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Homework Help: "webassign" problem, CENTRIPETAL ACCEL

  1. Nov 6, 2007 #1
    [SOLVED] "webassign" problem, CENTRIPETAL ACCEL

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

    The Emerald Suite, a revolving restaurant at the top of the Space Needle in Seattle, Washington, makes a complete turn once every hour. What is the magnitude of the centripetal acceleration of the customer sitting 14.5 m from the restaurant's center?

    2. Relevant equations

    4*pi*R(radius) / T^2

    where T=time

    3. The attempt at a solution

    so then it would be

    4*pi*14.5/3600^2, which gives me: 1.41E-5

    but it's showing me that its incorrect...but i dont see how!

    EDIT: also, is there a difference between the magnitude of centripetal acceleration as opposed to mere centripetal acceleration?
    Last edited: Nov 6, 2007
  2. jcsd
  3. Nov 6, 2007 #2
    The tangential velocity "v" is [tex] v = \frac{2 \pi r}{T} [/tex] if I remember correctly. Can you tell me what equation you used for the centripetal acceleration?
  4. Nov 6, 2007 #3
    i used 4*pi*R/T^2, as my teacher directed in class for this particular problem.

    for T i converted it to seconds
  5. Nov 6, 2007 #4
    The equation you posted will give you the units for acceleration, but I've never seen that equation before. Try using the equation for tangential velocity that I gave you, and plug it into [tex] a_{centrip} = \frac{v^2}{r} [/tex].
  6. Nov 6, 2007 #5
    I tried your equation, and got 4.41E-5, which gave me the correct answer. i don't understand why my teacher gave me that equation though...

    i actually used your equation yesterday, and got the same answer. it must have been a webassign glitch?

    thank you anyways.

    my mistake was that i didnt do the tangential velocity first.
  7. Nov 6, 2007 #6
    You're welcome. It'd be interesting to know where your teacher got that equation, but I'm glad it all worked out :)
  8. Nov 6, 2007 #7


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    Homework Helper

    There is an error here, the pi term should be squared. That's why you didn't get the correct answer with it.

    That equation is just the result of substituting [tex]v = \frac{2 \pi R}{T}[/tex] into [tex]a=\frac{v^2}{r} [/tex]
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