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Rotational motion question

  1. Nov 11, 2004 #1
    Q: What is the linear acceleration of a point on the rim of a 30-cm-diameter record rotating at a constant angular speed of 33.5 rev/min?

    I seem to have all the variables and equations in hand -
    r = .15m and [itex]\omega[/itex] = 3.49 radians/second;
    [itex]v = \omega r[/itex] ;
    (radial component of linear acceleration) [itex]a_r = \frac {v^2} {r} \omega^2 r[/itex]

    ...but I get 3.29 [itex]\frac {m} {s^2}[/itex] (pointing inward), and the book's answer is 1.8 [itex]\frac {m} {s^2}[/itex]. Am I wrong? Is the book wrong? Did I just convert something wrong somewhere? I'm very confused.
  2. jcsd
  3. Nov 11, 2004 #2
    I got 1.8. You need to use the equation for centripetal acceleration (since the record is spinning) and that is ac= angular velocity^2 times radius. That should give you your answer!
  4. Nov 11, 2004 #3
    *facepalm* Figured it was something simple like that. Thanks!
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