shubhamsrg 2 years ago 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

1. shubhamsrg

|dw:1357204606212:dw|

2. shubhamsrg

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

3. Diwakar

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.

4. 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 ?

5. Diwakar

Are you sure you are getting the I right?

6. shubhamsrg

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

7. shubhamsrg

:P

8. Diwakar

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

9. shubhamsrg

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

10. yrelhan4

11. Diwakar

@yrelhan4 Did you use Mux/2=Iw

12. yrelhan4

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

13. Vincent-Lyon.Fr

Use conservation of angular momentum, not of kinetic energy.

14. Vincent-Lyon.Fr

Moment of inertia about edge is correct.

15. shubhamsrg

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

16. shubhamsrg

thanks all..