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Homework Help: Homework Help. Rotating platform. difference of acceleration when close to center

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A person is on a horizontal rotating platform at a distance of 4.3 m from its centre. This preson experiences a centripetal acceleration of 56m/s^2. What is the centripetal acceleration is experienced by another person who is at a distance of 2.5 m from the centre of the platform?

    2. Relevant equations


    3. The attempt at a solution
    centripetal acceleration= [itex]\frac{v^2}{r}[/itex]

    centripetal acceleration= [itex]\frac{v^2}{r}[/itex]
    = [itex]\frac{4.9^2}{2.5}[/itex]
    = 9.6m/s^2
    I know that is not the right answer because it should be lower when closer to the center.
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 11, 2012 #2
    uhh guys i dont know why it looks like that. Only read the first three stuff please.

    Thank you
  4. Apr 11, 2012 #3
    disregard that message please. i fixed the first post
  5. Apr 11, 2012 #4
    Okay guys I figured this out, but don't know why this works.

    4.9m/s*1/4.3m = 1.14s

    1.14*2.5= 2.84m/s

    (2.84m/s)^2/2.5m = 3.246 = centripetal acceleration

    the answer is 3.3m/s

    Btw im not doing this to bump the thread, that is if it's possible.
    Last edited: Apr 11, 2012
  6. Apr 11, 2012 #5
    the RED needs your attention.

    the linear velocity changes based on the radius. ANGULAR (ω) velocity remains constant though.

    the first Centripetal Acc is correct the second one is wrong.

    formula to use v = ω * r

    use the initial v and r to find ω.

    then use ω and new r to find new v.

    then find the correct second Centripetal Acc
    Last edited: Apr 11, 2012
  7. Apr 11, 2012 #6
    its called the EDIT button. please learn to use it.
  8. Apr 11, 2012 #7
    K sorry for being a noob, I thought it would be better if I made another post instead of editing the other one.
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