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Rotations, Angular Momentum

  1. Apr 14, 2008 #1
    Question:
    A fighter pilot is being trained in a centrifuge of radius 15m. it rotates according to theta=0.25(t^3) + ln(t+1) befor it stabilises (theta in radians). what are the magnitudes of the pilots:

    a) angular velocity: d(theta)/dt at t = 4 (12.2 rad/sec)

    b) linear velocity: 12.2 * 15

    c radial acceleration: 2nd derivative of theta=0.25(t^3) + ln(t+1) at t = 4 (5.96rad/sec^2)

    d) tangential acceleration: 5.96 * 15

    Are b and d just the angular values times the radius thats if a and c are right in the first place??


    Thanks
     
  2. jcsd
  3. Apr 15, 2008 #2

    tiny-tim

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    Yes! :smile:

    (v = rω and dv/dt = (d/dt)(rω) = r(dω/dt), since r is constant.)

    But why was that worrying you? :confused:
     
  4. Apr 15, 2008 #3
    Because all the angular equivalents scare me!
     
  5. Apr 15, 2008 #4

    tiny-tim

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    … just keep differentiating rθ …

    Hi forty! :smile:

    Just remember the definition of a radian: the angle whose arc-length = r.

    And therefore generally:
    tangential length = rθ.​

    So (if r is constant), differentiate once for:

    tangential speed v = rθ´ = rω

    tangential acceleration a = rθ´´ = rω´. :smile:

    (and of course radial acceleration = -rω² = -vω = -v²/r.)
     
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