Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vertical velocity and integration

  1. Mar 21, 2007 #1


    User Avatar

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

    a list of data is given for a seat on a ferris wheel that is traveling in a circle. the data is time vs its vertical velocity. the question is to estimate the diameter of the ferris wheel.

    2. Relevant equations


    3. The attempt at a solution

    what i did was integrate the function from 0 to 30 using a reimann sum (which gave me just an approximation because there was no function given...just the vertical velocities every 5 seconds).

    i chose o to 30 because at 30 seconds, the vertical velocity became 0 again. this implied to me that the ferris wheel would be at the top because the vertical velocity is 0 at the top.

    so i integrated by adding up all the rectangles from 0 to 30 (each with equal base length) and got some value. and since im integrating a velocity function, i should come out with a displacement function for the seat on the ferris wheel.

    now my question is, does this value represent the actual distance traveled of the ferris wheel (ie, how far the seat traveled along the circumference) OR does it represent the vertical displacement of the seat (ie, how far it traveled upwards)?

    if its the first, then i can find the diameter by doubling that value to get the ful circumference, then setting that equal to (pi)d and getting my diameter.

    if its the second, then my answer should automatically be the approximation of the diameter since the vertical displacement from the bottom to the top is the diameter of the ferris wheel.

    which one is correct? am i logically justifying it correctly?

  2. jcsd
  3. Mar 21, 2007 #2


    User Avatar

    You say that at 30 seconds the velocity is zero again. Does it mean that it started at zero?
    If so, in 30 seconds the seat travelled from up to down to up again, so the vertical displacement is twice the diameter.
  4. Mar 21, 2007 #3
    Well... I'm not sure I understand fully but I don't think this requires to jump to integrals... If the vertical velocity goes to 0 at 30 seconds interval, and the actual velocity around the circle is constant, what happens mid way through these 30 seconds? That's all you would need to think about in order to solve this...
    Last edited: Mar 21, 2007
  5. Mar 22, 2007 #4


    User Avatar

    it started at zero, then hit 0 again at 30 seconds and 60 seconds. which i believe means it was zero to begin with at the bottom, then 30 seconds later it hit the top (0 vertical velocity), then 60 seconds again at the bottom (where it started).

    so, if my assumption is correct, and the integral of this velocity function is simply the vertical displacement only, then integrating from 0 to 30 is the diameter itself.
  6. Mar 22, 2007 #5


    User Avatar

    yeah but how could i possibly get the diameter of the ferris wheel from that? many different wheels of all different diameters could all have the same period. the question asked for the diameter of the wheel (hence i tried using integrals).
  7. Mar 22, 2007 #6


    User Avatar

    Integration from 0 to 30 is twice the diameter. It is the time for going from up to down to up again.
  8. Mar 23, 2007 #7


    User Avatar

    why is that? isnt the vertical velocity at the top also 0? therefore from 0 to 30 should be from bottom to top once - hence its the diameter.

    but again thats using the assumption that integrating the vertical velocity gives you only the vertical displacement and not the actual distance traveled by the ferris wheel (which would correlate more with the circumference) - thats my main question.
  9. Mar 23, 2007 #8


    User Avatar
    Science Advisor

    You've said that twice now. Can you explain why? Presumably the vertical velocity coming down would be negative and the OP has said nothing about that: he has a list of vertical velocities that starts at 0, becomes positive and then, 30 seconds later, is 0 again. Sounds to me like the second "0" is at the top.

    dnt, yes, what you are saying will work. I wouldn't really consider it "integrating". What you are doing is treating each velocity as an "average velocity". Multiply that average velocity by the time interval for which it applies (from one time to the next) and you get the (vertical) distance traveled in that time interval. Add them all together to get the total distance traveled.

    You might get better results by averaging the velocities at the beginning and end of each time interval and using that as the average velocity during that time interval. That would be the same as using the trapezoid approximation to the integral.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook