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Torque and Angular Momentum

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    At t = 15 s, a particle has angular momentum <5, 8, -3> kg · m2/s relative to location A. A constant torque <12, -13, 20> N · m relative to location A acts on the particle. At t = 15.2 s, what is the angular momentum of the particle relative to location A?


    3. The attempt at a solution

    i know that torque = R x Fnet , but am not sure how to use it in this problem here. any help please?
     
  2. jcsd
  3. Dec 5, 2008 #2

    LowlyPion

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    Isn't the torque equal to the first derivative of the angular momentum?

    So wouldn't that mean that the rate of change of the angular momentum is expressed by the Torque given relative to point A?
     
  4. Dec 5, 2008 #3

    i still dont get how to solve the problem
     
  5. Dec 5, 2008 #4

    rl.bhat

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    Torque =( L2-L1)/t
    Torque*t = L2 - L1.
    Find L2.
     
  6. Dec 5, 2008 #5

    oh u meant deltat thanks
     
  7. Dec 6, 2008 #6

    LowlyPion

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    Yes. Like that.

    The Δt in this case is .2 s and is a scalar to the Torque Vector which as rl.bhat indicated should yield a difference vector ΔLa that added to the original La vector should yield the New La at T + Δt.
     
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