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Find the initial acceleration of the block

  1. Sep 17, 2015 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations


    3. The attempt at a solution

    The question says block B has negligible mass ,which means particle will be effectively in free fall i.e acceleration would be 'g' . But then it is given a horizontal velocity i.e it has centripetal acceleration ac about the pulley v2/r .

    The forces acting on the particle are Mg and tension T .

    So, Mg -T = M(√(g2+ac2))

    I am really clueless with this problem .

    Please help me .

     

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  2. jcsd
  3. Sep 17, 2015 #2

    Orodruin

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    If the mass of B is negligible, how is B going to affect the movement of A? (I.e., what is the tension in the string going to be?)
     
  4. Sep 17, 2015 #3
    B would not affect the movement of A . The tension would be zero ??
     
  5. Sep 17, 2015 #4

    Orodruin

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    Yes, so how can you use this to find the acceleration of B?
     
  6. Sep 17, 2015 #5
    Acceleration of B = acceleration of the particle = √(g2+ac2) = 2√65 ??
     
    Last edited: Sep 17, 2015
  7. Sep 17, 2015 #6

    Orodruin

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    No, the acceleration of A is not the acceleration of B. Furthermore, the acceleration of A is just g in the down direction. It is also unclear why you would add the accelerations in quadrature.

    I suggest that you think about how the motion of B relates to the motion of A. The expression v^2/r will appear, but not as you have stated.
     
  8. Sep 17, 2015 #7

    BvU

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    You have an idea about ac ? Does the particle follow a circular path (in view of what you said in post #3 ?)
     
  9. Sep 17, 2015 #8
    After the particle has moved, if it has moved horizontally x and has moved down vertically y, what is its distance from the pulley in terms of H, x, and y?

    Chet
     
  10. Sep 17, 2015 #9
    o:)

    I agree,what I did doesn't make sense.But then if acceleration of A is just g , isn't A undergoing circular motion about the pulley ?? Why don't we account for v^2/r in the net acceleration of A ??

    Sorry, I am bereft of any more ideas .
     
  11. Sep 17, 2015 #10
    ##2H+\sqrt{x^2+y^2}##
     
  12. Sep 17, 2015 #11
    Not exactly. Think Pythagorean theorem.

    Chet
     
  13. Sep 17, 2015 #12
    2H is the initial distance of A from pulley . It moves an additional distance ##\sqrt{x^2+y^2}## . Isn't it now at a distance ##2H+\sqrt{x^2+y^2}## from the pulley ??
     
  14. Sep 17, 2015 #13
    No. One leg of the big right triangle is (2H+y) and the other leg of the triangle is x. The distance from the pulley to the particle is the hypotenuse of the triangle. Try again.

    Chet
     
  15. Sep 17, 2015 #14
    Should it be ##\sqrt{x^2+(2H+y)^2}## ??
     
  16. Sep 17, 2015 #15
    Yes. Good.

    Now how much has the distance increased relative to what it was at time zero? Given that the string is of constant total length, in terms of x, y, and H, how much has mass B moved to the right?

    Chet
     
  17. Sep 17, 2015 #16
    ##\sqrt{x^2+(2H+y)^2} - 2H##
     
  18. Sep 17, 2015 #17
    Good. Now what is x as a function of t and y as a function of t? Substitute these expressions into the above equation for the distance d that the mass B has moved to the right.

    Chet
     
  19. Sep 17, 2015 #18
    ##x=vt## and ##y=\frac{1}{2}gt^2##

    ##d = \sqrt{(vt)^2+(2H+(\frac{1}{2}gt^2))^2} - 2H##
     
  20. Sep 17, 2015 #19
    OK. Now next use the expression for v given in the problem statement and expand out the expression in parenthesis. Let's see what you get.

    Chet
     
  21. Sep 17, 2015 #20
    ##d = \sqrt{\frac{1}{4}g^2t^4+10Hgt^2+4H^2} - 2H##
     
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