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Rotational momentum conservation

  1. Nov 13, 2007 #1
    b]1. The problem statement, all variables and given/known data[/b]
    A small block of mass .250kg is attached to a string passing through a hole in a frictionless horizontal surface. The block is originally revolving in a circle with a radius of .800 m about the hole with a tangential speed of 4 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30N. What is the radius of the circle when the string breaks?


    2. Relevant equations

    KE = 1/2mv^2
    MVoRo=MVR

    3. The attempt at a solution

    I am not sure how to get started. I know that momentum must be conserved.

    This is what I did from a similar problem in a book. I am not sure if it is right.
    1/2mv^2(Ro^2/R^2) and I got r=.57m If it is wrong can someone point me in the right direction.

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 13, 2007 #2

    rl.bhat

    User Avatar
    Homework Helper

    The breaking strength of the string is 30N. Breaking strength is due to centrfugal reaction and is equal to mv^2/R. From this find v^2/R.
    According to conservation of angular momentum we have MVoRo=MVR
    or VoRo=VR. Squaring both the sides we get Vo^2Ro^2 = V^2R^2. To the right hand side multiply and divide by R. So we have Vo^2Ro^2 = V^2R^3/R = (V^2/R.)R^3.Substitue the appropreate values and find R
     
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