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Angular Acceleration: finding the coefficient of friction

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.80 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. Determine the coefficient of static friction between the car and track from these data.


    2. Relevant equations

    [tex]\Delta[/tex][tex]\theta[/tex]=1.57 rad
    Fc = mv2/r
    f= Fc
    f=uN

    3. The attempt at a solution

    I really don't understand how to go at this problem. I was told to find Vf, but how do you do that? Knowing only the tangential acceleration, how would I use that to find Vf. And not knowing what the radius is how do I find the centripetal force and connecting that with the normal force? Without the mass being given?
     
  2. jcsd
  3. Feb 18, 2009 #2

    tiny-tim

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    Hi vipertongn! :smile:

    You don't need the mass because it cancels.

    You can forget the tangential acceleration once you've calculated Vf (for which you use the usual constant acceleration equations … call the radius r, it'll cancel out later).

    Then just use good ol' Newton's second law in the radial direction. :wink:
     
  4. Feb 18, 2009 #3
    I was wondering which one though? I'm not sure if it starts from rest or not and i believe Vf=vi+at might be one equation i could use...but i don't know the time! so how would i calculate for Vf?
     
  5. Feb 18, 2009 #4

    tiny-tim

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    Yes, Vi = 0 … it "accelerates uniformly from rest".

    You have Vi Vf a and s …

    so the equation to use is … ? :smile:
     
  6. Feb 18, 2009 #5
    OHHHH i would use change in Vf^2=Vi^2-2as

    However, where would s come from? I only know the angular distance thingy. i do know that theta is equal to s/r ,but like... i don't know what r is.
     
  7. Feb 18, 2009 #6

    tiny-tim

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    The constant acceleration equations work for any constant double derivative (rate of rate of change) …

    we normally use it for distance speed and acceleration, but we can also use it for angle angular speed and angular acceleration (θ θ' and θ''). :smile:
     
  8. Feb 18, 2009 #7
    so does that mean it's interchangable w/ S?

    Vf^2=Vi^2-2a[tex]\Delta[/tex][tex]\theta[/tex]
     
  9. Feb 18, 2009 #8

    tiny-tim

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    Sorry :redface:

    I was making it too complicated :rolleyes:

    I should just have said that you can treat distance round the track as if it was straight …

    even though s doesn't represent a straight distance, it's still a variable with constant second derivative, so Vf2=Vi2-2as is still valid :smile:
     
  10. Feb 18, 2009 #9
    but what would s be then?

    sould i assume its like 1/4 2pi then?
     
  11. Feb 19, 2009 #10

    tiny-tim

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    just got up … :zzz:

    I think I've confused you :redface:

    No, s is just ordinary distance, as is shown on the car's "mileometer" … it just isn't in a straight line. :smile:

    (you could use combinations of r θ θ' and θ', but in this case it isn't necessary)
     
  12. Feb 19, 2009 #11
    yes ! i got hte correct answer! i just had to use variables nad plugi n the equations to cancel them out! haha yea you did confuse me abit ther but its ok you helped me ^^
     
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