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Rock in a tire

  1. Feb 14, 2006 #1
    Just curious if this was the right way to do this problem.

    A tire is going at 180 rev/min. There is a rock in the tire (on the outside) and the tire has a radius of .5 meters. What is the speed and acceleration of the rock?

    180 rev/min = 3 rev / s
    3 rev/s means its going 3 circumferences every second so thats 2(3.14)(.5) which is 3.14*3. So the rock is going 9.42 m/s for the speed. Then the acceleration is centripetal which is 9.42^2/.5?
     
  2. jcsd
  3. Feb 14, 2006 #2

    Astronuc

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    Staff: Mentor

    You are correct.

    http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq

    At a constant angular velocity, the tangential angular acceleration is zero, but there is still centripetal acceleration.

    The rotational frequency, f, in rpm or rps is related to the angular frequency, [itex]\omega[/itex] = 2 [itex]\pi[/itex] f, and the period of rotation, T = 1/f.
     
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