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



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shubhamsrg
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dw:1357204606212:dw

shubhamsrg
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i know angular momentum will be conserved here but unable to apply it..

Diwakar
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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.

shubhamsrg
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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 ?

Diwakar
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Are you sure you are getting the I right?

shubhamsrg
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2/3 mx^2 ..i googled it!
though i can derive that also..

shubhamsrg
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:P

Diwakar
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what is the correct answer? I will try to solve it at some other time?

shubhamsrg
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though that was an option which came by this method,
correct ans is
3u/2x

yrelhan4
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answer is 3v/4x..

Diwakar
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@yrelhan4
Did you use Mux/2=Iw

yrelhan4
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yup.. I would be icm + ma^2/2

VincentLyon.Fr
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Use conservation of angular momentum, not of kinetic energy.

VincentLyon.Fr
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Moment of inertia about edge is correct.

shubhamsrg
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ohh yes,,3u/4x..got it,,was easy..hmm

shubhamsrg
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thanks all..