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alexeis_nicole

  • 3 years ago

describe the motion of a point on a wheel that has a centre 4m off the ground, has radius of 15 cm, makes a full rotation every 10 seconds and starts at its highest point. use y=asink(t-b)+c model or cosine function.

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  1. Algebraic!
    • 3 years ago
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    c=4 and a=.15 ... is that part clear?

  2. alexeis_nicole
    • 3 years ago
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    a=15 not .15..

  3. Algebraic!
    • 3 years ago
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    false. :) continuing... \[f=\frac{ 1 }{10}\] \[\omega = 2 \pi f\] \[w= \frac{ \pi }{ 5 }\]

  4. Algebraic!
    • 3 years ago
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    agree so far?

  5. alexeis_nicole
    • 3 years ago
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    yes i got that part

  6. Algebraic!
    • 3 years ago
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    so the phase...

  7. Algebraic!
    • 3 years ago
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    we want it to be at the highest point when t=0... sin =1 when theta = pi/2 pi/5 *(t-b) = pi/2 pi/5(-b) = pi/2 b= 5/2

  8. Algebraic!
    • 3 years ago
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    \[y = .15 \sin (\frac{ \pi }{ 5} (t+\frac{ 5 }{2})) +4\]

  9. alexeis_nicole
    • 3 years ago
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    would it be appropriate to keep(t-b) it at pi/5 *(t) + pi/2 ?

  10. Algebraic!
    • 3 years ago
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    you can... that's not the form they ask you to use though...

  11. Algebraic!
    • 3 years ago
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    whoops, should say b=-5/2 up there... typo

  12. alexeis_nicole
    • 3 years ago
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    oh okay. & also for cosine model it would just be \[y=.15\cos(\frac{ \pi }{ 5 }t) +4\] right??

  13. Algebraic!
    • 3 years ago
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    yes

  14. alexeis_nicole
    • 3 years ago
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    why would it be -5/2 and not 5/2 ? :S

  15. Algebraic!
    • 3 years ago
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    why is b= -5/2? is that what you're asking?

  16. alexeis_nicole
    • 3 years ago
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    OH WAIT.. nvm.. you were referring to your reply before the equation. nvm. lol

  17. alexeis_nicole
    • 3 years ago
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    thanks!

  18. Algebraic!
    • 3 years ago
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    sure:)

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