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Question on particle dynamics

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks A and B , each of same mass m are attached by a thin inextensible string through an ideal pulley .
    Initially block B is held in position as shown in figure . Now the block B is released . Block A will slide to right and hit the pulley in time t(a) . Block B will swing and hit the surface in time t(b).
    Assume the surface as frictionless.
    Then prove that ,
    t(a)<t(b)



    3. The attempt at a solution

    Let the tension in the string be T .
    L=1/2(T/m)[t(a)^2]
    How to get the time t(b) .
     

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  2. jcsd
  3. Mar 22, 2012 #2

    tiny-tim

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    hi nik jain! :smile:

    the tension isn't constant, is it? :wink:

    call the tension T(t),

    do F = ma for block A, and F = ma and τ = Iα for block B …

    what do you get? :smile:
     
  4. Mar 23, 2012 #3
    Yes Tension is not constant.

    Let the block moves a distance (x) .
    1/2m[v(a)^2] = Tx (By energy conservation)
    1/2m[v(a)^2] = mAx (A is the acceleration of block A)
    1/2m[v(a)^2] = mv(a)(dv/dx)x
    integrating it
    v(a) = (2x)^1/2
    again integrating it
    t(a) = (L/2)^1/2

    How to get the t(b) ?
    Please make the equation as above .
    And how the acceleration of block A relates to the acceleration of block B ?
     
  5. Mar 23, 2012 #4

    tiny-tim

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    (try using the X2 button just above the Reply box :wink:)

    But T is not constant, so it's 1/2m[v(a)^2] = ∫ T dx
    But now you've lost T. :redface:

    The whole point is to connect the A and B equations, using T.

    For B, you will need two parameters, length and angle. :wink:
     
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