :wink:)
     
  8. Jun 24, 2012 #7
    So if they give u period in terms of revolutions per sec, you have to change it to radians per sec??
     
  9. Jun 24, 2012 #8

    tiny-tim

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    revolutions per second isn't a period, it's an angular velocity

    and yes, you must convert angular velocity to radians (not revolutions) per time, or the formulas won't work
     
  10. Jun 24, 2012 #9
    Wrong post, sorry.
     
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