## ksaimouli 3 years ago a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calculate the speed of the mass

1. ksaimouli

@henpen

2. henpen

|dw:1352058857282:dw| Where does the 0.25m circle come into this? I don't quite get the question

3. henpen

Sorry, I get it now

4. ksaimouli

:-)

5. ksaimouli

i found angle

6. henpen

|dw:1352059091446:dw|$T_y=mg$

7. ksaimouli

|dw:1352059120155:dw|

8. ksaimouli

and the angle is 12

9. ksaimouli

after that i dont get

10. henpen

Using MY angle theta (so I can copy/paste pictures easier later), $0.05g=Tsin(\theta)$

11. henpen

Get it?

12. henpen

|dw:1352059242987:dw|

13. henpen

As it does not move up or down

14. henpen

The mass, that is

15. ksaimouli

mass is .05kg

16. ksaimouli

what is the equation actually

17. henpen

Do you get why the equality is correct?

18. ksaimouli

|dw:1352059579894:dw|

19. henpen

$T_y-mg=0$I think, as the NET force in the vertical axis is 0, as it doesn't move up or down.

20. ksaimouli

it is circular motion so that must be= mv^2/r

21. henpen

In the horizontal axis yes, but we are resolving in the verical axis now|dw:1352059672600:dw|

22. henpen

$T=\frac{0.05 *9.8}{\sin78}\approx0.5 N$ $T_x=Tcos(\theta)=0.5\cos(78)\approx0.1N$

23. henpen

$T_x=\frac{mv^2}{r}$

24. henpen

$v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s$

25. henpen

|dw:1352059964786:dw|

26. henpen

Yes I know that centrifugal is fictitious, but it's required if you want to use Newton's Laws in rotating systems. http://xkcd.com/123/

27. ksaimouli

now i will explain my method can u correct me

28. ksaimouli

if i go wrong

29. ksaimouli

plz

30. henpen

开始

31. henpen

OK

32. ksaimouli

|dw:1352060338398:dw|

33. ksaimouli

Tcosx - mg =0

34. ksaimouli

T=mg/cos x

35. ksaimouli

T=6N

36. henpen

$0.05*9.8/\cos12=0.5$

37. henpen

Calculator error it seems

38. ksaimouli

You need to provide dimensions along with your numbers.

40. ksaimouli

where did the .1 came from

41. henpen

0.5 cos78

42. ksaimouli

i got .5N for tension

43. ksaimouli

why did u find that one

44. ksaimouli

because horizontal force?

45. henpen

T=0.5 I got when resolving vertically$Tsin \theta=mg$

46. henpen

|dw:1352061108396:dw|

47. ksaimouli

got u thx