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Solved: non-uniform circular motion problem

  1. Nov 2, 2014 #1
    The problem statement, all variables and given/known data
    A 200g ball on a 55-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. What was the tension in the string an instant before it broke?

    Given variables:
    initial height: 0.2 m
    final height: 0.6 m
    radius: 0.55 m
    mass of ball: 0.2 kg
    gravity: -9.8 m/s^2

    The attempt at a solution
    My answer: The ball is rotating around a point that is 200 cm (0.2 m) above the ground, and snaps off when the string is parallel to the ground- i.e. when the ball is the same height as the center-point of its rotation. It reaches a height of 0.6 m or 600 cm, with a total vertical displacement of (0.6-0.2) m, or 0.4 m, upward. During this displacement of 0.4 m, it has a negative acceleration of 9.8 m/s^2 due to gravity. Using this information, and assuming that the velocity at maximum height is equal to zero, I can use the following equation to figure out the tangential velocity at the instant the ball broke off the string:

    Vf^2 = Vi^2 + 2a(y2-y1)
    => 0^2 = Vi^2 + 2(-9.8)(0.6-0.2)
    => 0 = Vi^2 + 2(-9.8)(0.4)
    => -Vi^2 = -7.84
    => Vi = sqrt(7.84) = 2.8

    Now I know that the tangential velocity of the ball the instant before it broke off the string was 2.8 m/s upward.

    Since centripetal acceleration is "velocity squared over radius", and radius is 55 cm or 0.55 m, so

    ac = v^2 / r
    => ac = (2.8)^2 / (0.55) = 14.25

    With an acceleration of 14.25 m/s^2 toward the center, the force toward the center should be mass times acceleration, and the mass of the ball is 0.2 kg so

    Fc = ac * m
    => 14.25 * 0.2 = 2.85

    So the tension on the string should be 2.85 N.

    Where did I go wrong?
    Did I neglect some vertical component of tension that counteracts gravity while the ball is still attached to the string?
     
  2. jcsd
  3. Nov 2, 2014 #2

    BvU

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    The conversion of cm to m involves a factor of 100, not 1000 !

    Apart from that: my compliments for a very clear posting !

    And: welcome to PF, PET :)
     
    Last edited: Nov 2, 2014
  4. Nov 2, 2014 #3
    I would try using conservation of energy and see if that gets you the right answer..
     
  5. Nov 2, 2014 #4
    Oops...
    Thank you so much!
    One oversight that gave me a week of stress!
     
  6. Nov 2, 2014 #5
    I got the answer, so I'm going re-title the thread as "Solved:..."
    Thank you BvU and Newton!
     
  7. Nov 2, 2014 #6

    BvU

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    c from centum, Latin for 100. You'll never forget any more... century, centurion, dollarcent, eurocent, and so on and so forth :)
     
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