1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pendulum collision problem

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A stick of mass m and length l is pivoted at one end. It is held horizontal
    and then released. It swings down, and when it is vertical the free end elastically collides.
    Assume that the ball is initially held at rest and then released a split second before the
    stick strikes it

    a. Which conservation laws apply here?
    b. If the stick loses half of its angular velocity during the collision, what is the mass of
    the ball, M?
    c. Determine the speed V of the ball right after the collision

    2. Relevant equations

    Moment of inertia of the stick:

    [tex]\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

    For (a), I said that conservation of energy, and conservation of linear/angular momentum also apply

    For (b) I used conservation of energy just before the collision:

    [tex]mg\frac{L}{2}=\frac{1}{2}I\omega^2[/tex]

    [tex]mg\frac{L}{2}=\frac{1}{2}(\frac{1}{3}mL^2)\omega^2[/tex]

    This gives me an omega of [tex] \omega=\sqrt{\frac{3g}{L}}[/tex]

    I then used conservation of energy before and after collision:

    [tex]\frac{1}{2}I{\omega^2}=\frac{1}{2}I\frac{\omega^2}{4}+\frac{1}{2}M{V_{b}}^2[/tex]

    I reduce this down to: [tex] M{V_{b}}^2=\frac{3}{4}mLg[/tex]

    Then I used conservation of linear momentum (which is the part I'm not entirely sure about). I assumed the momentum of the mass of the stick before the collision was equal to the sum of the momentum of the CM and ball after the collision:

    [tex]mv_0=mv_1 + MV_b[/tex]
    [tex]V_{cm}=\frac{L}{2}\omega=\frac{L}{2}\sqrt{\frac{3g}{L}}=\sqrt{\frac{3Lg}{4}}[/tex]

    ==> [tex]\frac{mL\omega}{4}=MV_b[/tex]

    Isolating v and putting it into the [tex]M{V_{b}}^2[/tex] expression I get [tex]M=\frac{1}{4}m[/tex]

    Then the velocity of the ball comes out to be [tex]\sqrt{3gL}[/tex]

    Is this correct? For example, I did not use angular momentum, when my gut is telling me that I should have. But assuming that conservation of linear momentum holds, I don't need it right?

    Many thanks for your help guys.
     
  2. jcsd
  3. Feb 16, 2014 #2
    This is not conservation of energy. I recognize the right hand side as kinetic energy of the rod. The left hand side is probably the potential energy of the rod. Their sum would be the rod's total energy. But the equality makes no sense at all.
     
  4. Feb 16, 2014 #3
    Hi Voko, why is that? I figured the potential energy of the rod when horizontal, is equal to the rotational kinetic energy of the rod when vertical, just before it strikes the ball. How is this not conservation of energy?
     
  5. Feb 16, 2014 #4
    I was misled by the "just before the collision" bit. Is the left hand side the initial potential energy?
     
  6. Feb 16, 2014 #5
    The LHS is the initial potential energy, yes. The RHS is the rotational kinetic energy just before the stick collides with the ball.
     
  7. Feb 16, 2014 #6
    Then that part is correct.
     
  8. Feb 16, 2014 #7
    But do I have the correct solutions for the velocity of the ball, and the mass M?
     
  9. Feb 16, 2014 #8
    I have not checked your algebra, but the general method appears correct.
     
  10. Feb 16, 2014 #9
    As long as its correct conceptually, I'm on the right track. Thanks for your help!
     
  11. Feb 16, 2014 #10
    Regarding your doubt about linear vs angular momentum, you could redo the problem using the latter. You should get the same result.
     
  12. Feb 16, 2014 #11
    Just for the sake of argument, how would you go about doing it for angular momentum? For example, I get the following expression for angular momentum:

    [tex] I\omega=I\frac{\omega}{2} + MV_bL[/tex]

    But when I work this out, I get:

    [tex]\frac{1}{6}mL\omega=MV_b[/tex]

    But when I used linear momentum, I got:

    [tex]\frac{1}{4}mL\omega=MV_b[/tex]

    What went wrong?
     
  13. Feb 16, 2014 #12
    I have just realized that conservation of linear momentum does not hold here. This is because of the pivot, which prevents any linear motion of the rod.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pendulum collision problem
  1. Pendulum collision (Replies: 8)

Loading...