Looks difficult, but ain't:


The system shown consists of a disc and a ring, of equal mass 1kg, and equal radius 1m. Their tops are connected by a smooth mass less thread which winds over them.(as usual)

A time varying force F=kt is applied to the center of the disc. What is the force of friction acting on the ring (in terms of k and assuming that it is rolling without slipping) and in which dirn. does it act???

12 comments:

  1. Sambhav GuptaMarch 17, 2010 at 1:56 PM

    ring k case main jab force top par lagta hai...to friction 0 nhi aata rolling w/o slipping main?

    ReplyDelete
  2. ankush sachdevaMarch 17, 2010 at 2:11 PM

    haa,and we can even check that from previous ques
    "ZTL"

    ReplyDelete
  3. somay jainMarch 17, 2010 at 2:49 PM

    disc pe 0 friction thodi ayega..??

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  4. ankush sachdevaMarch 17, 2010 at 2:52 PM

    sambhav was talking about ring

    disk pe
    i m gettng
    kt/(3-(k^2/r^2))
    not very sure of this

    ReplyDelete
  5. AmitMarch 17, 2010 at 4:47 PM

    yes on ring friction would be zero
    I have a doubt that if ring is moving rolling without slipping then is it right to say that the disc is also moving like the same???

    if that's true then I am getting the answer as friction on disc=(3/7)kt

    practically it will cease at (mu)(1g) but that has not been mentioned in the question.

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  6. Sambhav GuptaMarch 17, 2010 at 5:10 PM

    yup....f=3kt/7......f=friction on the disc...opposite to the direction of motion

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  7. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○March 17, 2010 at 5:11 PM

    Yes, on the ring friction acting on it will be zero. Ring does not require friction for rolling.

    ReplyDelete
  8. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○March 17, 2010 at 6:42 PM

    I HAVE A DOUBT SIR PLEASE VISIT IN THE LINK GIVEN BELOW AND PLEASE SOLVE IT COMPLETELY PLEASE HELP ME....... http://www.goiit.com/posts/list/mechanics-please-solve-it-in-the-link-http-www-imagina-1011464.htm

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  9. AnujMarch 17, 2010 at 7:17 PM

    i claim that extension in both springs is 10cm. next, you can apply cons. of eng. to get one eqn. in terms of v1 and v2.

    for the second eqn. :
    let the distance moved by blocks 1 and 2 are x1 and x2. then the extensions in the springs are: x1 and (x2-x1).

    as the thread is massless:
    3x1=1(x2-x1).
    differentiate: 3v1=v2-v1.


    does this help???

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  10. ankush sachdevaMarch 17, 2010 at 8:22 PM

    ohh yup...i assumed alpha wrongly...me too getting (3/7)F

    ReplyDelete
  11. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○March 17, 2010 at 9:03 PM

    SIR THE ANSWER IS 1.8m/sec ,if you get it . I cant get it. SIR if you get it please right it completely please as comment to help me.

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  12. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○March 17, 2010 at 10:41 PM

    SIR CHECK BELOW EQUATION . I DONT THINK IT IS CORRECT 3x1=1(x2-x1).

    ReplyDelete

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