Fear Factor


A spool of radius 100m and mass 100kg has 100 grooves of radii 1m, 2m, ..., 100m. We use pulleys to suspend masses from the grooves as shown. The mass (i)kg is associated with the groove of radius (100-i)m.

Find the angular acceleration of the spool.

18 comments:

  1. Pulkit MendirattaOctober 26, 2011 at 12:32 PM

    Angular acceleration of spool ?

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  2. RohinOctober 26, 2011 at 5:52 PM

    there can be many questions in a way.
    acc of COM of spool;its angular acc.;accelerations of the blocks,their relative acc.;max relative acc of blocks ; acc. of COM of the spool,blocks system ;time when all blocks are at same height,if possible(in case their y coordinates are given)....many more

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  3. AnujNovember 5, 2011 at 5:25 PM

    wasssup?

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  4. Pulkit MendirattaNovember 6, 2011 at 1:34 AM

    This comment has been removed by the author.

    ReplyDelete
  5. shivam gargshyaNovember 9, 2011 at 10:18 AM

    bahiya can u explain centre of rotation method i can not get this by your post u created in march 2010

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  6. SumantraNovember 10, 2011 at 6:15 PM

    pulkit,MOI 577600 kaise aaya???

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  7. Pulkit MendirattaNovember 11, 2011 at 11:22 PM

    This comment has been removed by the author.

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  8. Pulkit MendirattaNovember 11, 2011 at 11:44 PM

    found an error in it..err.calculation...
    I had read somewhere that moi of a spool is the rms of radii, though i myself doubt this one...:P

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  9. AnujNovember 13, 2011 at 1:21 PM

    sorry if you got confused...usually we assume:
    spool~cylinder...

    i.e. MOI of a spool of mass 100Kg and radius 1 (with any number of grooves) is 100*1*1/2=50 units.

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  10. AnujNovember 13, 2011 at 1:22 PM

    hi pulkit...no need to delete your comments..

    they let ppl know the "Chain of Thought" which is fun.

    anyway does anybody have the answer yet?

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  11. MayankNovember 13, 2011 at 6:58 PM

    8.4*10^(-3)rad/s^2 ?

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  12. RohinNovember 14, 2011 at 11:29 AM

    1.69*10^(-3) ?

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  13. anshulNovember 27, 2011 at 11:13 PM

    Iα=2*g*(1*100 + 2*99 + ....49*50)+ g*(50)*(50)

    I= 1/2*1*(100)^2

    m nt gettin how to solve the series....:/

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  14. AnujNovember 30, 2011 at 12:37 AM

    Hi anshul...you CAN solve the series. its quite simple.

    Well the answer will be:
    I*alpha=(1*100+2*99+...+1*100)*g

    'I' is (100*(100^2)/2)+(100*(1^2)+99*(2^2)+...+1*(100^2))

    Obviously I didn't calculate the answer. Its based on an observation I made about 3.5 years ago. WOW SUCH A LONG TIME HAS PASSED!!!

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  15. anshulDecember 5, 2011 at 2:35 PM

    bhaiyaa plz tell me how did u get dat I?
    how can u consider the masses of blocks in I?

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  16. AnujDecember 5, 2011 at 3:42 PM

    Hi Anshul. If you read the latest 3-4 posts:

    http://anujkalia.blogspot.com/2011/11/lets-learn-something-new.html

    http://anujkalia.blogspot.com/2011/11/read-previous-post-first.html#comment-form

    http://anujkalia.blogspot.com/2011/12/and-now-try-these.html#comment-form

    you will "understand" it.

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  17. anshulDecember 10, 2011 at 1:52 AM

    bhaiya how can u apply dis method at the centre ..... as we can only apply dat method abt d pt instantaneously at rest....

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  18. AnujDecember 10, 2011 at 8:47 AM

    yes I made a mistake there. here is the correction:
    I*alpha=(1*200+2*199+...+100*101)*g
    {torques about the Center of Rotation}

    'I' is (100*(100^2)/2+100*(100^2))+
    (100*(101^2)+99*(102^2)+...+1*(200^2))
    {Moment of Inertia about the center of rotation}

    ReplyDelete

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