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Finding Length with rpm and static friction

  1. Jul 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A bug crawls outward from the center of a compact disc spinning at 190 revolutions per minute. The coefficient of static friction between the bug's sticky feet and the disc surface is 1.4. How far does the bug get from the center before slipping?

    2. Relevant equations

    (static friction)(normal force)=(mv^2)/r
    v=2pir/T

    3. The attempt at a solution

    I converted 190 rev/min into rad/sec, assuming that is how to find velocity. I got 19.89 rad/sec. I put (static friction)(mg)=mv^2/r equal to each other and got 14 meters as my r. However, this answer is wrong and I have no idea how else to solve this problem. Help please!
     
  2. jcsd
  3. Jul 23, 2009 #2
    Welcome to PF. :)

    You are very very close, I think you just made some small mistake along the way. You were correct in realizing the critical condition is when the static friction can no longer supply the bug with the centripetal acceleration required for him to rotate with the disc.
    (Conversely, from the bug's accelerated frame of reference, the static friction can no longer counter-balance the centrifugal force)


    Try it like this instead:
    Centripetal acceleration= [tex]\omega^2 r[/tex]
    [tex]\omega=2\pi f[/tex]
    [tex]f[/tex]=The frequency of the rotations.

    What's the textbook's answer, by the way?
     
  4. Jul 23, 2009 #3

    djeitnstine

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    How is velocity rad/sec? I thought the definition of velocity was [tex]\frac{length}{time}[/tex] and it seems that you are using meters as length....

    Also from class you should know that radians are dimensionless.

    So if angles are measured in radians, what is the quantity 19.89 rad/sec you calculated called?
     
  5. Jul 23, 2009 #4
    would f=190 rev/min? and how would i find centripetal acceleration?

    I think I got confused because the only factor I was missing was velocity so I just made 19.89 rad/sec into velocity but I think it's angular velocity.
     
  6. Jul 23, 2009 #5
    That is correct.

    For circular motion, what is the centripetal acceleration?

    If you wish to prove to yourself the relation I suggested you use, then the following will prove helpful:
    [tex]\omega=\tfrac{V}{r}[/tex]
    For every one radian the bug rotates, he moves across an arc length equal to his distance from the center times the one radian he's rotated (Draw this out and remember your circle geometry).
     
  7. Jul 23, 2009 #6
    Okay, thank you! I will try to work this out. I don't have the answers bc the homework is online.
     
  8. Jul 23, 2009 #7
    Okay, just for reference, I got 0.03469m (I used [tex]g\approx 9.81 \tfrac{m}{s^2}[/tex])
     
  9. Jul 23, 2009 #8

    djeitnstine

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    Gold Member

    You are correct however you should in future note that you can't make anything into something else without doing something to it first!

    Meaning, angular velocity is NOT linear velocity. that's like saying radians (angle) equals meters (length). Which makes no sense!

    So in order to convert you have to use the formula Royal Cat gave, [tex] v= r \omega [/tex] (velocity) = (distance) x (angular velocity)
     
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