## anonymous 5 years ago A cubical block of side "a" is moving with velocity v on a horizontal smooth plane. It hits a ridge at point O. Find the angular speed of the block after it hits the point O. Please view the attached image for more details

1. anonymous

This is the attached diagram

2. anonymous

So this thing is just heading towards O which then acts as a point of rotation?

3. anonymous

Yes

4. anonymous

Okay...I just to do something, then I'll take a look.

5. anonymous

*just go to do something*

6. anonymous

Okay, I will be waiting

7. anonymous

Still waiting?

8. anonymous

Yes

9. anonymous

I'm thinking you can attack the problem by considering conservation of energy, assuming only internal, conservative forces are acting.

10. anonymous

You have two regimes - translational (before the ridge), then rotational.

11. anonymous

The kinetic energy of the block before hand is $K_i=\frac{1}{2}Mv^2$

12. anonymous

The kinetic energy of the block just after impact, as it enters the rotational regime, is$K_f=\frac{1}{2}I \omega^2$

13. anonymous

Is this making sense so far?

14. anonymous

Yes

15. anonymous

If I am not mistaken then the angular speed is coming to be (v/a)(sqrt6)

16. anonymous

Now, your problem is now about finding the moment of inertia for a block whose axis of rotation is about one of its edges. A solid, rectangular parallelpiped is given by...

17. anonymous

I get something close, but different.

18. anonymous

Oh I am sorry, I calculated the I around the Cm

19. anonymous

Which is wrong

20. anonymous

Yes

21. anonymous

So how do you calculate it around that corner?

22. anonymous

Parallel Axis Theorem

23. anonymous

Oh yes, you are right

24. anonymous

The new moment of inertia will be$I=I_{cm}+Mh^2$ where h is the distance from the CM axis to the new, parallel axis.

25. anonymous

$I_{cm}=\frac{Ma^2}{6}$

26. anonymous

$Mh^2=M \left( \frac{a}{2} \right)^2=M \frac{a^2}{4}$

27. anonymous

I think it is $1/6(Ma^2)+2M$

28. anonymous

So $I=\frac{5Ma^2}{12}$

29. anonymous

You see the corner wise distance will be sqrt 2

30. anonymous

I mean sqrt 2 times a

31. anonymous

$w^2=\frac{12v^2}{5a^2}$

32. anonymous

Yes, you're right there.

33. anonymous

So change h to a.sqrt(2)

34. anonymous

a/sqrt(2)

35. anonymous

$h=\frac{a}{\sqrt{2}}$

36. anonymous

So I am supposed to solve the equation $0.5 Mv^2=0.5[(1/6)Ma^2+2M]\omega$ Am I right

37. anonymous

am i right?

38. anonymous

I think the moment of inertia should be$I=\frac{Ma^2}{6}+Mh^2=\frac{Ma^2}{6}+M \left( \frac{a}{\sqrt{2}} \right)^2=\frac{Ma^2}{6}+\frac{Ma^2}{2}$

39. anonymous

$=\frac{4Ma^2}{6}=\frac{2Ma^2}{3}$

40. anonymous

Then$\frac{1}{2}Mv^2=\frac{1}{2}\frac{2Ma^2}{3} \omega ^2$

41. anonymous

i.e.$v^2=\frac{2a^2}{3} \omega ^2 \rightarrow \omega ^2 = \frac{3v^2}{2a^2}$

42. anonymous

Ok I understand that

43. anonymous

Are you agreeing? I'm doing this on-the-fly, so if you pick up anything, interrupt.

44. anonymous

Any how I have got the thing

45. anonymous

46. anonymous

Lokisan, I have got what I expected, I can handle the rest

47. anonymous

There was some problem with the site, and some of our converstation got split up

48. anonymous

Okay. Happy physics.

49. anonymous

lol, thanks for the genius comment. we'll just see how this question pans out.