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Resultant linear acceleration at the rim of a disk

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform disk with radius 0.390m and mass 27.0kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to θ(t)=( 1.50rad/s)t+( 9.00rad/s2)t2 .

    What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100rev ?
    2. Relevant equations
    tangential acceleration = r * α

    3. The attempt at a solution
    The velocity equation should be the derivative of the position equation, so
    θ(t) = 1.5t + 9t2 → ω(t) = 1.5 + 18t
    Derivative of velocity equation should be acceleration so
    ω(t) = 1.5 + 18t → α(t) = 18 rad/s2

    Now at = rα = (0.39m)(18 rad/s2) = 7.02 m/s2

    Where did I go wrong?
     
    Last edited: Oct 21, 2014
  2. jcsd
  3. Oct 21, 2014 #2

    BvU

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    What is the exercise asking for, precisely ?
     
  4. Oct 21, 2014 #3
    Oh! My fault! It's asking:
    What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100rev ?
     
  5. Oct 22, 2014 #4

    BvU

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    Can't find a flaw in your work. Why do you think it's wrong ?
     
  6. Oct 22, 2014 #5
    Because mastering physics says it is incorrect :/
     
  7. Oct 22, 2014 #6

    BvU

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    Well, then it definitely must be wrong... So either we both make the same calculation error or we both have a wrong perception of what's going on.

    Let's bet on the latter: we both thought the tangential acceleration would be the answer. In other words: we calculated the acceleration of a particle that thus far has been attached to the disk and isn't any more. That can well have been our misconception:
    perhaps the writer of the exercise wants the linear acceleration of a particle that stays in the same place on the rim of the disk ...

    I can't help but feel a litlle bit tricked :(
     
  8. Oct 22, 2014 #7
    What would be the difference? Isn't the angular acceleration constant and the radius constant, thus making the linear acceleration constant for any point on the disk at that radius?
     
  9. Oct 22, 2014 #8

    BvU

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    Tangential acceleration is constant indeed. But apparently, linear acceleration ##\ne## tangential acceleration.

    The particle experiences a force to undergo this angular acceleration and also a force to stay in a circular trajectory on the rim of the disc.

    What I suspect the exercise means is: the linear acceleration squared is radial squared plus tangential squared

    And I must admit that linear acceleration is defined as ##\ddot { \vec x }##
     
  10. Oct 23, 2014 #9
    Ah I see what you mean now. We both misinterpreted the problem, thanks for working through it with me!
     
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