Let Tension lead the way..



Relate the accns. of the masses in the diagrams.

In the 1st diagram, the thread goes through a very small ring, free to slide on a fixed beam..You have to relate the accn. when the angle (specified in dig.) is A.

23 comments:

  1. ankush sachdevaApril 4, 2010 at 8:23 PM

    for the second one
    its a2=a1 cosA

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  2. ankush sachdevaApril 4, 2010 at 8:38 PM

    for the second ques

    a1=a2=a3 / 2

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  3. piyush gargApril 4, 2010 at 8:53 PM

    a1cosA=a2sinA

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  4. AnujApril 4, 2010 at 9:02 PM

    0/3..

    no one's got anything right yet...
    You can use the following approach:
    1)Take all variable distances relative to fixed points.
    2)Express all constraints in terms of fixed lengths and these variable distances.
    3)Differentiate...

    OR...
    Let tension guide the way....

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  5. Sambhav GuptaApril 4, 2010 at 9:21 PM

    for the ring one, the acc of the block 'A' is related to acc of the ring 'a' as..

    a + A(1-cosQ)=0
    obviously, when of them has to be decreasing for the other increasing...

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  6. UnknownApril 4, 2010 at 10:08 PM

    This comment has been removed by the author.

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  7. UnknownApril 4, 2010 at 10:25 PM

    m getting it as a(cosQ-1)-AcosQ=0

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  8. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○April 4, 2010 at 10:41 PM

    is the answer is , if the acceleration of m3=a,then acceleration of m2=m1=a/6

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  9. Sambhav GuptaApril 4, 2010 at 10:59 PM

    for the pulley block one...
    A3 + 4A2 + 2A1=0

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  10. ankush sachdevaApril 4, 2010 at 11:45 PM

    fr the ring problem

    a1(1-cosA)+a2=0

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  11. AnujApril 5, 2010 at 1:04 AM

    there are a few right answers and many wrong ones here..
    the last two posts look ok..

    anyways, let tension lead the way..

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  12. piyushApril 5, 2010 at 11:32 AM

    for the pulley one i am getting it as 2a1+3a2-a3=0

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  13. piyushApril 5, 2010 at 11:46 AM

    for the 2nd one it is a1(1-cosa)+a2=0

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  14. AnujApril 5, 2010 at 7:53 PM

    correct answers:
    1)a1(1-CosA)+a2=0;

    2)2a1+4a2+a3=0;...assuming all accn. to be in the downwards direction.

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  15. Sambhav GuptaApril 5, 2010 at 8:14 PM

    ring vale ka solution de do bhaiya....i m getting => a2(1-CosA)+a1=0

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  16. AnujApril 5, 2010 at 10:42 PM

    Hey sambhav..i'm confused myself.
    I was trying to solve the ring one but was stuck...

    I'll post the solution (if it exists) soon.

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  17. AnujApril 5, 2010 at 10:44 PM

    and i'm nearly sure that a1(1-CosA)+a2=0 isn't the right answer...

    this expression gives the velocity constraint ie.

    v1(1-CosA)+v2=0...but while differentiating this you can't assume CosA to be const..

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  18. ankush sachdevaApril 5, 2010 at 11:24 PM

    phir to bohot complicated sa answer aayega

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  19. SHUBHAMApril 5, 2010 at 11:53 PM

    please explain the solution

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  20. rajeev guptaApril 5, 2010 at 11:55 PM

    This comment has been removed by the author.

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  21. Sambhav GuptaApril 5, 2010 at 11:59 PM

    even i thought of differentiating the angle as well, but then thought that the answer shouldn't be that complicated..

    i would still be waiting for the answer..

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  22. piyush gargApril 6, 2010 at 5:58 PM

    why don't we equate along the string like we do for most such cases???

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  23. AnujApril 6, 2010 at 8:25 PM

    you can do that if you know the accn. of the string in contact with the ring (in terms of velocity of the ring)...that's not simple.

    anyways, i found out my mistake..turns out we can't relate the accns. after all...

    we can relate the velocities only..sorry

    ReplyDelete

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