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Question about conservation of angular momentum around an axis

  1. Jun 11, 2010 #1
    suppose i have a solid rod of length 2L, connecting two identical point masses on either side, and this system is spining around the center of mass at a steady angular speed of [tex]\omega[/tex]. another identical point mass is placed so that it collides with one of the point masses when the rod is horizontal. lets say you want to find [tex]\omega[/tex] at the center of mass after the collision, using conservation of angular momentum.

    this is where i am confused: I know the center of mass changes after the collision, and, assuming the origin is on the initial center of mass and the right and up directions are positive, calculated it to be: L/3.
    Now, the way i understood this in class, To answer this question i must compare the angular momentum right before and after the collision.
    but i recall that you cant directly compare the angular momentum at the axis that goes throught the initial center of mass and the angular momentum at the axis of the new center of mass after the collision, so i figured out the angular momentum at the axis of the new center of mass before the collsion and then using:
    [tex]I_{before}\omega=I_{after}\omega_{needed}[/tex]
    found out [tex]\omega_{needed}[/tex].
    The porblem is that the lecture notes compare the axis of the initial center of mass before collision directly to the axis of the new center of mass after colision. Is this a mistake or am i missing something?
     
  2. jcsd
  3. Jun 11, 2010 #2
    (sorry, I made a complete analysis for you but that's against forum's rules)
    I think there are 2 points to be made clear:
    1- The law is applied to only 1 axis or point. Thus, considering one axis throughout the process as you did is correct. But be careful that the rod rotates around different axes before and after the collision. Hint: Find the speed of each mass and apply the original formula of angular momentum: [tex]L=mvr[/tex] , or its vector form: [tex]\vec{L}=m\vec{r}\times\vec{v}[/tex]
    2- The solution of the lecture note is particularly right in this problem. The angular momentum of the system at a point in time doesn't depend on the axis, which results from that the total linear momentum of the system is always zero. You may try to prove it by using the vector formula of the angular momentum.
     
  4. Jun 12, 2010 #3
    ok thank you, and the the angular velocity around any axis is equal, right?
    oh, and about the rotation around different axis before and after the collision, if i calculate the angular momentum around the center of mass
    before the collision will the axis shift to the new center of mass after the collission, in other words can i directly compare the angular momentum of the two center of masses?
     
    Last edited: Jun 12, 2010
  5. Jun 12, 2010 #4
    The angular velocity around any axis in the reference frame of the axis is equal. But when you apply the conservation law of angular momentum, you must stay in the same reference frame. Thus, your [tex]\omega _{needed}[/tex] is wrong, and that's why you should use the original formula (the result you get is different from the one in the lecture note, right?)

    Well, that is the solution of the lecture note, which happens to be true in this case. This method is incorrect in general. I think you should derive this result and you will see why.
     
  6. Jun 12, 2010 #5
    i think i got it, so because i changed the axis before the collision from the given one at the center of mass, to the location of the center of mass after the collision, i have to account for it in the angular velocity (which changes from the given value). and this is where the original formula comes in. so does that mean that linear velocity of the masses stays the same in the reference frame of every axis in this problem?
     
  7. Jun 12, 2010 #6
    Uhm... I'm not sure if you really get it or not. So I'll make it clearer.

    _ In the reference frame of the lab/room/ground, before the collision, the motion of the rod is described simply by the angular velocity [tex]\omega [/tex] around the 1st axis, and that's why you can use the formula L=Iw here for the 1st axis. The same thing happens when you calculate L around the 2nd axis after the collision.

    _ But in the reference frame of the lab, before the collision, the rod doesn't rotate around the 2nd axis, which means its motion cannot be described as rotating with [tex]\omega [/tex] around the 2nd axis in this reference frame (only in the reference frame of the 2nd axis does the rod rotate with [tex]\omega [/tex] before the collision). Thus, to calculate L around the 2nd axis before the collision, you must use the original formula.

    The linear velocities change in general. For example, before the collision, in the reference frame of the lab (which is also the reference frame of 1st axis only before the collision), the masses have the speed of [tex]v=L\omega[/tex]; but in the reference frame of the 2nd axis, the masses have the speeds [tex]v_1=4L\omega /3[/tex] and [tex]v_2=2L\omega /3[/tex].

    In this problem, before the collision, in the reference frame of the lab, due to the fact that the linear momentum of the center of mass is zero, the angular momentum before the collision is the same regardless of the axis chosen. The same thing happens to the angular momentum after the collision. Of course, the angular momentums before and after the collision are different.

    In short, when applying the conservation law of angular momentum, you must stay in only 1 reference frame and consider only 1 axis throughout the process. You may choose any reference frame and any axis for convenience, but only 1 reference frame and only 1 axis.
     
    Last edited: Jun 12, 2010
  8. Jun 12, 2010 #7
    So is this correct: [tex]\omega_{needed} = \omega *((4/3*L)^2+(2/3*L)^2)/((4/3*L)^2+2*(2/3*L)^2) [/tex]
     
  9. Jun 12, 2010 #8
    Sorry, no.
    I guess you deduced the equation as in one of the 2 cases below:

    1/ You choose the reference frame of the lab:
    You got the right angular momentum around the 2nd axis after the collision: [tex]L_{after}=\omega_{needed}((4l/3)^2+2(2l/3)^2) [/tex]
    but wrong angular momentum before collision, as the rod doesn't rotate around 2nd axis before collision in this reference frame.

    2/ You choose the reference frame of the 2nd axis:
    You again got the right angular momentum around the 2nd axis after collision, but wrong angular momentum around 2nd axis before collision, because you forgot to include the motion of the third mass before collision in this reference frame.
     
    Last edited: Jun 12, 2010
  10. Jun 13, 2010 #9
    ok, so is this good: [tex]\omega_{needed}= \omega ((2L^2)/(4/3*l)^2+2*(2/3*L)^2)[/tex]?
     
  11. Jun 13, 2010 #10
    It is a correct one. And if you try the solution of the lecture note, you will find the same equation eventually.
     
  12. Jun 13, 2010 #11
    thank you so much for your help and patience:).
     
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