Who stole my L???

A cubical block is given an initial velocity V on a rough horizontal floor...After some time, it is found to stop.
The angular momentum of the block about any point on the ground is non-zero initially but zero finally. Which force is single-handedly responsible for this loss:

A)Normal reaction.
B)Friction
C)Weight of the Block
D)Coriolis force.

7 comments:

  1. ankush sachdevaApril 8, 2010 at 9:45 AM

    normal reaction??

    ReplyDelete
  2. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○April 8, 2010 at 12:42 PM

    FRICTION????? CORIOLIS FORCE IS OUT OF SYLLABUS

    ReplyDelete
  3. akash rupelaApril 8, 2010 at 4:11 PM

    friction

    ReplyDelete
  4. AnujApril 8, 2010 at 10:50 PM

    ans. is normal reaction according to me.

    Akash, why sayest thou friction??

    ReplyDelete
  5. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○April 9, 2010 at 12:45 AM

    give reasoning sir please...

    ReplyDelete
  6. ♥◘○µ╦┬░Γ╪╬└↑φΦ╫╛◙«½╣↔⌐☺♠♦♣◘○April 9, 2010 at 1:06 PM

    where are you sir?

    ReplyDelete
  7. AnujApril 9, 2010 at 4:11 PM

    as you know, the normal reaction shifts(its line of action) away from the horizontal centre of the block, to avoid toppling.

    this shift also produces a (say) clockwise torque abt any pt on the ground, more than the anticlockwise torque produced by the weight, hence causing the loss in L...

    and don't call me sir. i'm 1 yr your senior.

    ReplyDelete

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