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Homework Help: Circular motion (2 particles on a string)

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A light inextensible string of length 0.6m has one end fixed to a point A on a smooth horizontal plane. The other end of the string is attactched to a particle B, of mass 0.4kg. A particle P of mass 0.1kg is attatched to the mid-point of the string. The line APB is straight and rotation continues at 2 rad s^-1 on the surface of the plane.

    1) Calculate the tension in the section of the string AP.
    2) Calculate the tension in the section of the string BP.

    2. Relevant equations

    Centripetal force = mass x radius x (angular velocity)^2

    3. The attempt at a solution

    I imagined that the tension in AP would simply be:
    0.1 x 0.3 x 2^2 = 0.12N

    however the answer is given as 1.08N. So my question really is how does two particles on the same string affect the tension?

    I placed in q.2 for curiosity.

  2. jcsd
  3. Jun 15, 2010 #2


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    You must draw free body diagrams . You should first start with question 2 to determine the string tension BP by looking at particle B alone. Then look at particle P alone. There are 2 tension forces acting on particle P.
  4. Jun 15, 2010 #3
    Excuse the poor diagram but:


    where > and < indicates direction of the force
    from left to right forces are: T1, T1, T2, T2

    Considering B alone, resolving horizontally:
    T(at B) = 0.4 x 0.6 x 2^2 = 0.96N

    Considering P alone, resolving horizontally:
    T(at P) = 0.1 x 0.3 x 2^2 = 0.12N

    So I suppose this says that the resultant force on P is 0.12N.
    So if we take left as positive at P,

    T1 - T2 = 0.12N
    T1 = 0.12 + 0.96
    = 1.08N

    I believe this is the correct line of thinking?

    Also, if 2 particles hang on a string (light, inextensible) vertically, one above another. Will the tension in the top string be the sum of the weight of the two particles?

    Thank you.
  5. Jun 15, 2010 #4


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    The net (resultant) horizontal force at B is 0.96N; since the only horizontal force acting at B is T1, then T1 = 0.96N
    again, this is the resultant horizonatal force at P
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