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Angular Momentum - Find angular velocity of a hanging bar after a collision

  1. Mar 2, 2010 #1
    I think I've gotten it but figured it would be best to get a second opinion because I feel like I made a few leaps of faith. I sincerely appreciate your time. <3

    1. The problem statement, all variables and given/known data
    A 1.0 kilogram object is moving horizontally with a velocity of 10m/s, when it makes a collision with the lower end of a bar that is hanging vertically at rest. For the system of the bar and object, linear momentum is not conserved but kinetic energy is. The bar, of length l = 1.2m and mass m = 3kg, is pivoted about its upper end. Immediately after the collision, the object moves with speed v at an angle (theta) relative to its original direction. The bar swings freely, reaching a maximum angle of 90 degrees with respect to the vertical. The moment of inertia for the bar about the pivot is [itex] I = (ml^2) / 3 [/itex]. Ignore all friction.

    Synopsis:
    1.0kg object moving horizontally at 10m/s.
    1.2m, 3kg bar [itex] I = (ml^2) / 3 [/itex] suspended about its upper end.
    Object hits the bottom of the bar in a glancing collision.
    Bar then pivots up to 90 degrees with respect to the vertical.
    Object then deflects to an angle [itex]\theta[/itex] below the horizontal at a velocity v.

    Diagram:
    29QQA.jpg

    Questions:
    2. Relevant equations
    The all-important: [itex]F = ma [/itex]
    Conservation of angular momentum: [itex]L_i = L_f[/itex]
    Conservation of kinetic energy: [itex]K_o = K_ol + K_b[/itex]
    Angular momentum around a point: [itex]L = mvr sin(\theta)[/itex]
    [itex]\tau = F(lever arm) = I\alpha [/itex]
    [itex]L = I\omega[/itex]


    3. The attempt at a solution
    Enclosed in quotes to make it easier to see.
    Thanks so much!
     
    Last edited: Mar 2, 2010
  2. jcsd
  3. Mar 2, 2010 #2

    ideasrule

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    Homework Helper

    I haven't checked all your numbers, but your method seems good to me.
     
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