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Rotation of a Pivoted Rod

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane . Let [tex]T_1 [/tex] and [tex]T_2 [/tex] be the tensions at points L/4 and 3L/4 from the pivoted end, then whats the relation between [tex]T_1 [/tex] and [tex]T_2 [/tex] ?

    Help Needed - :

    Ok, this is a question from circular motion. I know that the net force must be mrw^2 towards the center at both points, but I am having trouble figuring what [tex]m[/tex] I must use at r=L/4 and r=3L/4 . Also, can somebody explain how tension acts in this problem...since the rod has a mass, tension is different along both directions at the same point, but there is only a single value provided ?
    Plz help I dont need the solution, just help me clear the concepts
  2. jcsd
  3. Oct 14, 2008 #2


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    Homework Helper

    Hi f(x),

    If you think about T1, what is T1 doing to the part of the rod that is past the point at L/4? What expression can you write for T1 based on that? Do you see how to get the answer?
  4. Oct 14, 2008 #3
    Hello sir,

    OK, I spent some more time and I feel that T1 at L/4 from pivot causes the rest of the rod to rotate, hence , T1 must provide the necessary centripetal force. But I am confused as to what should be the value of Mass and Radius in this case. Can you plz help with this ?
  5. Oct 14, 2008 #4


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    That's right.

    Remember that you are not looking for the values of T1 and T2, you are looking for how they are related. (Is one twice as much as the other, or one-third, etc.) This is essentially a ratio and so many of the common factors will cancel.

    They directly give you the radius values for each case: L/4 and 3L/4. Now if the entire rod has mass M, what is the mass that T1 accelerates? and T2?

    Then taking the ratio gives how they are related. What do you get?
  6. Oct 14, 2008 #5
    ok,T_1/T_2 = (m_1.r_1.w^2)(m_2.r_2.w^2) = (3M/4).(L/4)/(M/4).(3L/4) = 1
    But that is incorrect, because the book says ( Hint: T_1 > T_2 )
    I had a doubt, since the body was elongated, isnt it wrong to take radius as L/4 instead of the distance to COM from pivot ?
    Last edited: Oct 14, 2008
  7. Oct 14, 2008 #6


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    That's right; in my last post I was not just giving you the r values to plug into your equation. I was pointing out that since you know the radius of the point that each force is acting on, you can figure out everything you need for your equations.
  8. Oct 14, 2008 #7
    So , in general for any rotational problem with an elongated continuous mass, Magnitude of Radius Vector is equal to Distance from axis of rotation to the Centre of MAss of the body, right ?

    Also, I get 15/7 as the answer
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