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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
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i know angular momentum will be conserved here but unable to apply it..
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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Other answers:

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

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