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Newton's Third Law/ Circular Motion

  1. Oct 4, 2005 #1
    Question:
    Mass m1 on the frictionless table is connected by a string through a hole in the table to a hanging mass m2. With what speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?

    My answer:

    Fnetym2 = T - m2g
    For m2, a = 0
    Therefore,
    0 = T - m2g
    T = m2g

    Fnetxm1 = T
    m1a = T
    m1a = m2g
    Therefore,
    a = m2g / m1

    Since a = v^2 / r, v = sqrt(ra)

    Therefore,
    v = sqrt(rm2g / m1)

    Is this correct?
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2005 #2

    Päällikkö

    User Avatar
    Homework Helper

    Looks correct to me.
     
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