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Circular Motion 2

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data

    While at a country fair, you decide to ride the Farris Wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 14m, and you use your watch to find that each loop around takes 26s. a.) What is your speed? b.) What is the magnitude of your acceleration? c.) What is the ratio of your apparent weight to your true weight at the top of the ride? d.) What is the ratio of your apparent weight to your true weight at the bottom?

    2. Relevant equations

    v = wr
    a = w^2r = v^2/r

    3. The attempt at a solution

    I successfully solved part a by finding the frequency. I then plugged the frequency into w = 2(pi)radians(f). Then I used v = wr to find the velocity, which is 3.346. I then tried using both equations for acceleration, but I didn't get the right answer. I got it to be .8m/s^2.
     
  2. jcsd
  3. Oct 5, 2007 #2

    learningphysics

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    I think your trouble may be with rounding... for example... I get 3.383 for the speed... and 0.8176 for acceleration...
     
  4. Oct 5, 2007 #3
    Yeah, that was the problem. Normally I complete each equation separately and round off the answer after I find the solution to each equation....if that makes sense. How do I do the apparent weight thing?
     
  5. Oct 5, 2007 #4

    learningphysics

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    The apparent weight is the normal force exerted by the seat on the person... use centripetal acceleration to find the normal force at the bottom and top...
     
  6. Oct 5, 2007 #5
    Is it just the same as the acceleration from before?
     
  7. Oct 5, 2007 #6

    learningphysics

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    yes.

    use [tex]\Sigma{F} = ma[/tex] at the bottom and top to find the normal force.
     
  8. Oct 5, 2007 #7
    One more question before I start, what is the mass?
     
  9. Oct 5, 2007 #8

    learningphysics

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    you don't need the mass... just use m.

    later you'll be getting ratios... and the m's will cancel.
     
  10. Oct 5, 2007 #9
    So would the normal force = .818(mass)?
     
  11. Oct 5, 2007 #10

    learningphysics

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    no. 0.818*mass is the net force... not the normal force.

    use [tex]\Sigma{F} = ma[/tex]

    0.818 goes in the right side... what forces go in the left side of this equation?
     
  12. Oct 5, 2007 #11
    normal force-centripetal force = ma
     
  13. Oct 5, 2007 #12

    learningphysics

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    no... centripetal force is not an independent force in itself... a centripetal force is the result of other forces...

    what are the regular forces acting on the person in the ferris wheel? normal force is correct. what is the other force?
     
  14. Oct 5, 2007 #13
    normal force - weight = ma
     
  15. Oct 5, 2007 #14

    learningphysics

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    yes. that is correct. that's at the bottom of the loop.

    at the top it is the opposite... weight-normal force = ma.
     
  16. Oct 5, 2007 #15
    Yes, the centripetal "force" in a sense is not a true force, it is an acceleration that is a result of other forces. Forces shouldn't be confused for accelerations.

    I get a ratio of 0.846 while using a radial acceleration approximate value of 0.846, what do you get?
     
  17. Feb 27, 2008 #16
    I have this same problem but with different variables and found my acceleration to be 1.61 m/s^2. Can you further explain how the ratio of apparent weight to true weight is determined?
     
  18. Feb 27, 2008 #17
    I figured it out so that can previous post can be ignored.
     
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