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Circular Acceleration & Force

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A 610-kg racing car completes one lap in 14.3 s around a circular track with a radius of 60.0 m. The car moves at constant speed.

    (a) What is the acceleration of the car?
    (b) What force (newtons) must the track exert on the tires to produce this acceleration?

    2. Relevant equations
    v = 2*Pi*Radius / Time
    a = velocity^2 / Radius
    T = 2(Pi)R / Velocity


    3. The attempt at a solution
    (2)(Pi)(60.0)/(14.3)=26.36
    695.01^2 = 695.01 / 60 = Acceleration = 11.58
     
  2. jcsd
  3. Oct 23, 2008 #2

    alphysicist

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    Hi itsmarasilly,

    That looks right to me. Did you get part b?
     
  4. Oct 23, 2008 #3
    i think newtons are basically Kg m/s^2, if so, how do i figure that out from the information given? thanks a lot for your help
     
  5. Oct 23, 2008 #4

    alphysicist

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    Pick a certain point on the circle, and let the x direction be towards the center of the circle. Apply the equation Fnet,x=m ax. What forces are to the center of the circle? You already have the mass and accleration. What do you get?
     
  6. Oct 23, 2008 #5
    fnet = m*a, so
    fnet = 610 * 11.58
    fnet = 7063.8 ?
     
  7. Oct 23, 2008 #6

    alphysicist

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    That looks right for the horizontal direction.

    I was rereading my earlier post and I think it could possibly mislead you, so let me add this: They are asking for the total force from the track, and the track puts a horizontal and vertical force on the car. So you'll need to find both of those components to find the total force from the track. What do you get?
     
  8. Oct 14, 2009 #7
    No that is the final answer. We are not figuring with vertical force. All that is needed is to take the acceleration multiplied by the mass in kg. F=ma
     
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