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Homework Help: Circular Motion

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    A man sitting on the edge of his seat on a Ferris Wheel has a mass of 50.0 kg. The Ferris Wheel has a radius r=30m and the ferris wheel completes a single revolution every 20 seconds. Find the force between the man and the chair.


    2. Relevant equations

    FNET=ma
    Fg=mg

    [tex]\tau[/tex]=2[tex]\pi[/tex]r/v
    a=v2/r


    3. The attempt at a solution

    I found the force of gravity on the man to be 490N. I think the period is 0.05rev/sec so I set up the [tex]\tau[/tex] equation to be 0.05=2[tex]\pi[/tex](30)/v. I solved for V however and get 3769.91 m/s. It seems like too big of a number. This then gives me a very large acceleration of 473740.71m/s2.
    I need to solve for a to plug into my FNET equation in order to find the force on the chair on man.
    What am I doing wrong to get these large numbers?
     
  2. jcsd
  3. Mar 22, 2010 #2

    rock.freak667

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    The period is the time taken for one revolution, so T = ?

    then you can use ω=2π/T

    Your main error lies in the periodic time.
     
  4. Mar 23, 2010 #3
    But I don't have a velocity so how can I solve for T? I thought I had to solve for velocity first?
     
  5. Mar 23, 2010 #4
    Force and Circular Motion??

    1. The problem statement, all variables and given/known data

    A man sitting on the edge of his seat on a Ferris Wheel has a mass of 50.0 kg. The Ferris Wheel has a radius r=30m and the ferris wheel completes a single revolution every 20 seconds. Find the force between the man and the chair.


    2. Relevant equations

    FNET=ma
    Fg=mg

    [tex]\tau[/tex]=2[tex]\pi[/tex]r/v
    a=v2/r


    3. The attempt at a solution

    I set up FNET to be FGP-FCP=ma. I found the force of gravity on the man to be 490N and we know the person's mass so the equation is now:
    490-FCP=(50)a
    So in order to solve for the force between the chair and the person (FCP) you have to solve for the acceleration (a).
    I think the period is 0.05rev/sec so I set up the [tex]\tau[/tex] equation to be 0.05=2[tex]\pi[/tex](30)/v. I solved for V however and get 3769.91 m/s. It seems like too big of a number. This then gives me a very large acceleration of 473740.71m/s2.
    I need to solve for a to plug into my FNET equation in order to find the force on the chair on man.
    What am I doing wrong to get these large numbers?
     

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    Last edited: Mar 23, 2010
  6. Mar 23, 2010 #5

    kuruman

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    Gold Member

    Re: Force and Circular Motion??

    Try finding the speed directly. How much distance does he cover in 20 seconds?

    As for the force between the chair and the man, it looks like not enough information is given. That force depends on where the man is. For example at the "12 o' clock" position the force is different from the "6 o' clock" position.
     
  7. Mar 23, 2010 #6
    Re: Force and Circular Motion??

    Would it me 0.05rev/sec? And that is velocity?
    (I also just attached the diagram)
     
  8. Mar 23, 2010 #7

    kuruman

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    Re: Force and Circular Motion??

    You have posted duplicate threads. I will refrain from posting here until the admins merge the threads.
     
  9. Mar 23, 2010 #8

    Filip Larsen

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    Gold Member

    Re: Force and Circular Motion??

    It may help to think (or read in your textbook) about if there is a simple relationship between the centripetal acceleration, angular speed and radius for a steady circular motion. This should give you the [STRIKE]horizontal [/STRIKE]centripetal force acting on the man which, when vectorially added to the vertical gravity force acting on the man should give you the combined force acting on the man.

    Edit: I didn't know a Ferris wheel is a "vertical" wheel until a diagram was added (english being my 2nd language and all) and for that wheel the centripetal acceleration is of course not horizontal in general.
     
    Last edited: Mar 24, 2010
  10. Mar 23, 2010 #9

    rock.freak667

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    You can use T to get v.

    One revolution takes 20 seconds, so the period is what ?
     
  11. Mar 23, 2010 #10
    0.05rev/sec??
     
  12. Mar 23, 2010 #11

    rock.freak667

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    Periodic time is the time taken for one revolution. So T is ? (they gave it to you)

    Then you can find ω.
     
  13. Mar 23, 2010 #12

    berkeman

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    Staff: Mentor

    (two threads merged. please do not multiple post.)
     
  14. Mar 23, 2010 #13
    Oh that's what I forgot and what had me confused. Well that solves everything. Thank you!
     
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