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Conservation of Angular Momentum; angle of rotation

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A block of mass m slides down a frictionless ramp from height h above the floor. At the base of the ramp it collides and sticks to the lower end of a uniform rod, length L, mass 2m, that is suspended about a pivot at point O, about which it is free to rotate. Express answers in terms of m, g, h, and L.
    Find the angle θ through which the system rotates before momentarily coming to rest after the collision.

    2. Relevant equations
    I used conservation of energy to determine angular speed ω immediately after collision: Ei = Ef
    I used v = Rω to find kinetic energy of the system after collision.

    3. The attempt at a solution
    I found ω to be √6gh/5L^2 ; I found kinetic energy after the collision to be (9/5)(mgh). However, I'm at a loss at how to find the angle θ. I tried using conservation of angular momentum but became confused - if the initial point is immediately after collision and the final point is when the system is momentarily at rest, that means Li = 0. There has to be a mistake in my reasoning somewhere; I need help bridging this gap I have in my angular knowledge.
     
  2. jcsd
  3. Feb 2, 2016 #2

    TSny

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    Hello, and welcome to PF! :smile:

    Note:
    (1) "collides and sticks" ⇒ inelastic collision.
    (2) What quantity is actually conserved during the collision?
    (3) Angular momentum is not conserved for the "swinging up" after the collision. What quantity is conserved for the swing up?
     
  4. Feb 2, 2016 #3
    Ah, so

    1) and 2) Since it is an inelastic collision, energy is not conserved but momentum (angular) is during the collision
    3) Would energy be conserved for the "swinging up" after the collision?
     
  5. Feb 2, 2016 #4

    TSny

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    Yes. Good.
     
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