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Kinematics/uniform circular motion.

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A ball on the end of a string is whirled around
    in a horizontal circle of radius 0.28 m. The
    plane of the circle is 1.44 m above the ground.
    The string breaks and the ball lands 2.66 m
    away from the point on the ground directly
    beneath the ball’s location when the string
    breaks.
    The acceleration of gravity is 9.8 m/s2 .
    Find the centripetal acceleration of the ball
    during its circular motion.
    Answer in units of m/s2.

    2. Relevant equations

    T= [tex]\sqrt{}2y/g[/tex]
    ac=4*pi2r/T2

    3. The attempt at a solution
    First I found time using this equation T= [tex]\sqrt{}2y/g[/tex]
    =T= [tex]\sqrt{}2*1.44m/9.8m/s[/tex]= .5421047s
    then I plugged my time into the acceleration equation ac=4*pi2r/T2
    = 37.61415924 m. My answer is wrong and I would like to know what is it that I did wrong, I tried many ways to do it, but non of the answers I get are right, if you can please calrify. Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 25, 2010 #2

    kuruman

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    Science Advisor
    Homework Helper
    Gold Member

    The symbol T in the expression for the centripetal acceleration is not meant to be the time it takes the ball to hit the ground. It is the time the ball requires to complete one revolution, i.e. the period. T is what it is regardless of whether the string breaks or not. It is related to the speed v at which the ball goes around before the string breaks.

    After the string breaks, you have projectile motion. Answer this question first, "How fast must the ball be moving in the horizontal direction so that it hits the ground 2.66 m away in 0.542 s?" This will give you the speed of the ball v. Knowing v, you can then find the centripetal acceleration. There is an expression that relates the centripetal acceleration to the speed. What is it?
     
    Last edited: Jan 25, 2010
  4. Jan 25, 2010 #3
    Ok so basically to find the Velocity I used X/t = V
    V= 4.907749077 m/s
    centripetal acceleration = V^2/r
    = 86.021432 m/s^2 and that is the right answer, thank you so much for helping me.
     
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