The rod (weight=W) is hinged at A. It is released from rest from the position shown in black.

1)What is the reaction from the hinge at this instant??

2)What is the reaction when the rod has rotated by 45 degrees?

13 comments:

  1. somay jainNovember 10, 2010 at 11:30 PM

    1) W/4
    2) (root 34 /8) W

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  2. AmitNovember 10, 2010 at 11:59 PM

    1.W/4
    2.root(65/32)W

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  3. Sambhav GuptaNovember 11, 2010 at 4:04 AM

    1)W/4
    2)root(17/32)W

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  4. somay jainNovember 11, 2010 at 5:33 PM

    i meant ((root34)/8)W
    same as sambhav's

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  5. Pratik JainNovember 11, 2010 at 7:16 PM

    i am not able to understand why we have to take the acceleration of centre of mass when the rod is in pure rotation......

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  6. AmitNovember 11, 2010 at 7:25 PM

    ok, I solved it like this(I think I have done a calculation mistake but concept is right)
    Loss in P.E.=gain in K.E.
    from here we will get omega.
    Impulse(along rod)-mg/root2=m(omega)^2(l/2)
    find alpha at that position.
    mg/(root2)-I(perpendicular to rod)=m(accn of COM)

    Now take net of I

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  7. Sambhav GuptaNovember 11, 2010 at 8:22 PM

    me, (and probably somay too) have taken impluse(along rod) = mg/root2

    i have not considered radial acceleration, so, i must say your answer is correct

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  8. somay jainNovember 12, 2010 at 10:41 AM

    i took the most general case..
    impulse in X, Y lia..
    Acm lia perpendicular to the rod, alpha lia clockwise..

    then i set up 2 equations using newton's laws and one torque equation..
    also, Acm=L/2 alpha..

    isse a gya..
    interesting that i got the same ans as sambhav..

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  9. Sambhav GuptaNovember 12, 2010 at 11:10 AM

    arre yr somay dhang se pad, hum dono ne radial acceleration nhi lia, the CM of the rod is moving in a circle of radius l/2, so mw^2r vaali force bhi lagegi

    and btw, maine 3 tarah se kia, but since i didn't consider radial acc, answer vahi aaya

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  10. somay jainNovember 12, 2010 at 11:56 AM

    okay.. haan..
    maine to bina dekhe hi likh dia..
    1st part mein aisa kuch nhi tha but 2nd part mein hoga.. nice ques..!!

    ReplyDelete
  11. shouryaNovember 13, 2010 at 10:19 PM

    This comment has been removed by the author.

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  12. shouryaNovember 18, 2010 at 12:29 PM

    i am getting W root(58)/4 in part 2.

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  13. Tushar SinglaNovember 18, 2010 at 12:39 PM

    1. w/4

    2.
    after considering radial acc...along dir of rod....
    nd assuming components Fx Fy along nd Perpendicular to rod
    Fy = W/4root(2)

    Fx - W/root(2) = mv^2/(l/2)

    By COE.... v^2 = gl/root(2)
    Fx = 3W/root(2)

    Fnet = sqrt(Fx^2 + Fy^2) = root( 145/32)..

    ReplyDelete

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