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The linear mass density of a rod AB of length 'L' is proportional to the 'n'th power of the distance from end A....

Find the distance of the COM from the end 'B'

23 comments:

  1. speed thrills..and kills!

    go a bit easy man!

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  3. i know bhaiya.
    i know misfired.
    is it l-(l^n-(l/2)^n)^1/n??

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  4. na...yaar put n=0 for a uniform rod and check..

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  5. bhaiya shudn't it b n=1 fr unifrom rod??

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  6. na..
    linear mass density is constant and hence proportional to x^0..

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  8. This comment has been removed by the author.

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  9. what happened
    y'd u remove the post
    is my answwr correct

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  10. haan yaar..teri post rremove kar di taki sab answer na dekh lein!

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  11. hey kapil..use the result to find the COM of a solid and hollow cone..

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  12. Is the answer.
    COM=L/(n+2) from end B.

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  13. haan yaar

    for a solid cone,
    density d is proportional to (r^2) and x/H=r/R
    so d is proportional to x^2
    so, com is H/4 from bottom

    similarly for hollow cone
    we get d proportional to r[sqrt(x^2+r^2)]/x
    put x/H=r/R to get d proportional to x
    and so com is H/3 from bottom

    really gud ....

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  14. L/n+2
    nd ye method of finding fr cone .. bhi mast h !! :)

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  15. because then you don't have to integrate

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  16. L/n+2.
    AT KAPIL ,great method for finding com for cone

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  17. Can somebody explain me the post from kapil.
    that how is d proportional to them in both the cases.
    Thanx

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  18. the radius of the differential disc element that you consider on moving a distance 'x' from the apex is proportional to x^2.
    hence its mass is also proportional to x^2...

    similarly formulate for a hollow cone!

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  19. yess l/n+2

    we did this type of question in class( there n was 1). we also derived area of circle= pi r^2 ....i understood the importance of integration that day

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  20. but my answer is coming out to be L*(n+1)/(n+2)
    please tell me where am i wrong.

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  21. Vijay,you got the distance from A .For distance from B , subtract your answer from L.

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  22. SIR PLEASE SOLVE THE PROBLEM GIVEN BELOW PLEASE!! A child of mass m sits in a swing of negligible mass suspended by a rope of length l. Assume that the dimensions of the child are negligible compared with 1.
    His father pulls the child back until the rope makes an angle of one radian with vertical, then pushes with a force F=mg along the arc of the circle releasing at the vertical :
    a) How high up will the swing go?
    b) How long did the father push?

    ans:(a) theta=63degree above horizon (b) t=1.52(l/g)^1/2

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