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Instantaneous acceleration of blocks

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known data

    The blocks are of mass 2 kg shown is in equilibrium. At t = 0 right spring in fig (i) and right string in
    fig (ii) breaks. Find the ratio of instantaneous acceleration of blocks?


    2. Relevant equations



    3. The attempt at a solution

    In fig (i) before the string is cut,(2kx)sin37° = 2g , kx=g/sin37°

    In fig (ii) before the string is cut,(2T)sin37° = 2g , T=g/sin37°

    Is the instantaneous acceleration in the two cases same i.e ratio 1:1


    I would be grateful if somebody could help me with the problem .
     

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    Last edited: Mar 14, 2014
  2. jcsd
  3. Mar 14, 2014 #2

    BvU

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    I know physics isn't democracy, but I'm inclined to vote for 1:1 too!
    Net force after cutting will be minus the missing force, and these are equal.
    I'll be listening in to see if someone can prove me (us) wrong!
     
  4. Mar 14, 2014 #3

    PhysicoRaj

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    What is the state of the spring at t=0? Stretched? Equilibrium? Compressed?
     
  5. Mar 14, 2014 #4
    I don't think so. When the string is cut, the acceleration of the mass is purely tangential. When the spring is cut, the acceleration is not purely tangential.

    Chet
     
  6. Mar 14, 2014 #5
    Excellent thinking!!!

    Are you getting 25/24 ?
     
  7. Mar 14, 2014 #6
    I haven't actually worked the problem. Let me try it for real and see what I get.

    Chet
     
  8. Mar 14, 2014 #7
    Yes. I confirm that result.

    Chet
     
  9. Mar 14, 2014 #8
    Thanks Chet :smile:
     
  10. Mar 14, 2014 #9

    BvU

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    Impressed ! :redface:
     
  11. Mar 14, 2014 #10
    Tanya, can you please show how you solved this problem?

    Thanks!
     
  12. Mar 15, 2014 #11
    In the first case i.e having springs

    From equilibrium condition kx = 5g/3

    After string is cut , the acceleration of the block will have both radial and tangential components .

    at = 2gcos37° = 8g/5

    ar = kx - 2gsin37° = 5g/3 - 6g/5 = 7g/15

    ∴ net acceleration a1 = 5g/3

    In the second case i.e having strings

    From equilibrium condition T = 5g/3

    After string is cut , the acceleration of the block will be purely tangential i.e ar = 0

    at = 2gcos37° = 8g/5

    ∴ net acceleration a2 = 8g/5


    So a1/a2 = 25/24
     
  13. Mar 15, 2014 #12
    Excellent, thanks a lot Tanya! :smile:
     
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