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The Roundup, Circular Motion

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    In an amusement park ride called The Roundup, passengers standinside a 16 m diameter rotating ring.After the ring has acquired sufficient speed, it tilts into avertical plane, as shown in Figure

    knight_Figure_07_47.jpg

    Part A

    Suppose the ring rotates once every 4.70 s. If a rider's mass is 53.0 kg,with how much force does the ring push on her at the top of the ride?

    There are several parts, but I want to do one piece at a time please


    2. Relevant equations
    I think V(top)=r w = 2 pie r/ T

    T(top)=(m/r)(v^2 top)-mg

    3. The attempt at a solution
    I tried V(top)=r w = 2 pie r/ T
    =(2 pie (16)/(2(4.70s)=10.69 m/s
    then
    T(top)=(m/r)(v^2 top)-mg
    =(53/16)(10.69 m/s)^2-53(9.8 m/s^2)=
    -140.86 N ?
     
  2. jcsd
  3. Oct 19, 2011 #2
    Also to get the radius aka r, is it diameter/ 2 ? For my problem it would be 16/2=8=r?
     
  4. Oct 19, 2011 #3
    Also for the mass it would be .53m instead of 53 kg?
     
  5. Oct 19, 2011 #4
    With those two changes made my final answer is -3.2978N, is this correct?
     
  6. Oct 19, 2011 #5

    PeterO

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    Homework Helper

    Have you heard of the other formula for centripetal acceleration?

    a = 4∏2.R / T2

    this enables you to calculate the acceleration from the period and radius [quantities you were given, without having to calculate the speed.
     
  7. Oct 20, 2011 #6
    Does that give you the T(top)? I have 14.30 for that equation
    then do I plug it into the second equation i listed?
     
  8. Oct 20, 2011 #7

    PeterO

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    This device travels at constant speed - with a period of 4.70 seconds.
    The centripetal acceleration is the same at all times/positions.

    The only thing that changes is that at the top, the ring supplies force to supplement gravity in producing that acceleration, whereas at the bottom, the ring has to apply a force to overcome gravity as well as providing the acceleration.

    At all other places, friction comes in to play and it is just too hard to work out - so we are never asked.
     
  9. Oct 20, 2011 #8
    The answer at the top and bottom are 14.30N?
     
  10. Oct 20, 2011 #9

    PeterO

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    No,

    That formula calculates acceleration - after substituting in Radius (R) and Period (T)
     
  11. Oct 20, 2011 #10
    That's what I thought but I don't know what to do afterward.
     
  12. Oct 20, 2011 #11
    Well i'll post B if anyone wants to help, for B Suppose the ring rotates once every 4.70s . If a rider's mass is 53.0kg , with how much force does the ring push on her at the bottom of the ride?
     
  13. Oct 20, 2011 #12
    I got this now....i used 2 pie r / T to get the speed
    then to get the force i used F=mv^2/(r)-wg and got 237.68 N which is correct
    Thank you!
     
  14. Oct 20, 2011 #13

    PeterO

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    having found the acceleration is 14.3, the next step is to realise that acceleration is down. Gravity contributes enough force to create an acceleration of 9.8, but we need an extra 4.5 if we are to get the 14.3.

    F = ma

    F = 53 x 4.5 = 238.5 N

    Centripetal acceleration can be given by a = v2/R , meaning the Centripetal force needed is F = mv2/R

    The alternative formula is

    a = 4∏2R / T2 meaning F = 4∏2Rm / T2

    subtract wg from that force - as you did and the answer is there.
     
  15. Oct 22, 2011 #14
    Thank you PeterO for all your help and hard work!
     
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