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Circular motion- speed of rider, centripetal acceleration and coefficient of friction

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    One of the most popular rides at carnivals is the one that resembles a large circular drum that rotates on its axis. Once the ride is spinning fast enough, the floor drops down and the riders are left pinned to the inside wall of the drum (I believe they are describing a ride called the gravitron but im not sure).

    If the ride has a diameter of 5.2m and makes four complete rotations in 9.2s, determine:

    a) the speed of the rider.

    b) the centripetal acceleration of the rider.

    c) the coefficient of friction required to keep the rider from slipping downwards along the drum wall when the floor drops down.

    2. Relevant equations
    For part a) I used the following equations:

    Δd=2∏R
    Δt=T time interval for one cycle is equal to a period.
    v=Δd/Δt

    For part b) I used the following equations:
    ac=v2/r

    For part c) I used the following equation:
    v2=μgR

    3. The attempt at a solution

    a) From the diameter given d=5.2m, so from this we can divide by 2 to give R=2.6m
    With this value we can plug it into the Δd=2∏R equation to give 16.34m.

    We go on to find the time interval for one cycle by doing Δt=9.2s/4 rotations= 2.3s

    Then to find the speed of the rider we use v=Δd/Δt=16.34m/2.3s=7.10m/s and this is our final answer.

    b)For this part of the question I used v=7.10m/s which was found in the previous part a) and r=2.6m, and plugged these values into ac=v2/r= 19.4 m/s2 as our final answer.

    c) For the last part I simply plugged more values into v2=μgR and solved for μ. Which turned out to be μ=1.98

    I would like to just ask if anyone would be able to verify that I did this problem correctly and that all the significant figures are being respected! Thank you so much for your time :)
     
  2. jcsd
  3. May 5, 2012 #2

    ehild

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    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    Why did you use that equation? Force of friction equals the normal force times μ.
    From what does the normal force act on the man? Remember, there is no floor!

    ehild
     
  4. May 5, 2012 #3
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    because that was the only equation that I could think of that had μ... so if theres no floor, that means theres no coefficient of friction?
     
  5. May 5, 2012 #4
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    There is a coefficient of friction, but the frictional force is not due to the non-existent floor, instead because of the wall. Since gravity doesn't act towards the wall, it cannot be responsible for the normal force. So the force of friction, as ehild said, would be normal force(on the wall) times μ.
     
  6. May 5, 2012 #5
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    ohhh ok! thank you, ill try that!
     
  7. May 5, 2012 #6
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    in the meantime are my answers for part a) and b) correct?
     
  8. May 5, 2012 #7
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    i dont have the weight of the rider to calculate the normal force... Im having a mind block at the moment
     
  9. May 5, 2012 #8
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    Why would you need the weight? . What causes the normal force that 'pushes' him against the wall?


    Edit : The concepts for the other two parts seem right, though I haven't checked the calculations.
     
  10. May 5, 2012 #9

    tms

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    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    Don't forget Newton's third law.
     
  11. May 9, 2012 #10
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    i still don't understand what I can do without having the value of m... help please!
     
  12. May 10, 2012 #11

    ehild

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    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    You do not need the value of m. If it is easier to you, assume
    it 100 kg.


    ehild
     
  13. May 12, 2012 #12
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    how can you make an assumption like that though? :S
     
  14. May 12, 2012 #13

    tms

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    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    Because it ultimately does not matter. Write down the full equation for the normal force, and the full equation for the gravitational force, and you should see why.
     
  15. May 28, 2012 #14
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    I don't understand still :( part c) is giving me a headache
     
  16. May 28, 2012 #15

    ehild

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    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    The wall exerts a normal force N on the man, and that normal force keeps him in circular motion. The normal force is equal to the centripetal force.
    The force of friction is normal force times μ, and it acts along the wall, and upward, in the opposite direction as gravity. The man will not drop if the force of friction is equal to the force of gravity.

    ehild
     
  17. May 28, 2012 #16
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    and the normal force is just mv2/R... right? after I get the normal force and times that by the μ that I got from doing v2/gR; do then go on to do μ=Ff/Fn? Cause wouldnt that just end up giving me the same value for μ that I got in the first place? Thank you very much for your time by the way!
     
  18. May 28, 2012 #17

    ehild

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    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    The force of friction is Nμ, and N=mv2/r. So μmv2/r=mg. The friction acts against gravity. It is not the same equation you used.

    ehild
     
    Last edited: May 28, 2012
  19. May 28, 2012 #18
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric


    What you are doing μ=Ff/Fn=(mv2/R)/mg=v2/Rg
    Which is wrong.

    Ff is the value of frictional force which is perpendicular to the Fn which is equal to mg.
    Fn is the pressing force = centripetal force.

    The orientation is turn 90° from normal horizontal plane. Now its a vertical plane.
     
    Last edited: May 28, 2012
  20. Jun 5, 2012 #19
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    I think I got it finally.. so I did:

    μ=(9.8m/s2)*(2.6m)/ (7.10m/s)2
    μ=0.505

    does this seem like a reasonable answer? If someone could check for me, it would be greatly appreciated! Thank you!
     
  21. Jun 5, 2012 #20
    Re: circular motion- speed of rider, centripetal acceleration and coefficient of fric

    Yep, that is correct!! :approve:
     
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