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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

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  1. ksaimouli
    • 3 years ago
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    @henpen

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

  3. henpen
    • 3 years ago
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    Sorry, I get it now

  4. ksaimouli
    • 3 years ago
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    :-)

  5. ksaimouli
    • 3 years ago
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    i found angle

  6. henpen
    • 3 years ago
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    |dw:1352059091446:dw|\[T_y=mg\]

  7. ksaimouli
    • 3 years ago
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    |dw:1352059120155:dw|

  8. ksaimouli
    • 3 years ago
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    and the angle is 12

  9. ksaimouli
    • 3 years ago
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    after that i dont get

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

  11. henpen
    • 3 years ago
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    Get it?

  12. henpen
    • 3 years ago
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    |dw:1352059242987:dw|

  13. henpen
    • 3 years ago
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    As it does not move up or down

  14. henpen
    • 3 years ago
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    The mass, that is

  15. ksaimouli
    • 3 years ago
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    mass is .05kg

  16. ksaimouli
    • 3 years ago
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    what is the equation actually

  17. henpen
    • 3 years ago
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    Do you get why the equality is correct?

  18. ksaimouli
    • 3 years ago
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    |dw:1352059579894:dw|

  19. henpen
    • 3 years ago
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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.

  20. ksaimouli
    • 3 years ago
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    it is circular motion so that must be= mv^2/r

  21. henpen
    • 3 years ago
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    In the horizontal axis yes, but we are resolving in the verical axis now|dw:1352059672600:dw|

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

  23. henpen
    • 3 years ago
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    \[T_x=\frac{mv^2}{r}\]

  24. henpen
    • 3 years ago
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    \[v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s\]

  25. henpen
    • 3 years ago
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    |dw:1352059964786:dw|

  26. henpen
    • 3 years ago
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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/

  27. ksaimouli
    • 3 years ago
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    now i will explain my method can u correct me

  28. ksaimouli
    • 3 years ago
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    if i go wrong

  29. ksaimouli
    • 3 years ago
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    plz

  30. henpen
    • 3 years ago
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    开始

  31. henpen
    • 3 years ago
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    OK

  32. ksaimouli
    • 3 years ago
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    |dw:1352060338398:dw|

  33. ksaimouli
    • 3 years ago
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    Tcosx - mg =0

  34. ksaimouli
    • 3 years ago
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    T=mg/cos x

  35. ksaimouli
    • 3 years ago
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    T=6N

  36. henpen
    • 3 years ago
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    \[0.05*9.8/\cos12=0.5\]

  37. henpen
    • 3 years ago
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    Calculator error it seems

  38. ksaimouli
    • 3 years ago
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    lol it is radians

  39. katragaddasaichandra
    • 3 years ago
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    You need to provide dimensions along with your numbers.

  40. ksaimouli
    • 3 years ago
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    where did the .1 came from

  41. henpen
    • 3 years ago
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    0.5 cos78

  42. ksaimouli
    • 3 years ago
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    i got .5N for tension

  43. ksaimouli
    • 3 years ago
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    why did u find that one

  44. ksaimouli
    • 3 years ago
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    because horizontal force?

  45. henpen
    • 3 years ago
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    T=0.5 I got when resolving vertically\[Tsin \theta=mg\]

  46. henpen
    • 3 years ago
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    |dw:1352061108396:dw|

  47. ksaimouli
    • 3 years ago
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    got u thx

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