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Conservation of angular momentum on a frictionless table

  1. Nov 17, 2005 #1
    On a frictionless table, a glob of clay of mass 0.40 kg strikes a bar of mass 1.74 kg perpendicularly at a point 0.15 m from the center of the bar and sticks to it. If the bar is 1.14 m long and the clay is moving at 6.6 m/s before striking the bar, what is the final speed of the center of mass?
    I got this part by using conservation of momentum. The answer was 1.23m/s .
    At what angular speed does the bar/clay system rotate about its center of mass after the impact?
    I tried using conservation of angular momentum.
    [tex] L_o = L_f [/tex]
    [tex] I \omega= I \omega [/tex]
    [tex] \omega _ initial = v/r= 6.6/.15= 44 rad/s [/tex]
    [tex] I= (1/2)ML^2 = (1/2)(1.74)(1.14)^2 = 1.13 kgm^2 [/tex]
    so [tex] L_o= 1.13* 44 [/tex] = 49.72
    [tex]L_f= (1/2)(1.74+ .40)(1.14^2) * \omega = .452 *\omega [/tex]
    Solving for omega gave me 109.9 rad/s.. which doesn't really make sense. Can someone help me? Thanks in advance.
     
  2. jcsd
  3. Nov 17, 2005 #2

    Tide

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    The contribution of the clay to the moment of inertia is [itex]m r^2[/itex]. You divided it by 2.
     
  4. Nov 19, 2005 #3
    so if it's MR^2, it would be,
    [tex] L_0 =49.72 [/tex]
    [tex] L_f= MR^2 * \omega [/tex]
    [tex] MR^2= (1.74 +.40)(1.14^2)= 2.78 [/tex]
    Solving for omega gives me 17.8 rad/s.
    Is this the right way to do it?
     
  5. Nov 20, 2005 #4

    Tide

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    The 1.74 is still divided by 2!

    However, what you are doing is a bit of simplification. The problem is asking for the rotation rate about "the center of mass after impact" which is slightly different from the center of the bar. You need to take that into account.
     
  6. Nov 20, 2005 #5

    lightgrav

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    PunchLine, back up!

    The only thing moving initially is the clay, so
    L_i = ( I_clay * w_clay ) + ( I_rod * 0 ) .

    The rotational inertia of a uniform rod (around its center) is 1/12 ML^2
    but add the inertia of all (100%) of the clay at .15 m from that center.

    Fix these and you'll be in the ballpark. [do it before reading on]

    Now, the axis of rotation will be around the (new) center-of-mass
    [that's about .03 m toward the clay from the center of rod].
    So you SHOULD add the clay's inertia at .12 m from the axis,
    but add the entire rod mass (its c.o.m) as rotating at .03m from the axis.

    My bet is that these two corrections pretty much cancel, but make sure.
     
  7. Nov 20, 2005 #6
    Ok I think I'm a little confused.
    I understand that for the initial angular momentum, it's just the clay because the rod isn't moving so,
    [tex] MR^2 * \omega [/tex]
    (.40)(.15)^2 * 44 = .396
    Since the angular velocity of the rod is 0, the angular momentum would be 0.
    Then the final angular momentum is
    [tex] ((1/12)ML^2 + MR^2) *\omega [/tex]
    where L= .03 and r= .12 ?
    I don't think my final angular momentum is right.
     
  8. Nov 23, 2005 #7
    can anyone help?
     
  9. Nov 23, 2005 #8

    Tide

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    If you ignore the fact that the center of mass is (very) slightly displaced from the center of the rod then the initial angular momentum of the system is [itex]L_0 = m D v_0[/itex] where D is the distance of the clay from the center of mass (0.15 m).

    The final angular momentum is [itex]\left( \frac {1}{12}ML^2 + m D^2\right) \omega [/itex] where L is the length of the rod. Now apply conservation of anglar momentum to solve for [itex]\omega[/itex].

    Incidentally, in your calculation for the speed of the center of mass, you could have calculated its speed before the collision and note that the velocity of the center of mass doesn't change during the collision.

    EDIT: Corrected a typo
     
    Last edited: Nov 24, 2005
  10. Nov 24, 2005 #9
    Ok I did..
    [tex] mDv_o = ((1/2)ML^2 + MD^2) \omega [/tex]
    [tex] (.40)(.15)(6.6) = ( (1/2)(1.74)(1.14^2) + (.40)(.15^2)) \omega [/tex]
    Solving for omega gave me .348, which isn't right.. is my initial angular momentum right? Am I supposed to convert the linear velocity to angular velocity?
     
  11. Nov 24, 2005 #10

    Tide

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    Sorry about that - I mistyped the moment of inertia for a rotating rod. It should be

    [tex]I = \frac {1}{12} M L^2[/tex]

    so the final angular momentum should be

    [itex]\left( \frac {1}{12}ML^2 + m D^2\right) \omega [/itex]

    Let me know if that works out for you!
     
  12. Nov 24, 2005 #11
    Ok I plugged everything in again and got 2.01, which still isn't right.
    It doesn't make sense on why this is wrong! :devil:
     
  13. Nov 24, 2005 #12

    Tide

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    How do you know it's wrong and what is the "correct" answer?
     
  14. Nov 25, 2005 #13
    My homework is online and it tells me if I enter the correct answer. I have no idea what the right answer is.
    I did realize that I was entering my units as s^-1, which I thought was the same as rad/s, since rad isn't really a unit. Would that have something to do with it?
     
  15. Nov 25, 2005 #14

    lightgrav

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    You should've noticed that [tex]L_i = r x p = m r^2 \omega = .396[/tex] => 2.01 [rad/s]. Okay.

    NOW you should compute the location of the center-of-mass.

    Since the clay path is only .12 m from the NEW center-of-mass,
    for accuracy, you have to use L_i = .317 [kg m^2/s] => 1.61 [rad/s].

    Correctly calculate the accurate (new)
    I_final = ( 1/12 M L^2 + M(.03)^2 ) + (m r^2) , where now r = .12 m .
    Can you tell me the meaning of the first 2 terms in I_final?
     
    Last edited: Nov 25, 2005
  16. Nov 26, 2005 #15
    I know that the first term is the original moment of inertia of the rod. Since the rod has a different center of mass when the clay sticks to it, the second term is the moment of inertia of the difference in the center of mass... i think :uhh:
     
  17. Nov 26, 2005 #16

    siddharth

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    There is something I want to point out here in your calculations.
    Before the collision you calculated the Angular momentum about the center of the bar.
    After collision, you are calculating the Angular momentum of the system about the Center of Mass of the system (which as Tide pointed out, is different from the center of the bar).
    So when you say [tex] L_i = L_f [/tex] you are jumping frames (From the Center of the bar frame to the center of mass of the system frame).
    Is this valid? How can you equate the Angular momentum in two different frames?
    The way I would approach this problem is to calculate the angular momentum about the COM of the system(putty and rod) just before collision (mvr) . Then after collision the putty-rod system can be considerd as having pure rotation about the COM and translation of the COM.
     
  18. Nov 26, 2005 #17

    lightgrav

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    The first term is the Inertia of the rod around its center-of-mass,
    the second term is the Inertia OF the center-of-mass (hence 1 Mr^2) around the actual axis of rotation.

    The axis of rotation is the center-of-mass (with the clay attached!)
    So you'd better find out (for sure) where it is exactly
    ([tex]M_{total} X_{c.o.m} = \Sigma m_i x_i [/tex])

    You've finally nailed this problem. It has most of the possible subtleties
    (doesn't have changing shape, but your shrinking planet did).
    I usually prefer to learn on easier "exercises", then progress
    to challenging problems. If you're using Tipler, he DOES have exercises.
     
    Last edited: Nov 26, 2005
  19. Nov 28, 2005 #18

    siddharth

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    But lightgrav, you are still switching frames when you are equating the angular momentum.
    The initial angular momentum should be found about the Center of Mass of the system (putty and rod) just before collision.
    So the inital angular momentum should be [tex] mvr [/tex] where r should be the distance from the center of mass of the system(putty and bar) to the point where the putty strikes the bar.
     
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