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Pivot Putty Question

  1. Nov 20, 2011 #1
    Can someone please help me solve this: http://wug.physics.uiuc.edu/cgi/courses/shell/per/phys111/ie.pl?11/IE_putty_plus_rod [Broken]
    pivot putty question? Any help would be greatly appreciated!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 20, 2011 #2
    Try to use conservation of angular momentum of the whole system about the pivot of the rod.
     
  4. Nov 20, 2011 #3
    I tried that. I compared the energy lost to the initial kinetic energy with the equation: KEi-KEf/KEi and then substituted my equations and got:
    (1/2mv^2-1/2*1/3ML^2+mL^2*mvL)/1/2mv^2
    but that didn't work...
     
  5. Nov 20, 2011 #4
    Your formula is unclear and seems to have wrong dimensions in some terms.
    What is the final kinetic energy?
     
  6. Nov 20, 2011 #5
    I had the final kinetic energy as:
    (1/2)(1/3)ML^2+mL^2(MvL)
     
  7. Nov 20, 2011 #6
    And my initial kinetic energy was:
    (1/2)mv^2

    Sorry about the clarity issues!
     
  8. Nov 20, 2011 #7
    This is not a kinetic energy.
    Must be something wrong in your calculations.
    If you mean
    [tex] \frac{1}{2}\frac{1}{3}ML^2+mL^2 MvL [/tex]
    then the two terms have diferent dimensions.
    If you mean
    [tex] (\frac{1}{2}\frac{1}{3}ML^2+mL^2 ) MvL [/tex]
    then it is a valid formula but it's not kinetic energy.
    Try to use enough brackets to make the formulas unambiguous.
     
  9. Nov 20, 2011 #8
    The beginning part was supposed to be the moment of inertia and the second part is the omega.
     
  10. Nov 20, 2011 #9
    You mean MvL is the angular speed?
    You realize also that the final velocity is not the same as initial velocity, right?
     
  11. Nov 20, 2011 #10
    Yes and yes.
     
  12. Nov 20, 2011 #11
    Then you have some conceptual confusions.
    The angular speed is omega=v/R or in this case, v/L.
    MvL is not angular speed.
     
  13. Nov 20, 2011 #12
    How did you write conservation of angular momentum? You need this to find final velocity.
     
  14. Nov 20, 2011 #13
    Oh, I see what you mean. I tried it that way, with no luck. I think my formula is just wrong.
     
  15. Nov 20, 2011 #14
    I'm not quite sure. I said the moment of inertia for the rod is (1/12)ML^2
    And the moment of inertia for the putty is mL^2
     
  16. Nov 20, 2011 #15
    Try to find the velocity after collision.
    The equality of angular momentum before (just the putty ball) and after collision (putty ball + rod) will provide an equation to calculate this velocity. Let's call it vf. Do you know how to do this?
    Then the KE after collision will be [tex]\frac{1}{2} I \omega_{f}^2= \frac{1}{2} I (\frac{v_f}{L})^2[/tex] where I is the total moment of inertia. I see that you got the total moment of inertia right, by the way.
     
  17. Nov 20, 2011 #16
    No, I did not know how to do that... so then the final velocity would be equal to the final angular velocity times L?
     
  18. Nov 20, 2011 #17
    Yes, this is the relationship between angular and linear velocities.
    Angular momentum relative to the pivot is
    initial:Li= mvL
    final: [tex]L_f=I \omega_f = I \frac{v_f}{L}[/tex]
    where I is the total angular momentum.
    I hope now you can solve it.
     
  19. Nov 20, 2011 #18
    Okay, I think I got it. The only thing is,how do I find the final angular velocity?
     
  20. Nov 21, 2011 #19
    By equating the angular monetum before and after collision and solving the equation for final velocity (angular or linear).
     
  21. Nov 21, 2011 #20
    Okay, I got it! Thank you so much for all of your help!
     
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