1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rotational motion question
Loading...