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Pendulum: Generic Solution for Angle

  1. May 4, 2008 #1
    [SOLVED] Pendulum: Generic Solution for Angle

    Hey guys, I'm a little lost on this one, not sure where to start.

    I'm doing a lab involving a pendulum bob being hit by a ball and them sticking together (inelastic collision). Almost like a ballistic pendulum right? Anyway, I need to find the generic solution for the angle (theta) that the pendulum swings.

    From what I understand this is not simple harmonic motion in general, and can only use that as a model for very small angular displacements. Also momentum is only conserved right before the collision and right after the collision, right? Other than that I really don't know where to start. I figure that the angle has to relate to the velocity of the bullet, the force exerted on the bob by the bullet, and angular velocity of the two together.

    Any help to set me on track would be greatly appreciated!
     
  2. jcsd
  3. May 4, 2008 #2

    Nabeshin

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    Science Advisor

    Momentum conservation is useful before to immediately after the collision, after which energy conservation takes you the rest of the way home.
     
  4. May 4, 2008 #3
    But isn't energy not conserved because the collision is inelastic?
     
  5. May 4, 2008 #4

    Doc Al

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    Staff: Mentor

    Attack this problem in two steps:
    (1) The collision. What's conserved?
    (2) The rising of the pendulum after the collision. What's conserved?
     
  6. May 4, 2008 #5
    1. Ok at the collision, momentum is conserved p(i) = p(f)
    mv(i)(bullet) = [m(bullet) + m(bob)]v(f)

    2. As the pendulum is rising, energy is conserved? E(i) = E(f)
    so.....PE(i) + KE(i) = PE(f) + KE(f) --> ok here's where im a little confused, the initial conditions for PE and KE, they are for the instant after the collision has taken place? and the final KE and PE are for when the pendulum reaches its maximum height?

    the arc the pendulum makes equals (r)(θ) right? So angular velocity would be (r)(θ)/t? This all makes sense to me for the most part, but I don't understand how to put it all together.


    lol thank you though for making me think of it in two parts, it makes way more sense this way.
     
  7. May 5, 2008 #6
    what factors do I include in the equation of theta? can I use kinematics equations?
     
  8. May 5, 2008 #7

    alphysicist

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    Homework Helper

    Hi Return,

    That's right; and you have formula for PE and KE. Use them in the energy equation; once you see what you have you'll know what you still have to find. What do you get for the energy equation?
     
  9. May 5, 2008 #8
    ok so i end up getting

    V^2/2 = gh

    so I need to express h in terms of theta right? Lsin(theta)?
     
  10. May 5, 2008 #9

    Doc Al

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    Staff: Mentor

    Right, where V is your v(f) from the momentum equation.

    Yes.
    No, that gives you the horizontal distance; you want the vertical. (Draw yourself a diagram.)
     
  11. May 5, 2008 #10
    phew, ok thanks for the help, i got the generic solution earlier today!
     
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