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Merry Go Round (Linear Speed)

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A rotating merry-go-round makes one complete revolution in 3.80 s. What is the linear speed of a child seated 1.24 m from the center?


    2. Relevant equations
    vt = wr
    w = 2(pi)/T
    Vt = 2(pi)r/T


    3. The attempt at a solution
    I know that the further you are from the center, the higher velocity that you'll have, but I'm not sure where to go with that.
     
  2. jcsd
  3. Dec 8, 2008 #2

    tiny-tim

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    Hi kaylanp01! :smile:

    hmm :rolleyes: … let's try an easy case …

    how far does the child go in 3.80s? :wink:
     
  4. Dec 8, 2008 #3

    LowlyPion

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    You derived the equation, so just plug in your numbers.
     
  5. Dec 8, 2008 #4
    So thats the equation for linear velocity? I'm confused. I just grabbed that out of my notes.
     
  6. Dec 8, 2008 #5
    Wellll...technically...he doesn't go anywhere, as he ends up in the same place ;).
     
  7. Dec 8, 2008 #6

    LowlyPion

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    Well your notes were right.

    The trick now is to understand what you took notes about then isn't it.
     
  8. Dec 8, 2008 #7
    I'm just trying to get through it now...try to pass the CAPA part of the course even if I can't pass the exam. haha. It says in my notes that the equation I derived above was for tangential v...so are tangential and linear the same thing?
     
  9. Dec 8, 2008 #8

    LowlyPion

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  10. Dec 8, 2008 #9

    LowlyPion

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    Yes tangential V is linear V.
     
  11. Dec 8, 2008 #10
    Okay, great.
    I looked at that website, and it eliminates my next question (I was going to ask how to calculate acceleration, but its all there) so you just saved me (and yourself) from more pain. haha.
    Thank you (twice). :)
     
  12. Dec 8, 2008 #11
    Okay. I've been defeated. I still can't get acceleration.
    I used a = 2.05/1.24 = 1.65 and thats not correct.
     
  13. Dec 8, 2008 #12

    LowlyPion

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    The acceleration is the centripetal acceleration v2/r
     
  14. Dec 8, 2008 #13
    So how will I know when its centripetal and when it isn't? I'm so confused. :(
     
  15. Dec 8, 2008 #14

    LowlyPion

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    If it is undergoing circular motion its mass is undergoing centripetal acceleration.

    When the disk is changing ω at some rate of increase that is called α, and it is called angular acceleration.
     
  16. Dec 8, 2008 #15
    Okay, its a little more clear now. I'm going to go study until my eyes water and/or begin to blur. Any study tips to offer?
     
  17. Dec 8, 2008 #16

    LowlyPion

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    Next term read ahead before the lectures.
     
  18. Dec 8, 2008 #17
    Thankfully, I won't be doing another physics course after this term. I usually read ahead throughout the semester, but really got off track within the past few weeks. Luckily, I currently have an A, so I can afford to do not-so-well on the final. ha.
     
  19. Dec 8, 2008 #18

    LowlyPion

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    Good luck on the exam then.

    Cheers.
     
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