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ksaimouli Group Title

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

  • one year ago
  • one year ago

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  1. ksaimouli Group Title
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    @henpen

    • one year ago
  2. henpen Group Title
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    |dw:1352058857282:dw| Where does the 0.25m circle come into this? I don't quite get the question

    • one year ago
  3. henpen Group Title
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    Sorry, I get it now

    • one year ago
  4. ksaimouli Group Title
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    :-)

    • one year ago
  5. ksaimouli Group Title
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    i found angle

    • one year ago
  6. henpen Group Title
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    |dw:1352059091446:dw|\[T_y=mg\]

    • one year ago
  7. ksaimouli Group Title
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    |dw:1352059120155:dw|

    • one year ago
  8. ksaimouli Group Title
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    and the angle is 12

    • one year ago
  9. ksaimouli Group Title
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    after that i dont get

    • one year ago
  10. henpen Group Title
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    Using MY angle theta (so I can copy/paste pictures easier later), \[0.05g=Tsin(\theta)\]

    • one year ago
  11. henpen Group Title
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    Get it?

    • one year ago
  12. henpen Group Title
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    |dw:1352059242987:dw|

    • one year ago
  13. henpen Group Title
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    As it does not move up or down

    • one year ago
  14. henpen Group Title
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    The mass, that is

    • one year ago
  15. ksaimouli Group Title
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    mass is .05kg

    • one year ago
  16. ksaimouli Group Title
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    what is the equation actually

    • one year ago
  17. henpen Group Title
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    Do you get why the equality is correct?

    • one year ago
  18. ksaimouli Group Title
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    |dw:1352059579894:dw|

    • one year ago
  19. henpen Group Title
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    \[T_y-mg=0\]I think, as the NET force in the vertical axis is 0, as it doesn't move up or down.

    • one year ago
  20. ksaimouli Group Title
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    it is circular motion so that must be= mv^2/r

    • one year ago
  21. henpen Group Title
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    In the horizontal axis yes, but we are resolving in the verical axis now|dw:1352059672600:dw|

    • one year ago
  22. henpen Group Title
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    \[T=\frac{0.05 *9.8}{\sin78}\approx0.5 N\] \[T_x=Tcos(\theta)=0.5\cos(78)\approx0.1N\]

    • one year ago
  23. henpen Group Title
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    \[T_x=\frac{mv^2}{r}\]

    • one year ago
  24. henpen Group Title
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    \[v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s\]

    • one year ago
  25. henpen Group Title
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    |dw:1352059964786:dw|

    • one year ago
  26. henpen Group Title
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    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/

    • one year ago
  27. ksaimouli Group Title
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    now i will explain my method can u correct me

    • one year ago
  28. ksaimouli Group Title
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    if i go wrong

    • one year ago
  29. ksaimouli Group Title
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    plz

    • one year ago
  30. henpen Group Title
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    开始

    • one year ago
  31. henpen Group Title
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    OK

    • one year ago
  32. ksaimouli Group Title
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    |dw:1352060338398:dw|

    • one year ago
  33. ksaimouli Group Title
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    Tcosx - mg =0

    • one year ago
  34. ksaimouli Group Title
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    T=mg/cos x

    • one year ago
  35. ksaimouli Group Title
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    T=6N

    • one year ago
  36. henpen Group Title
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    \[0.05*9.8/\cos12=0.5\]

    • one year ago
  37. henpen Group Title
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    Calculator error it seems

    • one year ago
  38. ksaimouli Group Title
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    lol it is radians

    • one year ago
  39. katragaddasaichandra Group Title
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    You need to provide dimensions along with your numbers.

    • one year ago
  40. ksaimouli Group Title
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    where did the .1 came from

    • one year ago
  41. henpen Group Title
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    0.5 cos78

    • one year ago
  42. ksaimouli Group Title
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    i got .5N for tension

    • one year ago
  43. ksaimouli Group Title
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    why did u find that one

    • one year ago
  44. ksaimouli Group Title
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    because horizontal force?

    • one year ago
  45. henpen Group Title
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    T=0.5 I got when resolving vertically\[Tsin \theta=mg\]

    • one year ago
  46. henpen Group Title
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    |dw:1352061108396:dw|

    • one year ago
  47. ksaimouli Group Title
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    got u thx

    • one year ago
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