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Net Force at the Bottom of Circular Motion

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data
    An object is being swung vertically on a stick. What is the net force at the bottom?

    The mass of the object = 2kg.
    g= 9.81ms-1.
    v = 6ms-1 at the bottom.



    2. Relevant equations
    F= mg
    Fcentripetal = (m*v^2)/r


    3. The attempt at a solution
    I just need to know if i must add the force of gravity to the centripetal force.


    Thank you.
     
  2. jcsd
  3. Mar 2, 2012 #2
    Thanks anyways. I solved it.

    For anyone that is looking for an answer:

    Yes, you must add the centripetal force to the force of gravity. This is because the centripetal force in this example is found through the velocity, rather than through the acceleration due to gravity.
     
  4. Mar 2, 2012 #3

    PhanthomJay

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    But the net force at the bottom IS the centripetal force, mv^2/r. You don't add anything to it to get the net force, which is what the problem seems to be asking.
     
  5. Mar 2, 2012 #4
    Thank you PhantomJay.

    The centripetal force in this problem is provided by the tension of the string and also the force of gravity.

    So, Fnet = m*g + (m*v^2)/r
     
  6. Mar 2, 2012 #5

    PhanthomJay

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    yes, correct
    No, that is not correct. Fnet = mv^2/r

    The gravity force acts down, and the tension in the wood stick acts up. And the centripetal force acts up toward the center of the circle.

    Shoudn't it be

    [itex] F_{net} = T - mg = mv^2/r [/itex] ?
     
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