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Homework Help: Pendulum collision

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A stick of mass m and length L is pivoted at an end. It is held horizontal and then released. It swings down, and when it is vertical the free end elastically collides with a ball. (assume that the ball is initially held at rest and then released a split second before the stick strikes it.) If the stick loses half of its angular velocity during the collision, what is the mass of the ball? What is the speed of the ball right after the collision?


    2. Relevant equations
    moment of inertia for the stick: [tex]I=\frac{1}{3}*mL^{2}[/tex].
    [tex]L=I\omega[/tex].

    [tex]\omega =\frac{v}{r}[/tex]
    [tex]K_{rotational}=\frac{1}{2}I\omega^{2}[/tex]


    3. The attempt at a solution
    Alright so first I found the angular velocity the instant before it hit the ball by using conservation of energy:

    [tex]mgL = \frac{1}{2}(\frac{1}{3}mL^{2})\omega^{2}[/tex]
    and found that [tex]\omega=\sqrt{6g/L}[/tex].

    then I used E-conservation for before and after the collision:

    [tex]\frac{1}{2}I\omega^{2}=\frac{1}{2}I(\omega/2)^{2}+\frac{1}{2}m_{ball}v_{ball}^{2}[/tex]

    and substituting I and [tex]\omega[/tex], I reduced this for the mass and velocity of the ball:

    [tex]m_{ball}v_{ball}^{2}=\frac{3}{2}mLg[/tex]

    Then I used linear momentum conservation here. My reasoning here is that the momentum of the centre of mass of the stick before is equal to the momentum of the centre of mass of the stick afterwards plus the momentum of the ball afterwards.
    so before

    [tex]v_{cm}=\frac{L}{2}\omega[/tex]

    and just divide this by 2 for the velocity after the collision. So using

    [tex]mv=mv/2+m_{ball}v_{ball}[/tex]

    I found an expression for mv of the ball, which is

    [tex]\frac{m}{2}\sqrt{3/2gL}=m_{b}v_{b}[/tex]

    isolating v, and substituting it into the [tex]m_{b}v_{b}^{2}[/tex] expression, I got

    [tex]m_{b}=\frac{1}{4}m[/tex]

    and then putting this into the previous v expression, I get

    [tex]v_{b}=2\sqrt{3/2Lg}[/tex]

    So does what I did make sense? The only thing I can think of is that by using my method, the ball would have the same end velocity no matter where it struck the stick, as this does not matter in my p-conservation equations. I can't think of a way to use angular momentum, since the ball doesn't really have angular momentum.
     
  2. jcsd
  3. Apr 6, 2010 #2

    rl.bhat

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    Homework Helper

    When the stick falls, center of mass falls through height L/2.
     
  4. Apr 7, 2010 #3
    ok, so in the first E-conservation equation, I replaced mgL with mg(L/2). So I got

    [tex]\omega = \sqrt{\frac{3g}{L}}[/tex]

    for my new angular velocity. In the end, using the same procedure as before, I got my mass to equal the same, and my ball velocity became:

    [tex]v_{b} = \sqrt{3gL}[/tex]
     
  5. Apr 7, 2010 #4
    so other than that mistake, I did it right?
     
  6. Apr 7, 2010 #5

    rl.bhat

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    Homework Helper

    I think so.
     
  7. Apr 7, 2010 #6
    allright, thank you very much kind sir.
     
  8. Apr 1, 2011 #7
    I apologize for bringing up a dead topic, but It just came up in one of my assignments. I followed the same procedure and it makes sense to me and seems consistent, but I am having a hard time understanding why conservation of angular momentum is not used here (shouldn't we need it?). Id appreciate if anyone could help me understand this? It appears to me that we should need conservation of energy, momentum (linear) and momentum (angular)
     
  9. Apr 1, 2011 #8

    gneill

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    Staff: Mentor

    Note that the ball *does* have angular momentum with respect to the pivot point of the stick. Because the initial angle of its velocity vector is 90° to the radius vector, it's simply its linear momentum multiplied by the radius (L). But you could have gone the completely angular momentum route and arrived at the same result.
     
  10. Apr 1, 2011 #9
    I in fact ended up using angular momentum conservation directly with energy conservation (you dont have to solve for the angular velocity (at least not right away) ). This produced a mass of [tex]\frac{m}{9}[/tex], and a speed of [tex]\frac{3\omega l}{2}[/tex]. Where the angular velocity is that given by the OP (corrected one).
     
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