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  • Nov 25, 2007, 11:14 PM
    whitney42
    Angular acceleration
    Find the angular accleration of a thin ring of radius 0.300 m and 1.50 kg mass if a constant 20.0 N force acts on a string wrapped around the ring's circumference. Find the angular velocity of the ring after 5.00 seconds.
  • Nov 27, 2007, 06:29 AM
    terryg752
    Torque = 20 times .3 = 6

    Moment of Inertia of ring = Mr^2 = 1.5 times (.3)^2 = .135

    I = .135

    Torque = I times angular acceleration

    6 = .135 times angular acceleration

    Angular acceleration = 6/.135 = 44.44


    After 5 seconds, angular velocity = 5 times acceleration
    = 222.22

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