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Calculating tangential velocity of an air parcel circulating a tornado

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    If an air parcel 2000m from a tornado center has a tangential velocity of 10 m/s, what is the resultant tangential velocity if the parcel is 100m from the tornado center.

    2. Relevant equations

    v = rω, where r = radius, and ω = angular velocity
    ω = v/r

    3. The attempt at a solution

    I understand that the tangential velocity of the air parcel should be much faster as radius drops. But how am I supposed to calculate tangential velocity of it requires the calculation of angular velocity which requires velocity. I'm really lost.
     
  2. jcsd
  3. Jan 29, 2013 #2
    Angular Velocity is not dependant on radius, therefore, it will be the same for each radius. You can find angular velocity with the equation ω = v/r. Use that value to find the tangential velocity at the new radius.
     
  4. Jan 29, 2013 #3
    Okay, here's my work.

    ω = 10 (m/s)/ 2000m = .005 radians/sec

    Then I put that into the tangential velocity question:

    v = rω = 100m*.005 radians/sec = .5 m/s

    Why is it that the tangential velocity got slower as radius decreased? I thought it was the other way around as I mentioned in my first post.
     
  5. Jan 30, 2013 #4
    The tangential velocity is slower near the center of the disk because it doesn't have to circle around as big of a radius as the outside. Imagine running around a circular track... the people on the outside have to run faster to keep up with the people on the inside lanes. This is not true for planets though, because the inner planets are affected by the sun's gravity more and travel faster in the tangential direction to maintain their orbits.
     
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