shubhamsrg
  • shubhamsrg
A cube of side x is moving with velocity u on a horizontal smooth plane. There is a ridge at O. find angular speed of the cube after it hits O
Physics
chestercat
  • chestercat
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shubhamsrg
  • shubhamsrg
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shubhamsrg
  • shubhamsrg
i know angular momentum will be conserved here but unable to apply it..
anonymous
  • anonymous
Apply the conservation of energy. Mu^2/2=Iw^2/2 where I is the moment of inertia about the vertical edge about which it will rotate.

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shubhamsrg
  • shubhamsrg
so, mu^2 = 2/3 (mx^2) * w^2 w^2 = 3u^2 /2x^2 nopes , not the right ans why you used cons of energy ?
anonymous
  • anonymous
Are you sure you are getting the I right?
shubhamsrg
  • shubhamsrg
2/3 mx^2 ..i googled it! though i can derive that also..
shubhamsrg
  • shubhamsrg
:P
anonymous
  • anonymous
what is the correct answer? I will try to solve it at some other time?
shubhamsrg
  • shubhamsrg
though that was an option which came by this method, correct ans is 3u/2x
yrelhan4
  • yrelhan4
answer is 3v/4x..
anonymous
  • anonymous
@yrelhan4 Did you use Mux/2=Iw
yrelhan4
  • yrelhan4
yup.. I would be icm + ma^2/2
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Use conservation of angular momentum, not of kinetic energy.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Moment of inertia about edge is correct.
shubhamsrg
  • shubhamsrg
ohh yes,,3u/4x..got it,,was easy..hmm
shubhamsrg
  • shubhamsrg
thanks all..

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