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Rotating Ride

  1. Sep 17, 2010 #1
    Calculating an amusement park ride - the ride part that rotates weighs 10,000kg. It is approx 9,300mm diameter. The ride needs to turn at 26rpm. The acceleration time to 26rpm is 20 seconds and deceleration time is 20 seconds. The ride goes from a horizontal position at start to a vertical position when at full speed. What power is required to turn the wheel? Or what formula is required to calculate the torque required?
     
  2. jcsd
  3. Sep 17, 2010 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF, Jack.

    What is the context of your question? If it is for schoolwork, I can move it to the Homework Help forums.

    In any case, you need to list the relevant equations for your question, and show us the work you have done on it so far.
     
  4. Sep 17, 2010 #3
    When you say "turn the wheel" do you mean from a horizontal to a vertical position? Or turn it as in rotate it?
     
  5. Sep 17, 2010 #4
    You also need to know the moment of inertia "I"

    is the ride a solid disk? a hoop? all of this goes into account when solving this problem.

    If there's a picture, that would help too
     
  6. Sep 19, 2010 #5
    It is for Schoolwork - project.
    10,000kg x 9.81 x 4.650m x 0.05=22808kgm x9.81 =223,749Nm However I am not sure how to input the inertia or acceleration into the equation?
     
  7. Sep 19, 2010 #6
    Rotate the ride not lift from horizontal to vertical.
     
  8. Sep 19, 2010 #7
    [PLAIN]http://img685.imageshack.us/img685/3397/photoeu.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Sep 19, 2010 #8
    A simple rotating arm balanced centrally with a weight of 200kg on each end.
     
  10. Sep 19, 2010 #9
    So calculate the "I" and plug it into the equations I set up for you right there... If you still have trouble, I'll help but you have to try it on your own first.

    && moderators, please move the thread to the appropriate section
     
  11. Sep 20, 2010 #10
    Inertia :-
    0.5 x 10,400 x 4.650 x 4.650 =11243.7kgM^2 ?? does this look right?
     
    Last edited: Sep 20, 2010
  12. Sep 20, 2010 #11
    I =0.5 x 10400 x 4.650 x 4.650 =11,243.7kgM^2 -Does this look right?
     
    Last edited: Sep 20, 2010
  13. Sep 20, 2010 #12
    If the entire platform is just one big solid disk, then yes, you are correct. Plug it into the equation I set up now and what do you get as a final answer?
     
  14. Sep 20, 2010 #13
    T = 11243.7KGm^2 x 0.136 = 1,529kgm or 15,000Nm @26rpm = 40.8kw output power.
    How would this value change -if instead of a solid disc it was a tube?
     
  15. Sep 20, 2010 #14
    look up how the moment inertia of a hoop relates to that of a disk.

    It should be twice as great. What does that do to your final answer?
     
  16. Sep 21, 2010 #15
    22487.4 KGM^2 X0.136 =3,058.3kgm x9.81 =30,000Nm x26 /9550 =81.6kw
     
  17. Sep 21, 2010 #16
    This should read 0.5 x 10,400 x 4.650 x 4.650 =112,437kgm^2 - is the decimal point in the wrong place on previous answer?
     
  18. Sep 21, 2010 #17
    So this should read 224874KGM^2 x 0.136 =30,583kgm x 9.81 =300,000Nm x 26 /9550 =816kw - this is not correct there must be a mistake somewhere??
     
  19. Sep 21, 2010 #18
    Yeah you put the decimal place in the wrong spot. I didn't bother seeing if you did the math right because that can always be fixed up in a later check - the concept is the most important part.

    your final answer should be 816kW if it's a hollow cylinder.

    Anyone else wanna double check our work?
     
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