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Two blocks -which will hit in lesser time?

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Two blocks A and B ,each of same mass 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 t1.

    Block B will swing and hit the surface in time t2 . Assume the surface as frictionless,then which block will hit in lesser time

    i.e t1 or t2 ? Please explain mathematically??


    2. Relevant equations



    3. The attempt at a solution

    Intuitively it looks block A will hit first.But mathematically I am not able to relate.Block A does a linear motion covering distance L.Block B doing a circular motion with varying radius(as it falls the initial radius length L will increase ? How to relate displacement , accelaration ?
     

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    Last edited: Sep 2, 2012
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  3. Sep 2, 2012 #2

    TSny

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    Think about what causes the horizontal component of acceleration of block B.
     
  4. Sep 2, 2012 #3
    As block B falls ,at an instant if we assume

    1 the string makes angle θ with the vertical
    2 T is tension in the string

    then along the radial T and mgcosθ acts,whereas tangentially mgsinθ acts
     
  5. Sep 2, 2012 #4

    TSny

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    For an arbitrary θ, what is the net horizontal force acting on B? What is the net horizontal force on A?
     
  6. Sep 2, 2012 #5
    Forces in Vertical direction = Tcosθ - mg
    Forces in horizontal direction =Tsinθ
     
  7. Sep 2, 2012 #6
    net horizontal force on A=T
    net horizontal force on B=Tsinθ
     
  8. Sep 2, 2012 #7

    TSny

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    So, which block will have the greater horizontal component of acceleration?
     
  9. Sep 2, 2012 #8
    Okay...A will have greater accelaration...thank u very much....can we mathematically calculate the time taken by block B to hit the wall in terms of given variables.
    For A accelaration is T/m and distance to be covered is L .Here L=(1/2)(T/m)(t1)[2]
    from here we can calculate t1

    But for B, it will be undergoing circular motion with varying radius as when it falls,the string length acting as the circular radius increases

    Forces in Vertical direction = Tcosθ - mg
    Forces in horizontal direction =Tsinθ

    please explain??
     
  10. Sep 2, 2012 #9
    I think the horizontal force of the right mass is-
    Fr_to-left=Cosθ(mgSinθ)
    As at the top, the body has no horizontal force only downward force.
    At the bottom too, only downward force.
     
  11. Sep 3, 2012 #10

    TSny

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    Since the tension is not constant during the motion, the acceleration of block A will not be constant. So, L = (1/2)(T/m)t12 will not be valid.
    The position of block B is determined by the position, xA, of block A and the angle θ. So, the natural system variables are xA and θ. It is possible to derive differential equations for xA and θ but they are coupled, nonlinear equations. So, I don't see a way to actually get the time, t1, for A to reach the pulley except by numerical integration of the equations. After t1, B's motion will simplify to swinging along a circular arc of radius 2L until B hits the surface at t2. But, still, I think the time between t1 and t2 would have to be determined numerically.
     
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