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Circular Motion and calculating time

  1. Jun 24, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]


    2. Relevant equations

    Velocity =2PiR/T

    Time=2PiR/V

    Force Centripetal = MV^2/R

    Angular Displacement/Time interval = Average Angular Velocity

    Angular Velocity x Radius Of Circle Path = Linear Velocity

    Average Angular Acceleration = Change In Angular Acceleration/ Time Interval

    Linear Acceleration = Radius of Circular Path x Angular acceleration

    Force= Mass x Acceleration



    3. The attempt at a solution

    How do i calculate time? Once i know that i can use this formula: Velocity =2PiR/T

    Is T(Time) = 360 radians?

    Im not honestly sure where to start, ive been trying for 4 days now, constantly seeking infomation while at work.
     
  2. jcsd
  3. Jun 24, 2008 #2

    Hootenanny

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    Wouldn't it perhaps be easier to use a different formula, one in which you know all the required information?
     
  4. Jun 24, 2008 #3

    What do you mean.

    1) Transpose the equation again

    or

    2) Theres another formula that i have missed


    If so, any hint on where to look.
     
  5. Jun 24, 2008 #4

    Hootenanny

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    The formula is on your list :wink:. You know the centripetal force, the mass and the radius. You want to know the velocity.
     
  6. Jun 24, 2008 #5
    I so hope this is correct.

    Fc= MV^2/R -> Fc x R/M=V^2

    Velocity -> (1100x140)/1000 = 154^2 -> [tex]\sqrt{154}[/tex] = 12.4 (m/s ?)


    Acceleration -> V^2/R -> 12.4^2/140 -> 1.09 m/s^2


    Force Gravity -> M x g -> 1000x9.81 = 9810g
     
    Last edited: Jun 24, 2008
  7. Jun 24, 2008 #6

    Hootenanny

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    I'm not checking your arithmetic, but your method looks good :approve:
    I'm guessing by the amount of g acting on the car, they mean how many times the acceleration due to gravity is the acceleration of the car, rather than the actual force. Do you follow?
     
  8. Jun 24, 2008 #7
    Would i be correct in saying that:

    Acceleration x Gravity = The amount of g acting on the vehicle?

    1.09 m/s^2 x 9.81 = 10.69
     
  9. Jun 24, 2008 #8

    Hootenanny

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    Not quite. If you accelerate at 1g means you are accelerating at a rate that is equal to the acceleration due to gravity (i.e. 9.81 m.s-2). Similarly, an acceleration of 2g is equivalent to an acceleration of twice that of gravity (i.e. 19.62 m.s-2). That number of g's 'pulled' can be calculated thus:

    [tex]n = \frac{a}{g}[/tex]

    Do you follow?
     
    Last edited: Jun 24, 2008
  10. Jun 24, 2008 #9
    Ok that would mean that.

    Acceleration / Gravity = The amount of g acting on the vehicle

    1.09 m/s^2 / 9.81 = 0.1 (g)
     
  11. Jun 24, 2008 #10

    Hootenanny

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    Looks okay to me :approve:
     
  12. Jun 24, 2008 #11
    Hootenanny : It was a pleasure. :biggrin: Thanks for the help.
     
  13. Jun 24, 2008 #12

    Hootenanny

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    Not a problem :smile:
     
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