A 10 g dart traveling at 400 m/s sticks into a 10 kg, 1 m wide door at the edge opposite the hinge causing the door to swing open. What is the angular velocity of the door immediately after impact?

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  1. iniitial angular momentum about hinge = m v r = .010 *400 * 1 = 4 kg m^2/s
    omega = v/r = 400/1
    I = m r^2 = .010 (1)^2)
    so I omega = .01 * 400 = 4 agains

    now final angular momentum is the same
    I = .010 *1^2 + (1/3)(10)1^2
    = .01 + 10/3
    = 3.34 kg m^2
    I omega = 4 = 3.34 omega
    omage = 1.2 radians/sec

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