## experimentX Group Title Q. one year ago one year ago

1. experimentX Group Title

|dw:1350734665748:dw|

2. Callisto Group Title

Third case: T = g?

3. experimentX Group Title

well ... its g N

4. experimentX Group Title

first case??

5. Callisto Group Title

Wait. N = unit?

6. experimentX Group Title

yes ... tension always has unit of force ... except for surface tension.

7. Callisto Group Title

First case: Σ F = mg mg - T = mg T = 0

8. Callisto Group Title

Second case: Σ F = ma mg - T = ma T = m (g-a) = m(g-5) Unit: N for first and second case.

9. Callisto Group Title

But..... but.... but.... the reel.... Shouldn't I take it into consideration??

10. experimentX Group Title

you get acceleration after considering the rotational motion of reel. it shouldn't make any difference.

11. experimentX Group Title

in other words you have already taken care of rotational motion of reel. still if you need to verity then ... T*R = torque being applied on reel.

12. experimentX Group Title

sorry .. mg*R - TR = torque being applied on reel ... that makes it go round.

13. experimentX Group Title

mg*R - TR <--- this part creates motion.

14. Callisto Group Title

acc. of the reel = acc. of the mass?

15. experimentX Group Title

no ... the reel does not have linear acceleration. It's rotating ... it has angular acceleration.

16. Callisto Group Title

tangential acc. of the reel = acc. of teh mass?

17. experimentX Group Title

yes ... maybe ... worth checking it out.

18. Callisto Group Title

If we don't have the acc. of the reel, how can we suppose to find torque, so that we can find tension!?

19. experimentX Group Title

well .. we have to find the acceleration reel first.

20. experimentX Group Title

or acceleration of the mass falling down first.

21. Callisto Group Title

But.. how...

22. experimentX Group Title

using conservation of energy ${1 \over 2} mv^2 + {1 \over 2}I \omega^2 = mgh \\ \omega r = v,$ solving those two will give you final velocity, you know distance 'd', and initial velocity.

23. experimentX Group Title

solve for acceleration of the using it.

24. Callisto Group Title

Wow! We're doing it backward!

25. Callisto Group Title

(c) -> (b) -> (a) :O

26. experimentX Group Title

well ...

27. Callisto Group Title

That's not a problem :P

28. experimentX Group Title

the other way ... "tangential acc. of the reel = acc. of teh mass" ... at the circumference of reel ... they go foot to foot.

29. Callisto Group Title

Yes. Then we can find the angular acc. Using integration to find the moment of inertia, and then use it to find the torque. Is it possible?

30. experimentX Group Title

yep .. this is also possible.

31. Callisto Group Title

For the energy one, can we really use that way?

32. experimentX Group Title

sure ... the torque will always be constant ... hence there will be constant acceleration.

33. Callisto Group Title

I'm sorry, but how do I know if the torque is constant?

34. experimentX Group Title

the torque depends on two parameters $$r \times F$$ , F is part of mgh while r is radius of wheel. both are not changing ... neither is their angle between them.

35. Callisto Group Title

mgh?! the PE of the reel or..?!

36. Callisto Group Title

hmm?

37. experimentX Group Title

mass .. hold on ... there seems to be discrepancy with second method.

38. experimentX Group Title

|dw:1350737767080:dw|

39. experimentX Group Title

|dw:1350737955528:dw|

40. Callisto Group Title

Hmm.. So, you're just considering the mass m ..?

41. experimentX Group Title

|dw:1350738106593:dw|

42. Callisto Group Title

|dw:1350738529141:dw|