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Rotation and Tension in Strings

  1. Jun 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Here is another question I had:

    http://desmond.imageshack.us/Himg84/scaled.php?server=84&filename=problem2v.jpg&res=landing [Broken]

    3. The attempt at a solution

    I think the equation of motion for this system is:

    [itex]T- \ mg \cos 45 = I \frac{d^2 \theta}{dt^2} = I \frac{d \omega}{dt}[/itex]

    And the moment of inertia for each rod is I=1/3ML2 (since the rotation axis is through the end). So the moment of inertia for the whole system would be 2I? I'm not sure where to go from here. What do I have to do with the right hand side of this equation? :confused:

    Any help would be greatly appreciated.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 23, 2012 #2
    Use free-body diagram.
    Analyse all the forces and resultant forces, vertically and horizontally.
     
  4. Jun 24, 2012 #3
    If I worte down forces more precisely how does that help with the right hand side of the equation?
     
  5. Jun 24, 2012 #4
    Your equation is wrong.
    As you see it is a constant angular velocity ω
    Your right hand side of equation is equal to zero, means T=mgSin45° ?

    Think about vertical component. Does it move relative to ground.
     
  6. Jun 25, 2012 #5
    Hello roam
    Strings are supposed to be mass less so you don't have to worry about moment of inertia in this case.The problem requires you to make a free body diagram so as to firstly explain the vertical stability of the mass m and secondly to maintain it in a circular using the tension in each of the strings.As such your problem involves two unknowns one for the tension in upper string and second for that in lower string.They will not be the same .Try solving it now.
    regards
    Yukoel
     
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