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Uniform Ciruclar Motion (whirling cord)

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    A 200g object is tied to the end of a cord whirled around on top of a horizontal and frictionless table top in a 2.4 diameter circle at 3 rev/s. Determine a) the velocity of the object b)t he acceleration of the object and c)t he tension in the cord.

    2. Relevant equations

    Vc = (2*pi*r)/T

    Ac = (Vc^2)/R

    Fnet = MA

    Ft - Fg =MAc

    3. The attempt at a solution

    So for A) I got Vc = (2*pi*1.2)(1/3) which equals 22.3 m/s

    Then for B) I got Ac=(22.6)^2/(1.2) which equals 425.6 m/s

    Now heres where I get lost

    I tried doing this but it doesn't come out right...

    Fnet = MA

    Ft - Fg =MAc

    Ft - 196.2 = (20)(425.6)

    which means Ft = 8708.2N, I know the answer should be Force of tension = 85.1N, I'm not sure what I'm doing wrong.
     
  2. jcsd
  3. Oct 29, 2007 #2

    Dick

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    200g=.2kg. Change to kg if you want the answer in newtons. Watch your units.
     
  4. Oct 29, 2007 #3

    PhanthomJay

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    you did fine in a and b. In c, however, you included the weight, but since the object is being twirled on a horizontal tabletop, the centripetal force and acceleration acts horizontally. The weight acts vertically.
     
  5. Oct 29, 2007 #4

    Dick

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    Yes, there's that too. I missed it.
     
  6. Oct 30, 2007 #5
    Ahh that you so much. I, for some reason, was converting 200g to 20kg, now that I switched it to .2kg and took out the force normal I got the right answer. Thank youuu :)
     
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