## experimentX Group Title Q. 2 years ago 2 years ago

1. experimentX

|dw:1350734665748:dw|

2. Callisto

Third case: T = g?

3. experimentX

well ... its g N

4. experimentX

first case??

5. Callisto

Wait. N = unit?

6. experimentX

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

7. Callisto

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

8. Callisto

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

9. Callisto

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

10. experimentX

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

11. experimentX

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

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

13. experimentX

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

14. Callisto

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

15. experimentX

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

16. Callisto

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

17. experimentX

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

18. Callisto

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

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

20. experimentX

or acceleration of the mass falling down first.

21. Callisto

But.. how...

22. experimentX

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

solve for acceleration of the using it.

24. Callisto

Wow! We're doing it backward!

25. Callisto

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

26. experimentX

well ...

27. Callisto

That's not a problem :P

28. experimentX

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

29. Callisto

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

yep .. this is also possible.

31. Callisto

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

32. experimentX

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

33. Callisto

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

34. experimentX

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

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

36. Callisto

hmm?

37. experimentX

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

38. experimentX

|dw:1350737767080:dw|

39. experimentX

|dw:1350737955528:dw|

40. Callisto

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

41. experimentX

|dw:1350738106593:dw|

42. Callisto

|dw:1350738529141:dw|