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Acceleration on Lowest and Highest Points (Vertical Circular Motion)

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A boy rides a Ferris wheel with a radius of 10.0m. The wheel completes one turn every 30.0s. Find the boy's acceleration at the lowest and highest points on the circular path.

    2. Relevant equations


    [tex]\frac{m(V_{max})^{2}}{r}=T-F_{w}[/tex]

    3. The attempt at a solution

    I tried solving the highest point, meaning the part when the velocity is at its minimum value. I got [tex]9.80 m/s^{2}[/tex]. The value of g that I used was [tex]g = 9.80 m/s^{2}[/tex].
     
  2. jcsd
  3. Feb 5, 2010 #2

    ideasrule

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    Gravity is not accelerating the boy at 9.8 m/s^2 because the seat is pushing back.

    This question is easier than you might expect. The boy is moving in a circle. Therefore, acceleration is given by the equation a=___________
     
  4. Feb 5, 2010 #3
    So 9.80 m/s[tex]^{2}[/tex] is not the answer for the highest point?

    [tex]a=\frac{V^{2}}{r}[/tex] right? But I don't have the Velocity and I don't know how to get it with those givens.

    How about for the lowest point? Meaning at the bottom part of the ferris wheel.
     
  5. Feb 5, 2010 #4

    ideasrule

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    No, a=v^2/r. You can solve for velocity very easily. The circle is moving in a 10-meter radius in 30 seconds. How much distance does the boy cover in 30 seconds?
     
  6. Feb 5, 2010 #5
    The distance is the circumference of the circle right? Then I'll divide the answer by 30 secs to get the velocity?

    Ohhh. I really don't know. I got more confused. :( Please help.
     
    Last edited: Feb 5, 2010
  7. Feb 5, 2010 #6

    ideasrule

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    No, he doesn't move 300 m. He moves the circumference of the circle, which has a radius of 10 m.
     
  8. Feb 5, 2010 #7
    Yes I forgot. The 300 answer was very stupid. Sorry, was my edited post correct? Can you explain it to me fully? I really need. That's the only questions I wasn't able to answer properly on our test. :(
     
  9. Feb 5, 2010 #8

    ideasrule

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    Exactly.
     
  10. Feb 5, 2010 #9
    The answer is 0.437 meters per second squared? That's the acceleration to what? Lowest point? Meaning at the bottom of the ferris wheel?
     
  11. Feb 5, 2010 #10

    ideasrule

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    That's the centripetal acceleration, meaning acceleration towards the center. Whenever something moves only in a circle (and not perpendicular to the circle), it ALWAYS has a centripetal force of a=v^2/r. Whether or not the object is in a gravitational field doesn't matter.
     
  12. Feb 5, 2010 #11
    I'm a bit confused about that statement. Centripetal force is mv^2/r right? not v^2/r (this is the centripetal acceleration). Sir, so 437 meters per second squared is not the answer to any of my questions? The accceleration on the lowest and highest point? or is it the answer to the lowest point or highest point?
     
  13. Feb 5, 2010 #12

    ideasrule

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    Oops, I meant centripetal acceleration, not force. Sorry about that.

    0.437 m/s^2 is the answer to both questions. Whenever something is going in a circle, its acceleration is ALWAYS v^2/r. Whether it's at the top or at the bottom of the circle doesn't matter.
     
  14. Feb 6, 2010 #13
    Oh ok sir. I'm gonna trust you with that one. :) Thanks! :)
     
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